1
Probability
DEFINITION : PROBABILITY
Suppose an event E can happen in m ways out of a total of n possible equally likely ways.
Then the probability of occurrences of the event (called its success) is denoted by
p = P(E) = m .
n
The probability of non-occurrence of the event (called its failure) is denoted by
q = P(not E) = 1 –
m
= 1 – p.
n
The event not E is denoted by E or EC.
Note. If p is the probability that an event will occur, the odds in favour of its happening are p : q
(read “p to q”) ; the odd against its happening are q : p. Thus the odds against a 3 or 4 in a single toss
of fair die are q : p = 2 : 1 = 2 : 1 ; i.e., 2 to 1.
3 3
LIMITATIONS OF CLASSICAL DEFINITION
This definition breaks down in the following cases.
(i) If the various out comes of the trial are not equally likely or equally probable.
(ii) If the exhaustive number of cases in a trial is infinite.
For this reason a statistical definition of probability is followed in some situation.
STATISTICAL PROBABILITY
If in n trials an event E happens m times, then the probability ‘p’ of the happening of E is
given by
m
.
n
Permutation. A permutation of n different objects taken r at a time is arrangement of r
out of n objects with attention given to the order of the arrangement. The number of permutations of n objects taken r at a time is
p = P(E) = lim
n→∞
nPr = n(n – 1)(n – 2) ... (n – r + 1) =
C10\N-PROB\CHP1-1
n!
(n − r) !
2
Probability, Random Processes and Queueing Theory
In particular
nPn = n(n – 1)(n – 2) ... 2.1 = n !.
Example. The number of permutations of the letters a, b, c taken two at a time is 3P2 = 3
× 2 = 6. These are ab, ba, ac, ca, bc, cb.
Note. The number of permutations of n objects, in which n1 are alike, n2 are alike, ... is
n!
n1 ! n2 ! ...
, where n = n1 + n2 + ...
10 !
.
3!3!2!
Combinations. A combination of n different objects taken r at a time is a selection of r
out of n objects with attention not given to order of arrangements. It is denoted by nCr (or)
Example. The number of permutations of letters in the word STATISTICS is
n
C(n, r) or   and
r
nCr =
n( n − 1) ... ( n − r + 1)
n!
=
r!
r ! (n − r ) !
 n  is called binomial coefficient.
r
 
Fundamental rule. If an operation can be performed in m different ways and after this is
done in any one way, a second operation can be performed in n different ways, then both
operations can be performed in m × n different ways.
USEFUL FORMULAE CONNECTING BINOMIAL COEFFICIENTS
(i)
(ii)
(iii)
m
Cn = 0 if m < n or n < 0
Cn = (m–1)C(n–1) + (m–1)Cn, 0 ≤ n ≤ m
m
Cn = (m–1)C(n–1) + (m–2)C(n–1) + ... + nC(n–1) + nCn.
m
m −1
∑
=
k
k = n −1
(iv) (a + b)n =
n
∑
n
k
(m–k)
r =0
Cn–1
Cr an–r br
k
(v)
m
Cn =
∑
j =0
Cj
C(n–j),
where k and n are positive integers such that k ≤ n.
1. A bag contains 4 red, 5 white, 6 black balls. What is the probability that 2 balls drawn
are red and white ?
Solution. Out of 15 balls, 2 balls can be drawn in 15C2 ways. Out of 4 red balls 1 ball is
drawn in 4C1 ways and out of 5 white ball, 1 ball is chosen in 5C1 ways. Hence the total
number of favourable cases is 4C1 × 5C1. The required probability is
C10\N-PROB\CHP1-1
4C1 × 5C1
.
15C2
Probability
3
Some Properties of Probability
1. P(φ) = 0, where φ is an empty set.
2. P(S) = 1, where S is a sample space.
3. If A and B are mutually exclusive events, then P(A ∪ B) = P(A) + P(B).
Theorems. 1. Let E be any event. Then P(EC) = 1 – P(E), where EC is the complement of E.
2. Let E1, E2 be events such that E1 ⊆ E2. Then P(E2 – E1) = P(E2) – P(E1).
3. Addition theorem. Let E1, E2 be events. Then P(E1 ∪ E2) = P(E1) + P(E2) – P(E1 ∩ E2).
4. If {En} is a sequence of events such that E1 ⊆ E2 ⊆ ..., then
 ∞

P  ∪ En  = Lim P ( En ) .
=
1
n

 n →∞
5. If {Fn} is a sequence of events such that F1 ⊇ F2 ⊇ ... then
 ∞

P  ∩ Fn  = Lim P ( Fn ) .
=
1
n

 n→∞
6. Boole’s inequality. Let E1, E2, ... be sequence of events, then
∞
 ∞

P ( En ) .
P  ∪ En  ≤
=
1
n

 n =1
∑
Independent events. Let E1, E2 be two events. Then E1, E2 are said to be independent
events if
P(E1 ∩ E2) = P(E1) P(E2).
Mutually exclusive events. E1, E2, ... En are said to be mutually exclusive if Ei ∩ Ej = φ,
for i ≠ j.
Theorem 1. If E1 and E2 are independent and mutually exclusive events then either P(E1)
= 0 or P(E2) = 0.
2. If E1 and E2 are independent events, then E1 and E2C are also independent.
Independence of More than 2 events
Mutual independence. Let C = {Ei : i ∈ I} be a collection of events. These events are said to
be mutually independent if for every non-empty subset E1, E2, ..., En,
P(E1 ∩ E2 ∩ ... ∩ En) = P(E1) P(E2) ... P(En).
Pair wise Independence. Let C = {Ei : i ∈ I} be a collection of events. These events are
said to be pair wise independent if
P ( Eik ∩ Ei j ) = P ( Eik ) . P ( Ei j )
for each pair (ik, ij), where ik, ij ∈ I, and ik ≠ ij.
Note. Mutual independence of events implies pair wise independence of events. The converse is not
true.
Conditional Probability. Let E and F be events such that P(F) ≠ 0. The conditional
probability of E given F, denoted by P(E/F) is defined as
P(E/F) =
C10\N-PROB\CHP1-1
P (E ∩ F )
.
P( F )
4
Probability, Random Processes and Queueing Theory
Theorems. E and F are independent events if and only if P(E/F) = P(E), provided P(F) > 0.
Multiplication Theorem. If E and F are events such that P(F) > 0, then
P(E ∩ F) = P(F) . P(E/F).
Theorem. (Law of Total Probability).
Let E1, E2, ... En be a collection of events such that P(Ei) > 0, for each i,
 n

P  ∪ Ei  = 1 and Ei ∩ Ej = φ, for i ≠ j.
=
i
1


Let F be an event. Then
n
P(F) =
∑ P( F /E ) P( E ) .
i
i
i =1
Theorem. (Law of Compound probability).
Let E1, E2, ..., En be events, such that
P(E1 ∩ E2 ∩ ... ∩ En) > 0.
Then P(E1 ∩ E2 ∩ ... ∩ En) = P(E1) . P(E2/E1) . P(E3/E1 ∩ E2) ... P(En/E1 ∩ E2 ∩ ... ∩ En–1).
Baye’s Theorem. Let E1, E2, ... En be a collection of events such that P(Ei) > 0, for all i,
 n

P  ∪ Ei  = 1, Ei ∩ Ej = φ, for i ≠ j. Let F be an event such that P(F) > 0. Then
=
i
1


P(Ek/F) =
P ( F /E k ) . P ( Ek )
n
, k = 1, 2, ... n.
∑ P ( F /E ) P ( E )
i
i
i =1
Theorem. Let E1, E2, ... En be a collection of mutually independent events. Then
(a) the probability of the occurrence of atleast one of the events Ei is given by
n
1–
∏ (1 − P (E ))
i
i =1
(b) the probability of the occurrence of exactly one event, say Ei is
n
P(Ei)
∏ (1 − P ( E
k ))
.
k =1
k ≠i
n
Proof. (a) ∪ Ei denotes the occurrence of atleast one of the events E1, E2, ..., En. Then,
i =1
using De Morgan’s law,
 n

p  ∪ Ei  = 1 − P
 i =1 
C
 n
 
n

 ∪ Ei   = 1 − P  ∩ Eic 
 i =1  
i =1

n
=1–
∏ P (E
i
i =1
C10\N-PROB\CHP1-1
C
) , since EiC are mutually independent events.
Probability
5
(b) The occurrence of exactly one event Ei corresponds to occurrence of Ei and non
occurrence of each of the remaining events. So
P(Ei ∩ E1C ∩ E2C ∩ ... Ei–1C ∩ Ei+1C ∩ ... ∩ EnC)
n
= P(Ei)
∏ P (E
C
k )
k =1
k≠i
n
= P(Ei)
∏ (1 − P ( E
k ) ).
k =1
k ≠i
Examples
1. What is the probability of getting two heads in two flips of a balanced coin.
Solution. Since the probability of head is
1
for each flip, and two flips are independent,
2
the probability is 1 × 1 = 1 .
2 2 4
2. What is the probability of not rolling any 6’s in four rolls of a balanced die?
5 5 5 5
625
.
× × × =
6 6 6 6 1296
3. For the binary communication system
Solution. The probability is
0.9
1
x = 0, P(x = 0) = —
2
y=0
0.1
0.1
1
x = 1, P(x = 1) = —
2
0.9
y=1
Fig. 1.1
Find
P(Y = 0) and P(Y = 1),
Solution. P(Y = 0) = P[Y = 0/X = 0] P(X = 0) + P[Y = 0/X = 1] P(X = 1)
= (0.9)(0.5) + (0.1)(0.5) = 0.5
P(Y = 1) = 1 – P(Y = 0) = 0.5.
SOLVED PROBLEMS
1. Out of 50 students in a class, what is the probability of a single student to opt for a
picnic.
Solution. P (event of a single student opt for a picnic) = 1 = .02
50
2. What is the probability of obtaining two heads in two throws of a single coin?
Solution. The probability of obtaining a head in the first throw is
C10\N-PROB\CHP1-1
1
. The probability of
2
6
Probability, Random Processes and Queueing Theory
1
(it is not affected by the first throw of
2
the coin). Two throws being independent, the probability of obtaining head in both
throws is the product of probability of obtaining a head in the first throw and the
probability of obtaining a head in the second throw. Hence the required probability is
obtaining a head in the second throw is also
1 1 1.
× =
2 2 4
3. Find the probability of drawing two red balls in succession from a bag containing 3 red
and 6 black balls when
(i) the ball that is drawn first is replaced,
(ii) it is not replaced.
Solution. Let A1 be the event that the first ball drawn is red and A2 be the event that
the second ball drawn is red.
(i) If the first ball drawn is replaced, the events are independent. Then
P (A1 ∩ A2) = P (A1) P (A2) = 3 × 3 = 1 .
9 9 9
(ii) If the first ball is not replaced after taking a red ball the bag will contain only 8 balls
out of which 2 are red. The events are not independent. Therefore
3 2
6
× =
.
9 8 72
4. What is the probability of picking an ace and a king from a deck of 52 cards.
Solution. Let A be the event of picking an ace and B be the event of picking a king.
Probability of picking ace and king
= Probability of picking an ace first and a king second + probability of picking a king
first and an ace second
= P(A) × P (B/A) + P (B) × P(A/B)
P (A1 Ç A2) = P (A1) × P (A2/A1) =
8
4
4
4
4
×
+
×
=
.
52 51 52 51 663
5. A fair coin is tossed 5 times. What is the probability of having atleast one head?
Solution. The probability of getting no head in 5 tosses of a coin is given by P(A)
=
5
=  1  = 1 . Hence, the probability of getting atleast one head is given by
32
2
P(Ac) = P(S) – P(A) = 1 – 1 = 31 .
32 32
6. Prove that if P(A) > P(B), then P(A/B) > P(B/A).
Solution. P(A/B) =
P( A ∩ B)
P( A ∩ B)
implies P(B) =
P(B)
P( A/B)
P(B/A) =
P( A ∩ B)
P( A ∩ B)
implies P(A) =
P( A)
P(B/ A)
If P(A) > P(B), then
C10\N-PROB\CHP1-1
P(A ∩ B)
P ( B/ A )
>
P(A ∩ B)
P ( A/ B )
and hence P(A/B) > P(B/A).
Probability
7
7. One card is drawn from a deck of 52 cards. What is the probability of the card being
either red or a king?
Solution. Let A represents an event that the card drawn is red and B represents that
the card drawn is king. For a card to be either red or a king, it is required to find the
probability of A È B. Now
4
1
26 1
=
= ; P(B) =
.
52 13
52 2
There are two red coloured king cards. So
P(A) =
P(A Ç B) =
2
1
=
52 26
P(A È B) = P(A) + P(B) – P(A Ç B) = 1 + 1 − 1 = 7 .
2 13 26 13
8. A bag contains 12 balls numbered from 1 to 12. If a ball is taken at random, what is the
probability of having a ball with a number which is a multiple of either 2 or 3.
Solution. Let A be an event that the ball number is a multiple of 2 and B be an event
that the ball number is a multiple of 3. Then
A = {2, 4, 6, 8, 10, 12}, B = {3, 6, 9, 12}, A Ç B = {6, 12}
Therefore,
P(A) =
6 1
= ; P(B) = 4 = 1 , P(A Ç B) = 1 = 1 .
12 2
12 3
12 6
Hence
1 1 1 4 2
+ − = = .
2 3 6 6 3
9. If two events A and B are independent, show that
(i) AC and BC are independent
(ii) AC and B are independent
(iii) A and BC are independent.
Solution. Given A and B are independent. So
(i) P(A Ç B) = P(A) × P(B)
Now P(AC Ç BC) = P((A È B)C) = 1 – P(A È B)
P(A È B) = P(A) + P(B) – P(A Ç B) =
= 1 – [P(A) + P(B) – P(A Ç B)] = 1 – [P(A) + P(B) – P(A) P(B)]
= [1 – P(B)] – P(A) [1 – P(B)] = (1 – P(B)) (1 – P(A)) = P(BC) P(AC)
Hence A and B are independent.
C
C
S
AÇB
A Ç BC
AC Ç B
B
A
Fig. 1.3
C10\N-PROB\CHP1-1
8
Probability, Random Processes and Queueing Theory
(ii) We know (AC Ç B) È (A Ç B) = B. Hence by addition theorem
P(AC Ç B) = P(B) – P(A Ç B), as (AC Ç B) and (A Ç B)
are mutually exclusive. Now
P(AC Ç B) = P(B) – P(A Ç B) = P(B) – P(A) P(B)
= P(B) (1 – P(A)) = P(B) × P(AC)
Hence A and B are independent.
(iii) P(A Ç BC) = P(A) – P(A Ç B) = P(A) – P(A) P(B)
C
13. Given P(A) =
Solution.
S
B
A Ç BC
Fig. 1.4
S
AC Ç B
B
A
AÇB
P(A Ç B) £ P(A) £ P(A È B) £ P(A) + P(B).
Solution. By the Venn diagram it is clear that A Ç BC and
A Ç B are two disjoint event such that
(A Ç BC) È (A Ç B) = A. Hence P(A Ç BC) + P(A Ç B)
= P(A). Since P(A Ç BC) ³ 0, P(A) ³ P(A Ç B).
We know that P(A È B) = P(A) + P(B) – P(A Ç B). Since
P(B) ³ P(A Ç B), P(B) – P(A Ç B) ³ 0. Hence P(A È B)
³ P(A). Also since P(A Ç B) ³ 0, P(A È B) £ P(A) + P(B).
Hence
P(A Ç B) £ P(A) £ P(A È B) £ P(A) + P(B).
12. Prove that for any event A in S,
P(A Ç AC) = 0.
Solution. We know P(S) = 1 and A È AC = S. Hence
1 = P(S) = P(A È AC) = P(A) + P(AC) – P(A Ç AC)
Þ
1 = P(A) + (1 – P(A)) – P(A Ç AC)
Þ
1 = 1 – P(A Ç AC)
Þ
P(A Ç AC) = 0.
A
A Ç BC
= P(A) (1 – P(B)) = P(A) P(BC)
Hence A and BC are independent.
10. If B Ì A, prove that P(A Ç BC) = P(A) – P(B).
Solution. B È (A Ç BC) = A. Here B and A Ç BC are
mutually exclusive. Hence by addition theorem of probability we have
P(B) + P(A Ç BC) = P(A)
Hence P(A Ç BC) = P(A) – P(B).
11. For any two events A and B show that
Fig. 1.5
1
1
1
, P(B) =
, P(A Ç B) =
, find the following probabilities
6
4
3
P(AC), P(AC È B), P(AC Ç BC).
1 2
=
3 3
P(AC È B) = P(AC) + P(B) – P(AC Ç B) = P(AC) + P(B) – [P(B) – P(A Ç B)]
P(AC) = 1 – P(A) = 1 –
= P(AC) + P(A Ç B) =
C10\N-PROB\CHP1-1
2 1 5
+ = .
3 6 6
Probability
9
P(AC Ç BC) = P((A È B)C) = 1 – P(A È B) = 1 – [P(A) + P(B) – P(A Ç B)]
1 1 1  7
=1–  + − =
.
 3 4 6  12
14. Suppose from a pack of 52 cards one card is drawn at random what is the probability
that it is either a king or a queen.
Solution. Since the events are mutually exclusive (if a card is a king it cannot be a
4
queen and vice versa) the probability of drawing a king is
and similarly drawing a
52
4
queen is
. Hence the probability of drawing a king K or a queen Q is
52
P(K È Q) = P(K) + P(Q) =
4
4
2
+
=
.
52 52 13
15. One ticket is drawn at random from a bag containing 30 tickets numbered from 1 to 30.
Find the probability that it is a multiple of 5 or 7.
Solution. One ticket can be drawn out of 30 in 30C1 ways. The multiples of 5 are 5, 10,
15, 20, 25, 30 and multiples of 7 are 7, 14, 21, 28. Hence there are 6 multiples of 5 and 4
multiples of 7. None of these are common. So the events are mutually exclusive. The
probability of having a multiple of 5 or 7 is
6
10 1
4
+
=
= .
30 30 30 3
16. The probability that A will live upto 60 years is 3 and probability that B will live upto
4
60 years is 2 . What is the probability that both A and B will live upto sixty years.
3
Solution.
P(A Ç B) = P(A) × P(B) =
3 2 1
× = .
4 3 2
1
.
2
17. Find the probability of drawing two kings from a pack of cards in two successive draws,
the card drawn not being replaced.
Hence the probability that both A and B will live upto sixty year is
4
. The probability of
52
drawing a king in the second draw, given that the first draw has already given a king is
Solution. The probability of drawing a king in the first draw is
3
1 .
3/51. The combined probability of two events is 4 ×
=
52 51 221
18. A bag contains 3 red and 4 white balls. Two draws are made without replacement; what
is the probability that both the balls are red.
C10\N-PROB\CHP1-1
10
Probability, Random Processes and Queueing Theory
3
. Probability of
7
drawing a red ball in the second draw given that the first ball drawn is red is P(B/A) =
Solution. Probability of drawing a red ball in the first draw is P(A) =
2
(since only six balls are left and only two out of them are red). The combined probability
6
of the two events are
3 2 1
× = .
7 6 7
19. Three coins are tossed simultaneously. What is the probability that they will fall two
heads and one tail.
P(A Ç B) = P(A) × P(B/A) =
Solution. The probability P of getting a head is
1
. Now, the probability that out of 3
2
coins getting exactly 2 heads and one tail is
1 1 
1 3
´ ´ 1− = ,
2 2 
2 8
where Q denotes the probability of getting a tail and is Q = 1 – P.
3C2 P2Q = 3 ´
20. If A and B are mutually exclusive event, prove that P(A/BC) =
Solution. P(A/BC) =
P ( A ∩ BC )
=
P( A)
.
1− P(B)
P ( A ∩ BC ) P ( A ) − P ( A ∩ B )
=
1− P(B)
1 − P(B)
P ( BC )
Since A and B are mutually exclusive, P(A Ç B) = 0.
P( A)
.
1− P(B)
21. If A and B are two mutually exclusive events P(A È B) ¹ 0, prove that
Hence P(A/BC) =
P(A/A È B) =
Solution.
P(A/A È B) =
P( A)
.
P( A) + P(B)
P ( A ∩ ( A ∪ B ))
P( A ∪ B)
=
P(A)
P( A) + P( B) − P( A ∩ B)
P( A)
since P(A Ç B) = 0.
P( A) + P(B)
22. If A and B are two independent events, show that
P(A È B) = 1 – P(AC) × P(BC).
Solution.
P(A È B) = 1 – P(A È B)C = 1– P(AC Ç BC)
= 1 – P(AC) × P(BC),
since if A and B are independent, AC and BC are also independent.
=
23. Box 1 contains 1 white and 999 red balls. Box 2 contains 1 red and 999 white balls. A
ball is picked from a randomly selected box. If the ball is red, what is the probability
that it came from box 1?
C10\N-PROB\CHP1-1
Probability
11
Solution. Let A be the event that a ball picked is red and Ai be the event that ith box is
selected (i = 1, 2)
P(A/A1) =
1
999
and P(A/A2) =
1000
1000
1
. Hence by Baye’s theorem
2
1 × 999
P ( A1 ) ⋅ P ( A/ A1 )
2 1000
999
=
=
P(A1/A) = 2
.
1000
999
1 1

+


P ( A/ Ai ) P ( Ai )  1000 1000  2
and
P(Ai) =
∑
i =1
999
.
1000
24. Let A and B be boxes that contain 5 black, 6 white; 8 black, 4 white balls respectively.
Two balls are transferred from B to A and then a ball is drawn from A.
(a) What is the probability that this ball is white?
(b) Given that the ball drawn is white, what is the probability that atleast one white
ball was transferred to A?
Solution. Let E1 be the event that a white ball is drawn from A.
E2 be the event that two white balls are transferred from B.
E3 be the event that one white ball and one black ball are transferred from B.
E4 be the event that two black balls are transferred from B. Then
P(E1) = P(E2) × P(E1/E2) + P(E3) × P(E1/E3) + P(E4) × P(E1/E4)
Hence the required probability is
=
4 C2
12 C2
×
8 C1 × 4 C1
8 C2
8
7
6
+
×
+
×
13
12 C2
13 12 C2 13
6
8
32 7
28 6
136
×
+
×
+
×
=
.
66 13 66 13 66 13 429
25. A box contains 2000 components of which 5% are defective. Second box contains 500
components of which 40% are defective. Two other boxes contain 1000 components each
with 10% defective components. We select at random one of the above boxes and remove
from it at random a single component.
(i) What is the probability that this component is defective?
(ii) Finding that the selected component is defective, what is the probability that it was
drawn from box 2?
Solution. Let Ai be the event of selecting ith box. B be the event consisting of all
defective components. Therefore, we have
=
P(A1) = P(A2) = P(A3) = P(A4) =
P(B/A1) =
100
= 0.05
2000
P(B/A2) =
200
= 0.4
500
C10\N-PROB\CHP1-1
1
4
12
Probability, Random Processes and Queueing Theory
P(B/A3) =
100
= 0.1
1000
100
= 0.1
1000
Probability that the component is defective is given by
P(B) = P(A1) P(B/A1) + P(A2) P(B/A2) + P(A3) P(B/A3) + P(A4) P(B/A4)
P(B/A4) =
1
1
1
1
´ 0.05 +
´ 0.4 +
´ 0.1 +
´ 0.1 = 0.1625.
4
4
4
4
1
26. Of three independent events, the probability of first only happening is , the probability
4
1
of second only happening is
, and the probability that only third happens is 1 . Find
8
12
the marginal probability of each event.
Solution. Let A, B, C be events. Then
=
P(A Ç BC Ç CC) = P(A) × P(BC) P(CC) =
1
4
1
8
P(AC Ç BC Ç C) = P(AC) P(BC) P(C) = 1
12
P(AC) = x, P(BC) = y, P(CC) = z. Then
P(AC Ç B Ç CC) = P(AC) P(B) P(CC) =
Let
1
1
; xz (1 – y) =
; xy (1 – z) = 1
4
8
12
1
1
By remainder theorem, we get the solution of x as . Therefore yz =
and z (1 – y)
2
2
2
1
3
=
implies z = , y = . Hence
4
3
4
1
P(A) = 1 – P(AC) =
2
2 1
=
P(B) = 1 – P(BC) = 1 –
3 3
(1 – x) yz =
P(C) = 1 – P(CC) = 1 –
27. It is given that P(A È B) =
3 1
= .
4 4
1
5
, P(A Ç B) = 1 and P(BC) = . Show that the events A and
2
6
3
B are independent.
Solution.
Now
Þ
C10\N-PROB\CHP1-1
1 1
= .
2 2
P(A È B) = P(A) + P(B) – P(A Ç B)
P(B) = 1 – P(BC) = 1 –
1 1
5
−
= P(A) +
2 3
6
Probability
13
5 1 1 2
+ − =
6 3 2 3
2 1 1
⋅ =
Hence P(A) × P(B) =
= P(A Ç B). So A and B are independent.
3 2 3
28. A box contains four tickets with numbers 112, 121, 211, 222 and one ticket is drawn. Let
Ai (i = 1, 2, 3) be the event that the ith digit of the number of tickets drawn is 1. Discuss
the independent of the events A1, A2, A3.
Solution.
P(A) =
Þ
1
1
; P(A3) =
2
2
1
P(A2) =
2
1
P(A1 Ç A3) =
4
1
P(A1) P(A2) =
4
P(A1) =
; P(A1 Ç A2 Ç A3) = 0
1
4
1
; P(A2 Ç A3) =
4
; P(A1 Ç A2) =
; P(A1) × P(A3) =
1
1
; P(A2) P(A3) =
.
4
4
1
. But P(A1 Ç A2 Ç A3)
8
= 0. Hence P(A1) P(A2) P(A3) ¹ P(A1 Ç A2 Ç A3) and so A1, A2, A3 cannot be simultaneously independent.
29. A man has three coins A, B and C, A is unbiased; the probability that a head will result
Hence events are pairwise independent. Now P(A1) P(A2) P(A3) =
1
2
; the probability that a head will result when C is tossed is
. If
3
3
one of the coins chosen at random is tossed three times giving a total of two heads and
one tail, find
(i) the probability that the chosen coin was A
(ii) the probability that a fourth toss of the same coin will give a head.
Solution. Let D denote the probability of giving two heads and one tail. Then
when B is tossed is
2
P(D/A) = 3C2  1  1 = 3
2 2 8
2
2 1 = 4
P(D/B) = 3C2  
3 3 9
2
1 2 = 2
P(D/C) = 3C2  
3 3 9
(i) P(A/D) =
P ( A ) ⋅ P ( D/ A )
P ( A ) ⋅ P ( D/ A ) + P ( B ) ⋅ P ( D/B ) + P ( C ) ⋅ P ( D/C )
1×3
3 8
9
=
=
1 3 1 4 1 2 25
× + × + ×
3 8 3 9 3 9
C10\N-PROB\CHP1-1
14
Probability, Random Processes and Queueing Theory
(ii) The fourth toss of A will give head is
probabilities will be
probabilities
1
P(A) P(D/A). Similarly for B and C the
2
1
2
P(B) P(D/B),
P(C) P(D/C) respectively. Hence the required
3
3
241
1 3 1 2 1 4  1 1 2
.
=  × × + × × + × × =
 3 8 2   3 3 9  3 3 9 1296
30. Three candidates A, B, C are selected for the position of a general manager in a company whose chances of getting the appointment are in the proportion 4 : 2 : 3 respectively. The probability that A is selected will improve the office canteen is 0.3. The
probability of B and C doing the same are respectively 0.5 and 0.8. What is the probability that the office canteen will be improved.
Solution. Let H1, H2, H3, be the events that the candidates are selected for the position
of general manager. Let D be the event that the office canteen being improved. Then
from Baye’s Theorem,
P(D) = P(H1) × P(D/H1) + P(H2) × P(D/H2) + P(H3) × P(D/H3)
3
4
2
´ 0.3 +
´ 0.5 +
´ 0.8 = .13 + .11 + .27 = .51.
9
9
9
31. A box contains 10 white, 5 yellow and 10 black balls. A ball is chosen at random from
the box and it is noted that it is not one of the black balls. What is the probability that it
is yellow.
Solution. Let Y denote the event that the ball selected is yellow, and let BC denote the
event that it is not black. Now
=
P(Y/BC) =
P ( Y ∩ BC )
P ( BC )
But P(Y Ç BC) = P(Y) since the ball will be both yellow and not black if and only if it is
yellow. Hence, assuming that each of the 25 balls is equally likely to be chosen, we
obtain that
5
25 1
= .
P(Y/B ) =
15 3
25
32. For any three events A, B, C
P(A Ç BC/C) + P((A Ç B)/C) = P(A/C).
Solution.
P((A Ç BC)/C) + P((A Ç B)/C)
C
=
=
=
C10\N-PROB\CHP1-1
P ( A ∩ BC ∩ C )
P (C )
+
P( A ∩ B ∩ C)
P (C )
P [( A ∩ B C ∩ C ) ∪ ( A ∩ B ∩ C )]
P (C )
P [( A ∩ C ) ∩ BC ] ∪ [( A ∩ C ) ∩ B ]
P (C )
Probability
=
=
P ( A ∩ C ) ∩ ( BC ∪ B )
P (C )
15
by distributive law
P(A ∩C)
.
P (C )
Note. P((A2 È A3 È . . . ÈAn)/A1) = P(A2/A1) + P(A3/A1) + . . . + P(A n/A1 )
provided A2, A3, . . . An are pairwise disjoint sets.
33. If A Ç B = B, then show that P(A) £ P(BC)
Solution.
A = (A Ç B) È (A Ç BC) = B È (A Ç BC) = A Ç BC
Hence A Í BC which implies P(A) £ P(BC).
34. Two urn contains 4 white and 6 black balls and 4 white and 8 black balls. One urn is
selected at random and a ball is taken out. It turns out to be white. Find the probability
that it is from the first urn.
Solution. Let E1 and E2 be events that the first and the second urn respectively were
selected. Since the urn was selected at random
1
= P(E2).
2
Let A be the event that the ball taken out is white. Then
A = (E1 Ç A) È (E2 Ç A) and (E1 Ç A) Ç (E2 Ç A) = f.
Now
P(A) = P(E1 Ç A) + P(E2 Ç A)
P(E1) =
= P(E1) × P(A/E1) + P(E2) × P(A/E2) =
By Baye’s theorem P(E1/A) =
P ( E1 ) P ( A/E1 )
P(A)
1 4 1 4 11
×
+ ×
=
2 10 2 12 30
1× 4
2 10
6
=
=
.
11
11
30
35. The contents of urns I, II, III are as follows.
1 white,
2 black,
3 red balls
2 white,
1 black,
1 red ball
4 white,
5 black,
3 red balls
One urn is chosen at random and two balls are drawn. They happen to be white and red.
What is the probability that they come from urns I, II or III?
Solution. Let E1, E2, E3 denote the events that the urns I, II, III are chosen respectively
and let A be the event that the two balls taken from the selected urn are white and red.
Then
P(E1) = P(E2) = P(E3) =
P(A/E1) =
P(A/E2) =
C10\N-PROB\CHP1-1
1×3
=1
6 C2
5
2×1
4 C2
=1
3
1
3
16
Probability, Random Processes and Queueing Theory
P(A/E3) =
Hence
P(E2/A) =
4×3
12 C2
= 2
11
1×1
3 3
55
=
=
1×1+1×1+1× 2
118
P ( Ei ) P ( A/Ei ) 3 5 3 3 3 11
P ( E2 ) P ( A/E2 )
3
∑
i =1
1× 2
3 11
30
=
P(E3/A) =
1 1 1 1 1 1
118
× + × + ×
3 5 3 3 3 11
55
30
33
.
−
=
118 118 118
36. A box contains five balls. Two balls are drawn and found to be white. What is the
probability that all of the balls being white?
Solution. Let B be the probability that two balls drawn are white. Let A1, A2, A3, A4 be
the probability that the box contains 2, 3, 4 or 5 white balls, since these are the only
possibilities after the knowledge that two white balls are drawn. Hence
P(E1/A) = 1 –
1
= P(A2) = P(A3) = P(A4)
4
P(B/A1) is the probability that the box contains 2 white balls and both been drawn is
3
1 3
1 4 1
1
1 5 1
2 1
× =
. Similarly P(B/A2) = × =
, P(B/A3) = × = , P(B/A5) = × = .
4
5
20
4
4
5
5
4 5 4
5 10
37. In a bolt factory machines A, B, C manufacture respectively 25%, 35%, and 40% of the
total. Of their output 5, 4, 2 percent are defective bolts. A bolt is drawn at random from
the product and is found to be defective. What are the probabilities that it was manufactured by machines A, B, C?
Solution. Let E1, E2, E3 be the events that a bolt selected at random is manufactured by
the machines A, B, C respectively and let E denote the event of its being defective. Now
P(A1) =
P(E1) = 0.25, P(E2) = 0.35, P(E3) = 0.40
The probability of drawing a defective bolt manufactured by machine A is P(E/E1)
= 0.05. Similarly we have
P(E/E2) = 0.04, P(E/E3) = 0.02
Hence the probability that a defective bolt selected at random is manufactured by
machine A is given by
P(E1/E) =
P ( E1 ) P ( E/E1 )
3
∑ P ( E ) P ( E/E )
1
1
i =1
=
C10\N-PROB\CHP1-1
0.25 × 0.05
0.25 × 0.05 + 0.35 × 0.04 + 0.40 ×0.02
=
25
69
Probability
17
Similarly we get
28
16
, P(E3/E) =
.
69
69
Baye’s Theorem for future events.
The probability of materialization of another event C given P(C/A Ç E1), P(C/A Ç E2), . . .
is
P(E2/E) =
n
∑ P ( E ) P ( A/E ) P ( C/A ∩ E )
i
P(C/A) =
i
i
i =1
.
n
∑ P ( E ) P ( A/E )
i
i
i =1
38. Three boxes of the same appearance have the following proportion of balls
I
2 black
1 white
II
1 black
2 white
III
2 black
2 white
One of the urn is selected and one ball is drawn. It turns out to be white. What is the
probability of drawing white ball again, if the first one drawn is not replaced.
Solution. Let E1, E2, E3 be the event of drawing I, II, III urns respectively. Let A be the
event of drawing white ball. Now
P(E1) = P(E2) = P(E3) =
1
3
2
1
2
; P(A/E2) =
; P(A/E3) =
3
4
3
Let C denote the future event of drawing other white ball from the urns.
P(A/E1) =
P(C/A Ç E1) = 0 ; P(C/A Ç E2) =
1
1
; P(C/A Ç E3) =
.
2
3
3
∑ P ( E ) P ( A/E ) P ( C/E
i
P(C/A) =
i
i
∩ A)
i =1
3
∑ P ( E ) P ( A/E )
i
i
i =1
1 ×1 ×0+ 1 × 2×1 + 1 ×1 ×1
3 3
3 3 2 3 2 3 1
= .
=
1 1 1 2 1 1
3
× + × + ×
3 3 3 3 3 2
39. A certain item is manufactured by three factories say 1, 2 and 3. It is known that 1
turns out twice as many items as 2, and that 2 and 3 turns out the same number of item
(during a specified production period). It is also known that 2 percent of the items
produced by 1 and 2 are defective, while 4 percent of those manufactured by 3 are
defective. All the items produced are put into one stockpile, and then one item is chosen
at random. What is the probability that this item is defective?
C10\N-PROB\CHP1-1
18
Probability, Random Processes and Queueing Theory
Solution. Let A be the event that the item produced is defective. Let B1, B2, B3 be the
event that the items come from factories 1, 2, 3 respectively. Let the factory 1 manufacture 50 number of items. Then 2 and 3 manufacture 25 number of items each. Hence
50 1
25 1
= ; P(B2) = P(B3) =
=
100 2
100 4
P(A/B1) = P(A/B2) = 0.02 ; P(A/B3) = 0.04
P(B1) =
Hence
P(A) = P(B1) P(A/B1) + P(B2) P(A/B2) + P(B3) P(A/B3)
1
1
1
´ 0.02 +
´ 0.02 +
´ 0.04 = 0.025.
2
4
4
40. Suppose that among six bolts, two are shorter than a specified length. If two bolts are
chosen at random, what is the probability that the two shorts bolts are picked?
Solution. Let A1, A2 be the events that the first and second bolt chosen are short,
respectively. Now
=
1 2 1
× =
.
5 6 15
41. A pair of dice is thrown twice. What is the probability of getting totals of 7 and 11?
Solution. Let A1, A2, B1, and B2, be respective independent events that a 7 occurs on the
first throw, 7 occurs on the second throw, an 11 occurs on the first throw, and an 11
occurs on the second throw. A1 Ç B2, B1 Ç A2 are mutually exclusive events and
P(A1 Ç A2) = P(A2/A1) P(A1) =
P[(A1 Ç B2) È (B1 Ç A2)] = P(A1 Ç B2) + P(B1 Ç A2)
= P(A1) P(B2) + P(B1) P(A2)
1  1   1 1 1
.
=   +  =
 6   18   18   6  54
42. Three cards are drawn in succession without replacement from a deck of cards. Find the
probability that the event A1 Ç A2 Ç A3 occurs, where A1 is the event that the first card
is a red ace, A2 is the event that a second card is a 10 or a Jack and A3 is the event that
the third card is greater than 3 but less than 7.
Solution. P(A1) =
2
8
12
; P(A2/A1) =
; P(A3/A1 Ç A2) =
52
51
50
Hence
P(A1 Ç A2 Ç A3) = P(A1) P(A2/A1) P(A3/A1 Ç A2)
8 12
8
2
×
×
=
.
52 51 50 5525
43. A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3
times, what is the probability of getting 2 tails and 1 head.
Solution. The sample space S = {HHH, HHT, HTH, THH, HTT, TTH, THT, TTH}. Since
=
2
1
and P(T) =
.
3
3
Let A be the event of getting 2 tails and 1 head in the 3 tosses of the coin. Then
A = {TTH, THT, HTT}
a coin is biased and a head is twice as likely to occur as a tail, P(H) =
C10\N-PROB\CHP1-1
Probability
19
And since the outcomes of the 3 tosses are independent,
P(TTH) = P(T Ç T Ç H) = P(T) P(T) P(H) =
P(THT) = P(HTT) =
Hence
P(A) =
1 1 2
2
× × =
3 3 3 27
2
27
2
2
2
2
+
+
= .
27 27 27 9
44. A1, A2, ..., An are independent events with respective probabilities P(Ai) = 1 –
1
, i = 1,
2i
2, ..., n. Find the probability P(A1 + A2 + ... + An).
Solution. We know P(A1 + A2) = P(A1) + P(A2) – P(A1A2). Let f(n) = P(A1 + A2 + ... + An)
and so
f(n + 1) = P(A1 + A2 + ... + An+1)
= P(A1 + A2 + ... + An) + P(An+1) – P(A1 + A2 + ... + An) P(An+1)
⇒
⇒

1
= f(n) + 1 −
n+1
2

1
f(n + 1) = f(n) + 1 –
n+1
2
f(n + 1) – 1 = 1 [f(n) – 1]
2n+1
=
=
⇒
1
n +1
2
...
1


1 
 – f(n) 1 − n+1 
2



– f(n) + f(n) .
1
n+1
2
 1

.
[ f ( n − 1 ) − 1 ]
n
2

...
... by recursion
1
1
1
[ P ( A1 ) − 1]
2 2
22
1
1
1 
1

= n +1 . n ... 2 . 1 − − 1
2


2
2
2
1
1
=1 −
f(n + 1) = 1 – n +1 n
2
1
1 + 2 +... + n +1
2
. 2 ... 2 . 2
2
2
n +1
.
n
.
n −1
...
Hence
P(A1 + A2 + ... + An) = f(n) = 1 –
1
=1 −
21+ 2+...+ n
1
n ( n +1 )
2 2
.
45. In a company, 5% defective components are produced. What is the probability that
atleast 5 components are to be examined in order to get 3 defectives?
Solution. Given p = 0.05, q = 0.95,
Required probability = P(X = 5) + P(X = 6) + ...
4
=
∑ (x – 1) C (0.05)
x =5
C10\N-PROB\CHP1-1
2
3
(0.95)x–3
20
Probability, Random Processes and Queueing Theory
4
=1–
∑ (x – 1) C (0.05)
2
x =3
3
(0.95)x–3
= 1 – [(0.05)3 + 3C2 . (0.05)3 (0.95)]
= 0.9995.
46. If you twice flip a balanced coin, what is the probability of getting atleast one head?
Solution. 3 .
4
47. Let X and Y be integer valued random variable with P(X = m, Y = n) = q2 pm+n–2, n = m
= 1, 2, ..., and p + q = 1. Are X and Y independent?
Solution. Let P(X = m) = qpm–1 and
P(X = n) = qpn–1. Then
P(X, Y) = P(X) P(Y).
Hence X and Y are independent.
48. A manufacturer of airplane parts knows that the probability is 0.8 that an order will be
ready for shipment on time, and it is 0.7 that an order will be ready for shipment and
will be delivered on time. What is the probability that such an order will be delivered on
time given that it was also ready for shipment on time?
Solution. Let A = order is ready for shipment on time,
B = order is delivered on time.
Hence P(A) = 0.8 and P(A ∩ B) = 0.7. So
P(B/A) =
P ( A ∩ B ) 0.7
=
= 0.875.
P( A)
0.8
49. There are three unbiased coins and one biased coin with head on both sides. A coin is
chosen at random and tossed 4 times. If head occurs all the 4 times, what is the probability
that the biased coin has been chosen?
Solution. P(Head appears in 4 tosses in a coin) = 3
4
Let A = All heads in 4 tosses.
Let P(A biased coin is chosen) = P(F). Then P(F) =
P(F/A) =
1
. Now by Baye’s Theorem,
4
P ( A/F ) P ( F ) 16
=
.
ΣP ( A/F ) P ( F ) 19
50. Two dice are thrown (i) What is the probability that sum 8 comes up? (ii) What is the
probability that sum 8 comes up if it is given that the sum is an even number?
Solution. (i) Total number of cases = 36. Let A : Sum 8
Then from the following table, it is evident that
P(A) =
C10\N-PROB\CHP1-1
5
.
36
Probability
21
Sum 7
(1, 1)
(1, 2)
(1, 3)
(1, 4)
(1, 5)
(1, 6)
Sum 8
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(2, 5)
(2, 6)
Sum 9
(3, 1)
(3, 2)
(3, 3)
(3, 4)
(3, 5)
(3, 6)
(4, 1)
(4, 2)
(4, 3)
(4, 4)
(4, 5)
(4, 6)
(5, 1)
(5, 2)
(5, 3)
(5, 4)
(5, 5)
(5, 6)
(6, 1)
(6, 2)
(6, 3)
(6, 4)
(6, 5)
(6, 6)
Sum 10
(ii) Further, let B = Sum is even. Then the total number of cases is 18. Hence
P(B) =
18 1
=
36 2
5
.
18
51. What is the probability of throwing 8, 9 or 10 with two dice?
Solution. With reference to distribution table in the above example 50,
and
P(A/B) =
No. of favourable cases 12 1
=
= .
Total number of cases 36 3
52. An urn contains a white and b black balls. If α + β balls are drawn from this urn, find
the probability that among them there will be exactly α white and β black balls.
Solution. Total number of balls = a + b and hence the total number of cases is C(a + b,
α + β). We choose α white balls from a white balls in C(a, α) ways and β black balls in
C(b, β) ways. The required favourable cases is C(a, α) C(b, β). Therefore required probability is
P{8 or 9 or 10} =
C ( a , α ) C (b, β )
C ( a + b, α + β)
.
53. Six dice are rolled. What is the probability of getting three pairs.
Solution. We select three numbers out of six to form pairs in 6C3 ways. Consider one
such set as {1, 2, 3} say. Then the total number of pairs is the total number of ways of
dividing six dice into groups (1, 1), (2, 2), (3, 3) which is equal to
6!
.
( 2 !)3
The total number of all possible combinations = 66. Therefore, the required probability is
p=
C10\N-PROB\CHP1-1
 6 !
6 C3 
 ( 2 !)3
6
6



=
25
.
648
22
Probability, Random Processes and Queueing Theory
54. If W = {1, 2, 3, ..., 120}, find the probability of the set of numbers divisible by 4 or 6 or
both.
Solution. Let B = {w : w is a multiple of 4}
C = {w : w is a multiple of 6}
1
1
, P(C) =
. The set BC consists of those integers which are divisible by
4
6
1
both 4 and 6 namely, divisible by their least common multiple 12, hence P(BC) =
.
12
Then P(B) =
1
2
Hence
3
✓
5
4
✓
7
6
8
✓
9
10
11
✓
12
✓
P(B ∪ C) = P(B) + P(C) – P(BC)
1 1
1 1
+ −
− .
4 6 12 3
55. If a deck of 52 playing cards is thoroughly shuffled, what is the probability of getting all
the four aces in a row?
Solution. There are 4 aces among the 52 cards. Taking these 4 aces to be one card, we
have 49 cards in all and hence (49) ! permutations.
Now, if the 4 aces are arranged amongst themselves in 4 ! ways, the total number of
permutations with the four aces together is 4 ! (49) !.
Hence the required probability is
=
p=
4 ! ( 49 ) !
( 52 ) !
.
56. A lot consists of 10 good articles, 4 with minor defects and 2 with major defects. Two
articles are chosen from the lot at random (with out replacement). Find the probability
that (i) both are good (ii) both have major defects (iii) atleast 1 is good (iv) atmost 1 is
good (v) exactly 1 is good (vi) neither has major defects and (vii) neither is good.
Solution. Assume both articles are drawn simultaneously, as they are drawn without
replacement. From 16 articles, 2 articles can be chosen in 16C2 ways.
No. of ways of drawing 2 good artices
(i) P(both are good) =
=
Total no. of ways of drawing 2 articles
10 C2
16 C2
=3.
8
(ii) P(both have major defects) =
=
No. of ways of drawing 2 articles with major defects
Total no. of ways
2 C2
16 C2
= 1 .
120
(iii) P(atleast 1 good) = P(1 good or both good)
= P(exactly 1 good and 1 is bad or both good)
=
C10\N-PROB\CHP1-2
10 C1 × 6 C1 × 10 C2
16 C2
=7.
8
Probability
23
(iv) P(atmost 1 good) = P(none is good or 1 is good and 1 is bad)
10 C0 × 6 C2 + 10 C1 × 6 C1
=
16 C2
(v) P(exactly 1 good) = P(1 good and 1 bad)
=5.
8
10 C1 × 6 C1
=1.
16 C2
2
(vi) P(neither has major defects) = P(both are not of major defective articles)
=
14 C2
= 91 .
16 C2 120
(vii) P(neither is good) = P(both are defective)
=
6 C2
=1.
16 C2
8
57. A box contains 15 chips out of which 4 are defective. The chips are selected at random
one by one, and examined. The ones examined are not put back, what is the probability
that the 9th chip examined is the last defective?
Solution. Let A be the event of getting exactly 3 defectives in examination of 8 chips
and let B be the event that the 9th chip examined is a defective one. Since the chips are
examined without replacement,
=
We have
P(A) =
11 C5 × 4 C3
15 C8
P(B/A) = P(getting one defective chip out of left out chips)
=
Hence
1
.
7
P(A ∩ B) = P(A) . P(B/A) =
11 C5 × 4 C3
15 C8
×1
7
8
.
195
58. There are ten counters in a bag, 6 of them worth 5 rupees each while the other 4 are of
equal but of unknown value. If the expectation from drawing a single counter at random
is 4 rupees, find the unknown value.
=
6
= 0.6. The probability
10
of drawing a counter of unknown value, say Rs. x is 0.4. Hence the expectation from
drawing a single counter at random is 4 rupees is given by (0.6) × 5 + (0.4x) = 4
given ⇒ x = 2.5.
59. A toy is rejected if the design is faulty or not. The probability that the design is faulty is
0.1 and that the toy is rejected if the design is faulty is 0.95 and otherwise 0.45. If a toy
is rejected, what is the probability that it is due to faulty design.
Solution. Let D1, D2 denote the events that the design is faulty or not. Let A denote the
event that the toy is rejected.
Solution. The probability of drawing a counter worth Rs. 5 is
C10\N-PROB\CHP1-2
24
Probability, Random Processes and Queueing Theory
P(D1) = 0.1
and P(D2) = 1 – 0.1 = 0.9
⇒ P(A/D1) = 0.95
and P(A/D2) = 0.45.
P(rejection due to faulty design) = P(D1/A)
=
=
P (D1 ) . P ( A/D1 )
P (D1 ) . P ( A/ D1 ) + P (D2 ) . P ( A/ D2 )
( 0.1 ) × ( 0.95 )
= 0.19.
( 0.1 ) × ( 0.95 ) + ( 0.9 )( 0.45 )
60. A box contains 4 bad and 6 good tubes. Two are drawn out from the box at a time. One
of them is tested and found to be good. What is the probability that the other one is also
good?
Solution. Let D1 = One of the tubes drawn is good
D2 = The other drawn is also good.
P(D1 ∩ D2) = P(both the tubes drawn are good)
=
Hence
6 C2
10 C2
=
P(D2/D1) =
1
.
3
P ( D1 ∩ D2 )
P ( D1 )
1/3
5
= .
6/10 9
=
61. A bag contains 5 white and 3 black balls. Two balls are drawn at random one after the
other without replacement. Find the probability that both balls drawn are black.
Solution. P(Drawing a black ball in the first draw) =
3
8
P(Drawing the second black ball that the first ball drawn is black) =
2
7
3 2
3
× =
.
8 7 28
62. Four cards are drawn without replacement. What is the probability that they are all
aces?
P(both balls drawn are black) =
Solution.
P(A) = P(getting first ace) =
4
52
P(B) = P(getting second ace) =
P(C) = P(getting third ace) =
P(D) = P(getting fourth ace) =
3
51
2
50
1
.
49
P(all four cards are aces)
= P(A) × P(B) × P(C) × P(D)
=
C10\N-PROB\CHP1-2
3
4
2
1
×
×
×
.
52 51 50 49
Probability
25
63. Two fair dice are thrown independently. Three events A, B and C are defined as follows.
(a) odd face with the first die
(b) odd face with the second die
(c) Sum of the numbers in the 2 dice is odd.
Are the events A, B, C mutually independent?
3 1
3 1
= ; P(B) =
= .
6 2
6 2
The outcomes favourable to C are (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), ...
(see table of Example 50)
Hence
Solution. Given P(A) =
P(C) =
18 1
= , P(A ∩ B ∩ C) = 0, since C cannot happen when A and B happened. Now
36 2
1 1 1
× ×
≠ 0 = P(A ∩ B ∩ C)
2 2 2
Hence A, B, C are not mutually independent.
P(A) × P(B) × P(C) =
64. Two dice are thrown n times in succession. What is the probability of obtaining double
six atleast once?
Solution. There are 36 cases in every throw and two dice thrown n times gives (36)n
cases. In one throw there are 35 unfavourable cases and in n throws there will be (35)n
unfavourable cases.
Hence the total number of favourable cases will be (36)n – (35)n. The required probability
is
p=
( 36 )n − ( 35 )n
( 36 )n
n
35
= 1 −   .
 36 
65. What is the probability that among n people there are atleast two persons who have the
same birth day.
Solution. We consider a year with 365 days. Let all n people have different birth days.
The first person have any day of the year as his birthday with probability 1; The second
person can have any day but not the first man’s birthday, hence the birthday is
the third person can have any but not the two days, hence the probability is
on. Hence the probability
q=
=
=
C10\N-PROB\CHP1-2
365 364 363
.
.
... (n factors)
365 365 365
( 365 )( 364 )( 363 ) ... ( n factors )
( 365 )n
365( 365 − 1 )( 365 − 2 ) ... ( 365 − ( n − 1 ))
( 365 )n
364
;
365
363
and so
365
26
Probability, Random Processes and Queueing Theory
=
n −1
1 
2  

−
−
−
(365)(365)
... (365)
""" """
! 1 365  1 365  ... 1 365 
n times
(365)n
n − 1  

2  
p = 1 – q = 1 – 1 − 1 
1 −
 ... 1 −
 .
365
365
365  

 

66. A fair coin is tossed n times. What is the probability that exactly j heads appear?
Solution. When a coin is tossed n times we have a sequence of H and T. Total number
of sequences with j heads and n – j tails
=
n!
j ! (n − j ) !
The coin being fair, p = P(H) =
1
1
, q=
, and the trials being independent, each
2
2
sequence occurs with probability
1
.
2n
Hence the required probability
= P{exactly j heads in trials}
pjqn–j =
 1 
= nCj pj qn–j = nCj .  n  .
2 
67. A ball is drawn at random from an urn containing 7 red balls, 3 white balls and 5 blue
balls. Determine the probability that it is
(a) blue
(b) not blue
(c) red or blue.
Solution. Let A and B denote the events of drawing a red and a blue ball respectively.
Then
(a) P(B) =
5
1
=
7+3+5 3
(b) P(Not blue) = P(B′) = 1 – P(B) = 1 –
1 2
= .
3 3
7
5 4
+
= .
15 15 5
68. Two defective tubes get mixed up with 2 good ones. The tubes are tested, one by one,
until both defectives are found. What is the probability that the tube is obtained on (i)
the second test (ii) the third test and (iii) the fourth test?
Solution. Let D represent defective and N denotes the non-defective tube.
(i) P(Second tube is defective in the II test)
= P(First is defective in the I test and II is defective in II test)
= P(D1 ∩ D2) (say)
= P(D1) × P(D2)
(c) P(Red or Blue) = P(A ∪ B) = P(A) + P(B) =
=
C10\N-PROB\CHP1-2
2 1 1
× = .
4 3 6
Probability
27
(ii) P(Second tube is defective in the third test)
= P(D1 ∩ N2 ∩ D3 or N1 ∩ D2 ∩ D3)
2 2 1 2 2 1 1
× × + × × = .
4 3 2 3 3 2 3
(iii) P(Second tube is defective in the IV test)
= P(D1 ∩ N2 ∩ N3 ∩ D4) + p(N1 ∩ D2 ∩ N3 ∩ D4)
+ P(N1 ∩ N2 ∩ D3 ∩ D4)
=
2 2 1
2 2 1
2 2 1
× × ×1 + × × ×1 + × × × 1
4 3 2
4 3 2
4 3 2
1
= .
2
69. Four cards are drawn at random from a well shuffled pack of cards. Find the probability
that
(a) (i) all the four are queens
(ii) two cards are diamonds and two are spades
(iii) all the four cards are hearts and one of them is jack.
(b) Draw three cards at random. Find the chance that they are a king, a queen, and a
jack.
Solution. From a pack of 52 cards, 4 can be drawn in 52C4 ways.
=
(a) (i) P(all the four are queen) =
4 C4
52 C4
(ii) P(Two are diamonds and two are spades) =
13 C2 × 13 C2
52 C4
(iii) P(all the four cards are hearts and one of them is jack) =
1 C1 × 12 C3
52 C4
since out of 13
hearts, only one card is jack.
(b) Let three cards be drawn at random. So the exhaustive events = 52C3. A pack
contains 4 kings, 4 queens, and 4 jacks. A king, a queen, a jack can be chosen in 4C1
ways.
Therefore the required probability
=
4 C1 × 4 C1 × 4 C1
52 C3
.
70. A and B alternatively throw a pair of dice. A wins if he throws 6 before B throws 7 and
B wins if he throws 7 before A throws 6. If A begins, show that his chance of winning is
30/61.
Solution. Throwing 6 with 2 dice =
Getting 6 as the sum of numbers shown on the top faces of the two dice.
P(Throwing 6 with 2 dice) =
5
(see Example 50)
36
P(Throwing 7 with 2 dice) =
6
1
= .
36 6
C10\N-PROB\CHP1-2
28
Probability, Random Processes and Queueing Theory
Let X ≡ Event of A throwing 6
Y ≡ Event of B throwing 7.
A plays in the first, third, fifth, ... (2n – 1) trials
B plays in the second, Fourth, Sixth, ... 2n trials
Therefore A will win, if he throws 6 in the first trial or third trial or in subsequent (odd)
trials.
Hence
P(A wins) = P(X or XY X or
XYXYX or ...)
= P(X) + P( XY X) + ...
=
5  31 5
5   31 5 31 5
5 
+
×
×
×
×
×
×
+

36  36 36 36   36 36 36 36 36 
5
36
=
by using Geometric series expansion
1 −  155 
 216 
30
.
61
71. What is the chance that a leap year selected at random will contain 53 sundays.
Solution. In a leap year (which contains 366 days) there are 52 complete weeks and 2
days over. We have the following combinations of the left out days.
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(viii) Saturday and Sunday.
In order that a leap year selected at random will contain 53 sundays, one of the two over
days must be sunday. Since out of the above 7 possibilities, the favourable cases is 3.
2
Hence the required probability =
.
7
72. The sum of two non-negative quantities is equal to 2n. Find the chance that their
3
product is not less than
times their greatest product.
4
Solution. Let x, y be non-negative quantities such that x + y = 2n. We know that the
product of two positive quantities whose sum is constant is greatest when the quantities
are equal. Thus the product of x and y is maximum when x = y = n.
Maximum product = n × n = n2.
=
 | 3 n2  P  xy 3 n2  P  x ( 2 n x ) 3 n2 
−
≥
P  xy <
=  ≥
 = 
4
4
5




2
2
= P[(4x – 8nx + 3n ) ≤ 0]
= P[(2x – 3n)(2x – n) ≤ 0]
C10\N-PROB\CHP1-2
Probability
29
n
3n 

= P  x lies between and
2
2 

Favourable range
Total range
3n n
−
=n
2
2
= 2n
=
n
1
= .
2n 2
73. A committee of 4 people is to be appointed from 3 officers of the production department,
4 officers of the purchase department, two officers of the sales department and 1 chartered accountant. Find the probability of forming the committee in the following manner
(i) There must be one from each category
(ii) It should have atleast one from the purchase department
(iii) The chartered accountant must be in the committee.
Solution. Out of 3 + 4 + 2 + 1 = 10 persons, committee of 4 peoples are formed in 10C4
ways.
Required Probability
=
(i) P(one from each category) =
4C1 × 3C1 × 2C1 × 1
10
C4
(ii) p(atleast one from the purchase department) = 1 – P(committe has no purchase
department)
In order that the committee has no purchase officer, all the 4 members are to be
selected among officers of production department, sales department and chartered accountant.
i.e., out of 3 + 2 + 1 = 6 member, 4 persons can be chosen in 6C4 ways. So
P(Committe has no purchase department) =
6C4
.
10C4
6C4
1 13
=1–
.
=
10C4
14 14
(iii) P(The chartered accountant must be in the committe) = 1 × 9C3 = 9C3.
74. Out of (2n + 1) tickets consecutively numbered three are drawn at random. Find the
chance that the numbers on them are in A.P.
Solution. Since out of (2n + 1) tickets, 3 tickets can be drawn in (2n + 1)C3 ways.
To find the favourable number of cases we are to enumerate all the cases in which the
numbers on the drawn tickets are in A.P. with common difference d = 1, 2, 3, ..., n – 1, n
(say).
If d = 1, the possible cases are as follows.
1, 2, 3; 2, 3, 4; ... 2n – 1, n, 2n + 1. (i.e., 2n – 1 cases in all)
If d = 2, the possible cases are as follows.
1, 3, 5; 2, 4, 6; ... 2n – 3, 2n – 1, 2n + 1, (i.e., 2n – 3 cases in all)
and so on.
p(atleast one from the purchase department) = 1 –
C10\N-PROB\CHP1-2
30
Probability, Random Processes and Queueing Theory
If d = n – 1, the possible cases are as follows:
1, n, 2n – 1
2, n + 1, 2n
3, n + 2, 2n + 1 (3 cases in all)
If d = n, there is only on case viz. (1, n + 1, 2n + 1).
Hence the total number of favourable cases = (2n – 1) + (2n – 3) + ... + 5 + 3 + 1
n
[1 + (2n – 1)] = n2.
2
3n
n2
=
Therefore required probability =
.
2
n ( 4 n − 1 ) ( 4 n2 − 1 )
3
75. Show that the probability of getting 3 atleast once in 4 throws of a die is greater than
the probability of getting a double six atleast once in 24 throws with two dice?
1
Solution. The probability of getting 3 in one throw of a die =
6
5
The probability of not getting 3 in one throw of a die =
.
6
By compound probability theorem, the probability that in 4 throws of a die no ‘3’ is
=
4
5
obtained =   . Hence the probability of obtaining ‘3’ atleast once in 4 throws = 1 –
6
4
5
  = 0.516. In a trial of throwing two dice, the probability of getting double six i.e., (6,
6
1
35
. Probability of not getting (6, 6) is
. Probability of not getting (6, 6) in 24
6) =
36
36
24
 35 
throws of two dice = 
 .
 36 
P(getting (6, 6) atleast once in 24 throws) = 1 – P(not getting (6, 6) in 24 throws)
 35 
=1– 

 36 
24
= 0.491.
Hence the result follows.
76. Find the probability that atmost 5 defective fuses will be found in a box of 200 fuses if
experiences shows that 2% of such fuses are defective.
Solution. Given n = 200, p = 2% =
2
= .02, mean λ = np = 200 × .02 = 4. We know the
100
Poisson distribution:
P(x) =
e −4 . 4x
e− λ λx
=
.
x!
x!
e −4 . 4 x
, x = 0, 1, 2, 3, 4, 5, ...
x!
P(atmost 5 defective bulbs) = p(x ≤ 5).
= p(0) + p(1) + p(2) + p(3) + p(4) + p(5)
Hence P(x defective bulbs) =
C10\N-PROB\CHP1-2
Probability
=
e −4 . 40
0!
+
e −4 . 41
1!
+
e −4 . 4 2
2!
+
e −4 . 43
3!
+
e −4 . 4 4
4!
+
31
e −4 . 4 5
5!

4 42
43
44
45 
+
+
+
= e–4 1 + +

1 2! 3! 4! 5!

= e–4 × 42.87 ~
− 0.785.
77. If X and Y are independent Poisson variate such that P(X = 1) = P(X = 2) and P(Y = 2) =
P(Y = 3). Find the variance of X – 2Y.
Solution. We know P(X = x) =
e −λ λx
. Given
x!
P(X = 1) = P(X = 2)
e–λ λ =
⇒
and
⇒
e −λ λ2
2!
...(1.1)
P(Y = 2) = P(Y = 3)
e −µ . µ2 e −µ . µ3
=
2!
3!
...(1.2)
From (1.1) we get 2e–λ . λ = e–λ λ which implies λe–λ[λ – 2] = 0. Since λ > 0, λe–λ ≠ 0, the
only possibility is λ – 2 = 0, which gives λ = 2.
From (1.2), we get 6e–µ µ2 = 2e–µ µ3, which implies 3 = µ. Hence var (X) = λ = 2 and
var (Y) = µ = 3.
var (X – 2Y) = 12 var (X) + (– 2)2 . var (Y) = 2 + 4 × 3 = 14.
78. A box contains tags marked 1, 2, ..., n. Two tags are chosen at random
without replacement. Find the probability that the numbers on the tags will be consecutive integers.
Solution. If the numbers on the tags are to be consecutive integers, they must be
chosen as a pair from the following pairs (1, 2), (2, 3), (3, 4), ..., (n – 1, n).
Number of ways of choosing any one pair from the above (n – 1) pairs is (n – 1)C1 ways.
Total No. of ways of choosing 2 tags from the tags is nC2. Therefore
Required probability =
n −1
2
= .
n(n − 1) n
2
79. The probability that a company director will travel by train is
1
2
and by plane is
.
3
5
What is the probability of his travelling by train or plane?
Solution. Let E1 be the event of travelling by train and E2 be the event of travelling by
plane. The probability of travelling by train or plane = P(E1 or E2)
1 2 13
.
+ =
5 3 15
80. A is known to hit the target in 2 out of 5 shorts, whereas B is known to hit the target in
3 out of 4 shots. Find the probability of the target being hit when they both try.
= P(E1) + P(E2) =
C10\N-PROB\CHP1-2
32
Probability, Random Processes and Queueing Theory
2
3
1
3
, P(B) = , P ( A ) = ; P ( B ) = . As the events are independent,
5
4
4
5
3 1
3
the probability that the target will not hit = × =
. The probability that the target
5 4 20
3 17
will hit = 1 –
.
=
20 20
81. If two dice are thrown, find the probability that the sum is neither 6 nor 10.
Solution. Let A be the event of getting the sum as 6, B be the event of getting the sum
as 10. According to the table in Example 50, the cases favourable to the event A are
Solution. Given P(A) =
(1, 5), (2, 4), (3, 3), (4, 2), (5, 1). Hence P(A) =
5
3
. Similarly we have P(B) =
.
36
36
The probability that the sum is neither 6 nor 10
3 2
5
+
= P[(A ∪ B)C] = 1 – P(A ∪ B) = 1 – 
= .
 36 36  3
82. The odds in favour of A solving a mathematical problem are 3 to 4 and the odds against
B solving the problem are 5 to 7. Find the probability that the problem will be solved by
atleast one of them.
Solution. P(of A not solving the problem) =
P(of B not solving the problem) =
4
7
5
.
12
4 5 16
×
=
.
7 12 21
83. If from a lottery of 30 tickets numbered 1, 2, 3, ... 30, four tickets are drawn, what is the
chance that those marked 1 and 2 are among them?
Solution. Out of 30 tickets, 4 are drawn in 30C4 ways. In the 4 tickets 1 and 2 should be
present. The remaining 2 are chosen in 28C2 ways. Hence the required probability is
Probability of either one of them solving the problem = 1 –
28C2
.
30C4
84. If n pencils are distributed among m children. What is the probability that a particular
child receives r pencils, where r < n?
Solution. The first pencil can be given to any one of the m children that can be done in
m ways. The second pencil can be given in the same way in m ways. Therefore 2 pencils
can be given in m2 ways. Extending this concept, n pencil can be distributed in mn ways.
r pencils received by the particular child can be chosen from n pencils in nCr ways. If
this child receives r pencils, the remaining (n – r) pencils can be distributed among the
remaining (m – 1) children in (m – 1)n–r ways.
No. of favourable cases is nCr × (m – 1)n–r
Required probability =
C10\N-PROB\CHP1-2
nCr × (m − 1)n −r
mn
.
Probability
33
85. From 6 positive and 8 negative numbers, 4 numbers are chosen at random (without
replacement) and multiplied. What is the probability that the product is positive?
Solution. If the product is to be positive, all the 4 numbers must be positive or all the 4
must be negative or 2 of them must be positive and the other 2 must be negative.
No. of ways of choosing 4 positive numbers is 6C4 ways, and No. of ways of choosing 4
negative numbers is 8C4 ways, and No. of ways of choosing 2 positive and 2 negative
numbers = 6C2 × 8C2 ways. No. of ways of choosing 4 numbers from 14 numbers is 14C4
ways. Hence,
Probability that the product is positive
=
No. of ways by which the product is positive
Exhaustive cases
6C4 × 8C4 × 6C2 × 8C2
.
14C4
86. In a coin tossing experiment, if the coin shows head, 1 die is thrown and the result is
recorded. If the coin shows tail, 2 dice are thrown and their sum is recorded. What is the
probability that the recorded number is 6?
Solution. Let E1, E2 be the event of getting head and tail respectively. Let A be the
event of getting 6.
=
P(getting 6 when one die is thrown) = P(A/E1) =
1
6
P(getting a sum when two dice is thrown) = P(A/E2) =
5
36
Required probability = P(A)
= P(E1) . P(A/E1) + P(E2) . P(A/E2)
=
1 1 1 5
× + ×
2 6 2 36
11
.
72
87. A bag contains 50 tickets numbered 1, 2, ..., 50 of which 5 are drawn at random and
arranged in ascending order of the magnitude. What is the probability that the middle
one is 30?
Solution. Out of 50 tickets, 5 are drawn in 50C5 ways. Let the tickets be arranged in
ascending order say x1 < x2 < x3 < x4 < x5. If x3 = 30, then the two tickets with number x1
and x2 must come from 29 tickets numbered 1 to 29. They can be chosen in 29C2 ways.
Similarly x3, x4 come from 31 to 50 and it can be chosen in 20C2 ways.
The favourable cases = 29C2 × 20C2.
=
29C2 × 20C2
.
50C5
88. A problem in mathematics is given to five students A1, A2, A3, A4, and A5. Their chances
1 1 1 1 1
of solving it are , , , , respectively. What is the probability that the problem will
6 5 4 3 2
be solved?
Hence the required probability =
C10\N-PROB\CHP1-2
34
Probability, Random Processes and Queueing Theory
Solution. The probability that A1 fails to solve the problem = 1 –
that A2 fails to solve the problem 1 –
problem = 1 –
1 5
= , the probability
6 6
1 4
= , the probability that A3 fails to solve the
5 5
1 3
1 2
= , the
= , the probability that A4 fails to solve the problem = 1 –
3 3
4 4
probability that A5 fails to solve the problem = 1 –
1 1
= . The probability that the
2 2
problem is not solved by all five students (independent events) =
Hence, the probability that the problem will be solved = 1 –
5 4 3 2 1 1
× × × × = .
6 5 4 3 2 6
1 5
= .
6 6
89. Find the chance of throwing ten with four dice.
Solution. A die is a small cube marked with numbers 1, 2, 3, 4, 5, 6, in its faces. The
number of ways of throwing 10 with four dice
= Coefficient of x10 in (x + x2 + x3 + x4 + x5 + x6)4
= Coefficient of x6 in (1 + x + x2 + x3 + x3 + x4 + x5)4
1 − x 6 
= Coefficient of x in 

 1 − x 
4
6
= Coefficient of x6 in (1 – x6)4 (1 – x)–4


4 × 5 × x2
4 × 5 ... × 9x 6
+ ... +
+ ...
= Coefficient of x6 in (1 – x6)4 1 + 4x +
2!
6!



4 × 5 × x2
4 × 5 ... × 9x 6 
1
4
...
x
+
+
+
+

= Coefficient of x in (1 – x ) 
2!
6!


6
6 4
neglecting higher powers of x6.
=
4 × 5 × ... × 9
– 4 = 80.
6!
The total number of sums of numbers appearing on the faces of the four dice = 64.
5
.
81
6
90. If A1 and A2 are equally likely, mutually exclusive and exhaustive events and P(B/A1)
= 0.2, P(B/A2) = 0.3, find p(A1/B).
Solution. Since A1 and A2 are equally likely, we have P(A1) = P(A2). Further, they are
mutually exclusive and exhaustive
P(A1) + P(A2) = 1.
P(A1) = P(A2) = 0.5
The chance of throwing ten with four dice =
C10\N-PROB\CHP1-2
80
4
=
Probability
P(A1/B) =
=
35
P ( A1 ) . P ( B/ A1 )
P ( A1 ) . P ( B/ A1 ) + P ( A2 ) . P ( B/ A2 )
(0.5)(0.2)
= 0.4.
(0.5)(0.2) + (0.5)(0.3)
91. In a horse race, the odds in favour of the four horses are 1 : 3, 1 : 4, 1 : 5, 1 : 6
respectively. Assume that a dead heat is not possible, find the chance that one of them
wins the race.
Solution. Since it is not possible for all the four horses to cover the same distance in the
same time (a dead heat), the events are mutually exclusive.
Let p1 p2, p3, p4 be the probabilities of winning of the horses H1, H2, H3, H4 respectively,
we have p1 =
1
1
1
1
1
= , p3 = ,
= , (given odds in favour of H1 are 1 : 3), p2 =
6
1+4 5
1+3 4
1
1 1 1 1 638
. Thus, the chance that one wins the race = p1 + p2 + p3 + p4 = + + + =
.
7
4 5 6 7 840
92. The odds that person X speaks the truth are 3 : 2 and the odds that person Y speaks the
truth are 5 : 3. In what percentage of cases are they likely to contradict each other on an
identical point.
Solution. Let A : X Speaks the truth,
B : Y Speaks the truth.
p4 =
Given P(A) =
3 2
3
3
= and P( A ) = 1– P(A) = 1 – = ,
5 5
3+2 5
5 3
5
5
= .
and P( B ) = 1 – P(B) = 1 –
=
8 8
5+3 8
The event E that X and Y contradict each other on an identical point can happen in the
following mutually exclusive ways.
P(B) =
(i) X speaks the truth, Y tells a lie, i.e., A ∩ B
(ii) X tells a lie, Y speaks the truth i.e., A ∩ B
Hence by addition theorem,
P(E) = P(A ∩ B ) + P( A ∩ B)
= P(A) . P( B ) + P( A ) P(B) since A and B are independent
3 3 2 5 19
× + × =
= 0.475.
5 8 5 8 40
Hence A and B are likely to contradict each other on an identical point in 47.5% cases.
93. Why does it pay to bet consistently on seeing 6 atleast once in 4 throws of a die, but not
on seeing a double six atleast once in 24 throws with two dice?
Solution. Same as problem 75.
94. (Chebychev’s Problem). What is the chance that two numbers, chosen at random, will be
prime to each other?
=
C10\N-PROB\CHP1-2
36
Probability, Random Processes and Queueing Theory
Solution. If a number ‘a’ is divided by a prime number ‘r’ then the possible remainders
1
are 0, 1, 2, ..., r – 1. Hence the chance that ‘a’ is divisible by r is
(because the only case
r
favourable to this is remainder being 0).
1
. Since the numbers a
r
and b chosen at random, the probability that none of them divisible by r is given by
compound probability theorem by
Similarly that a number ‘b’ chosen at random is divisible by r is
2
1 
1 
1

1 − r 1 − r  = 1 − r  , r = 2, 3, 5, 7, ...


 

Hence the probability that the numbers chosen at random are prime to each other is
2
6
1

give by Π 1 −  , where r is a prime number, and by trigonometry this is equal to 2 .
π
r 
r
95. p is the probability that a man aged x years will die in a year. Find the probability that
out of n men, A1, A2, ..., An, each aged x, A1 will die in a year, and will be the first to die.
Solution. Let Ei = Event that Ai will die in a year, i = 1, 2, ..., n.
The probability that none of A1, A2, ..., An dies in a year
= P ( E1 ∩ E2 ... ∩ En )
= P ( E1 ) . P ( E2 ) ... P ( En )
by compound probability theorem
= (1 – P(E1)) . (1 – P(E2)) ... (1 – P(En))
= (1 – p)(1 – p) ... (1 – p) = (1 – p)n.
The probability that atleast one of A1, A2, ..., An dies in a year = 1 – P ( E1 ∩ E2 ... ∩ En )
1
, and since
n
this event is independent of the event that atleast one man dies in a year,
= 1 – (1 – p)n. The probability that among n men, A1 is the first to die is
required probability is
1
[1 – (1 – p)n].
n
96. A die is loaded in such a manner that for n = 1, 2, 3, 4, 5, 6, the probability of the face
marked n, landing on top when the die is rolled is proportional to n. Find the probability
that an odd number will appear on tossing the die.
Solution. Given P(n) is proportional to n. That is P(n) = kn, for some constant k. Also
1
P(1) + P(2) + ... + P(6) = 1 ⇒ k(1 + 2 + 3 + 4 + 5 + 6) = 1 or k =
.
21
1+3+5 3
= .
Required probability = P(1) + P(3) + P(5) =
21
7
97. A five-figured number is formed by the digits 0, 1, 2, 3, 4 (without repetition). Find the
probability that the number formed is divisible by 4.
C10\N-PROB\CHP1-2
Probability
37
Solution. The total number of ways in which the five digits 0, 1, 2, 3, 4 can be arranged
among themselves is 5 !. The number of arrangements that being with 0 is 4 !. The total
number of five digited numbers that can be formed from 0, 1, 2, 3, 4 is 5 ! – 4 ! = 96. The
number formed will be divisible by 4 it the numbers in the last two digits is divisible by
4. Such numbers are 04, 12, 20, 32, and 40. If the numbers end in 04, the remaining
three digits 1, 2, 3 can be arranged in 3 ! ways.
Similarly, the arrangements of 20, 40 is 3 ! in each case. If the numbers end with 12, the
remaining three digit 0, 3, 4 can be arranged in 3 ! – 2 ! = 4 ways. Similarly for 24, and
32 the number of ways of arranging the previous three letters is 4 in each case. Hence
the total number of favourable cases is:
3 × 3 ! + 3 × 4 = 18 + 12 = 30.
30
.
96
Hence the required probability =
2
98. A sportman chance of shooting an animal at a distance r(> a) is a . He fires when
r2
r = 2a, and if he misses he reloads and fires when r = 3a, 4a, ... If he misses at distance
na, the animal escapes. Find the odds against the sportman.
Solution. P[Sportsman shoots at a distance ia] =
a2
(ia )
P[Sportsman misses the shot at a distance ia] = 1 –
n
P[Animal escapes] =
1
i2
1
i2
1
2

n
1
2

∏ 1 − i
i =2
n
=
=
∏ 1 − i
i =2
=
2
∏
i =2
i −1
i
n
1

1 + i  .


∏
i =2
i+2
i
n − 1 3 4 5
n + 1
n +1
1 2
× × × ...
=  × × ...
=
.
2n
n   2 3 4
n 
2 3
n + 1
n + 1 
: 1 −
= 
= (n + 1) : (n – 1).

2n 
 2n  
99. Of three independent events, the chance that the first only should happens is a, the
chance of the second only is b and the chance of the third only is c. Show that the
independent chances of the three events are respectively
Required ratio
a
b
c
,
,
a+x b+x c+x
where x is the root of the equation (a + x)(b + x)(c + x) = x2.
C10\N-PROB\CHP1-2
38
Probability, Random Processes and Queueing Theory
Solution. P ( E1 ∩ E2 ∩ E3 ) = P(E1) . (1 – P(E2))(1 – P(E3)) = a
P( E1 ∩ E2 ∩ E3 ) = (1 – P(E1) P(E2)(1 – P(E3)) = b
P ( E1 ∩ E2 ∩ E3 ) = (1 – P(E1))(1 – P(E2)) P(E3) = c.
Multiplying (1.3), (1.4), (1.5), we get
P(E1) P(E2) P(E3)x2 = abc,
where x = (1 – P(E1))(1 – P(E2))(1 – P(E3))
Multiplying (1.3) by (1 – P(E1)), we get
P(E1)(1 – P(E1))(1 – P(E2))(1 – P(E3)) = a(1 – P(E1))
⇒
P(E1)x = a(1 – P(E1))
⇒
P(E1)(x + a) = a
...(1.3)
...(1.4)
...(1.5)
a
.
a+x
Similarly, using (1.4), (1.5), we prove
⇒
P(E1) =
P(E2) =
b
c
, P(E3) =
.
b+x
c+x
100. There is a series of n urns. In the ith urn there are i white and (n – i) black balls i = 1,
2, 3, ..., k. One urn is chosen at random and 2 balls are drawn from it. Both urn turn out
to be white. What is the probability that the jth urn was chosen, where j is a particular
number between 3 and n.
Solution. Let Ej denote the event of selection of jth urn, j = 3, 4, ... n and A denote the
event of drawing of 2 white balls, then
 j   j −1 
P(A/Ej) =   

n n −1
P(Ej) =
P(Ej/A) =
1
, P(A) =
n
n
1
i  i −1 
∑ n  n   n − 1 
i =1
1   j   j −1 
     −1
n n n

n
∑
i =1
1   i   i −1 
  

n  n   n −1
.
101. (i) Let the probability pn that a family has exactly n children be αpn, when n ≥ 1 and
p0 = 1 – αp(1 + p + p2 + ...). Show that all sex distributions of n children have the same
probability. Show that for k ≥ 1, the probability that a family contains exactly k boys is
2α . pk
.
(2 − p)k +1
(ii) Given that a family includes atleast one boy, show that the probability that there
are two or more boys is
C10\N-PROB\CHP1-2
p
.
2− p
Probability
39
Solution. Given pn = αpn, n ≥ 1,
p0 = 1 – αp(1 + p + p2 + ...)
Let Ej be the event that the number of children in a family is j and let A be the event
that a family contains exactly k boys. Then
P(Ej) = pj ; j = 0, 1, 2, ...
Now, since each child can have any of the two sex distributions (either a boy or a girl)
the total number of possibility for a family to have j children is 2j.
Therefore
P(A/Ej) =
jCk
2j
, j ≥ k.
∞
P(A) =
∞
∑ P(E ) P(A/E ) = ∑ p
j
j =k
∑
j=k
=α
∞
∑
∞
∑ αp
j
.
j =k
jCk
2j
j
k +r
 p
=α  
2
Hence
2
j
=
 p
 2  . jCk, j ≥ k ≥ 1.
 
r =0
We know
⇒
(– 1)n .
j
j =k
∞
=α
j
jCk
.
(–n)
Cr = (– 1)r .
–(k+1)
Cr = k+rCr.
 p
P(A) = α  
2
 p
=α  
2
k
k +r
 p
Ck  
2
∞
∑
k+r
r=0
(n+r–1)
k
∞
∑
, put j – k = r.
r
 p
Cr   , since nCr = nCn–r.
2
Cr ⇒ (– 1)r .
( − 1)r .
−( k +1)
r=0
k
∞
∑
Cr =
 p
Cr  
2
n+r–1
Cr
r
r
− p
Cr 

 2 
−( k +1)
r=0
k
–n
p
 p 
= α   1 − 
2
2 
− ( k +1)
by binomial expansion
k
2k +1
2 α pk
 p
=α   .
.
=
 2  (2 − p) k +1 (2 − p )k +1
(ii) Let B denote the event that a family includes atleast one boy and C denote the event
that a family has two or more boys. Then
∞
P(B) =
∑ P (family has exactly k boys)
k =1
=
∞
∑ (2 − p)
k =1
C10\N-PROB\CHP1-2
2α p k
k +1
2α
=
(2 − p)
∞
∑
k =1
 p 
 − p 
)
 (2
k
40
Probability, Random Processes and Queueing Theory


p


2α  2 − p 
=
by Geometric Series expansion
2− p 
 p 
1 − 


 2 − p  
p
2α
2− p
αp
×
×
=
2 − p 2 − p 2 − p − p (2 − p)(1 − p)
=
P(C) =
∞
∑ P [family has exactly k boys]
k=α
∞
=
2αpk
∑ (2 − p)
k =2
k +1
2α
=
2−p
∞
 p 


2 − p
k =2 
∑
k
2
 p 
2 − p
αp2
2α


=
=
.
(2 − p)2 (1 − p)
(2 − p )
 p 
1−

2 − p
Since C ⊂ B, and B ∩ C = C, P(B ∩ C) = P(C)
⇒
P(B) P(C/B) = P(C). Therefore
(1 − p)(2 − p)
P (C )
p
αp2
=
×
=
.
2
2−p
P ( B ) (2 − p) (1 − p)
αp
102. If n letters are randomly placed in correctly addressed envelopes, prove that the probability
that exactly r letters are placed in correct envelopes is given by
P(C/B) =
n −r
1
1
( − 1)k .
, r = 1, 2, ... n.
r ! k=0
k!
∑
Solution. Let Ei, i = 1, 2, ... n be the events that ith letter goes to the correct envelope.
The probability that none of the n letters goes to the correct envelope is
P ( E1 ∩ E2 ∩ En ) = 1 – P(E1 ∪ E2 ∪ ... ∪ En)


n
 n

n −1
P ( Ei ) −
P ( Ei ∩ E j ) + ... + ( − 1)
P ( E1 ∩ E2 ∩ ... ∩ En )
=1– 
 i =1

i, j =1


i< j
...(1.6)
by principle of inclusion and exclusion.
∑
We know P(Ei) =
1
1
1
.
, for all i, P(Ei ∩ Ej) = P(Ei) P(Ej/Ei) =
, for all i, j such that
n
n n −1
i < j, and P(Ei ∩ Ej ∩ Ek) =
C10\N-PROB\CHP1-2
∑
1
1
1
.
.
, for all i, j, k with i < j < k, and so on.
n n −1 n − 2
Probability
41
Substituting in (1.6), we get

1
1
1
+ nC3 .
P ( E1 ∩ E2 ∩ ... ∩ En ) = 1 – nC1 . − nC2 .
(
1)
(
1)(
n
n
n
n
n
n − 2)
−
−

+ ... + (– 1)n .

1

n( n − 1) ... 3 . 2 . 1 

1
1
1 
+
− ... + ( − 1)n −1 .
= 1 – 1 −

n
2
!
3
!
!

=
1
1
1
1
−
+
+ ... + ( − 1)n .
2! 3! 4!
n!
=
∑
n
k =0
( − 1)k
.
k!
The probability that each of the r letters is in the right cover is
=
1
1
1
.
. ...
n n −1
n − ( r − 1)
n−r
The probability that the remaining (n – r) letters goes to incorrect envelope is
∑
k =0
( − 1)k
.
k!
Hence by compound probability theorem, the probability that out of n letters, r letters
n−r
go to correct envelopes is
1
( − 1)k
.
, r ≤ n – 2.
n( n − 1) ... ( n − (r − 1)) k = 0 k !
∑
Since r letters goes to n envelopes in nCr mutually exclusive ways, the required probability of exactly r letters goes to correct envelopes is
n−r
1
( − 1)k
×
nCr ×
.
n( n − 1) ... ( n − ( r − 1)) k = 0 k !
∑
103. The chances of winning 3 out of 5 games and 4 out of 5 games are equal. What is the
chance of winning all the 5 games.
Solution. Let p, q be the probabilities of winning and losing a game. Then p + q = 1.
Probability of winning 3 out of 5 games = 5C3 q2p3 = 10q2 p3
Probability of winning 4 out of 5 games = 5C4 qp4 = 5qp4
Given
10q2p3 = 5qp4 ⇒ 2q = p ⇒ 2(1 – p) = p
⇒
p=
2
, and
3
1– p=
1
.
3
5
32
2
.
Hence probability of winning 5 out of 5 games = 5C5 p5 =   =
243
3
C10\N-PROB\CHP1-2
42
Probability, Random Processes and Queueing Theory
104. A pair of dice is rolled. If the sum of 9 has appeared, find the probability that one of the
dice shows 3.
Solution. Let A = the event that the sum is 9
B = the event the one of dice shows 3.
Exhaustive cases = 62 = 36.
Favourable cases for the event A = (3, 6), (6, 3), (4, 5), (5, 4), so P(A) =
4 1
= .
36 9
Favourable case for the event A ∩ B = (3, 6), (6, 3).
2
1
=
.
36 18
Since P(A ∩ B) = P(A) × P(B/A), where have
Hence P(A ∩ B) =
1
P ( A ∩ B ) 18 1
=
= .
P(B/A) =
1
P( A)
2
9
105. Out of 800 families with 4 children each, how many families would be expected to have
(i) 2 boys and 2 girls
(ii) atleast one boy
(iii) no girl
(iv) atmost 2 girls.
Solution. p = probability that a child is a girl =
1
2
1
2
n = total number of children in a family = 4.
q = probability that a child is a boy =
2
2
3
1 1
(i) P(exactly 2 boys) = 4C2q2p2 = 6     =
2
2
8
   
4
15
1 
(ii) P(atleast one boy) = 1 – P(no boy) = 1 – 4C0p4 = 1 –   =
.
16
2
4
1
1 
(iii) P(no girl) = P(all are boys) = 4C4q4 =   =
16
2
(iv) P(atmost 2 girls) = P(0 girls or 1 girl or 2 girl)
= P(4 boys or 3 boys or 2 boys)
= 4C2q2p2 + 4C3pq3 + 4C4q4
11
.
16
Total number of families = 800.
Therefore required number of families, in cases (i), (ii), (iii) and (iv) respectively, are
=
(i) 800 ×
3
= 300
8
C10\N-PROB\CHP1-2
(ii) 800 ×
15
= 750
16
Probability
43
1
11
= 50
(iv) 800 ×
= 550.
16
16
106. The odds against manager X setting the wage dispute with the workers are 8 : 6 and
odds in favour of manager Y setting the same dispute are 14 : 16.
(i) What is the chance that neither settles the dispute, if they both try, independently
of each other?
(ii) What is the probability that the dispute will be settled?
Solution. (i) Let E : The manager X will settle the dispute
F : The manager Y will settle the dispute
(iii) 800 ×
Then P ( E ) =
8
4
=
8+6 7
⇒
14
7
P(F) = 14 + 16 = 15
3
7
P(E) =
8
.
15
The probability that neither settle the dispute
⇒ P (F ) =
= P (E ∩ F ) = P (E ) . P ( F )
4 8
32
.
×
=
7 15 105
(ii) The dispute will be settled if atleast one of the managers X and Y settles the
dispute. Hence the required probability
P(E ∪ F) = 1 – P(none settles the dispute)
=
= 1 – P (E ∩ F )
=1–
32
73
=
.
105 105
EXERCISES
1. Two students A and B work independently on a problem. The probability that the first one will
2
3
and the probability that second one will solve it is
. What is the probability that
3
4
the problem will be solved?
2. For given three events A, B, C, verify that
solve it is
P((A È B)/C) = P(A/C) + P(B/C) – P((A Ç B)/C).
3. A bag contains 3 red, 6 white, 7 blue balls. What is the probability that two balls drawn are white
and blue.
4. If P(A1 È A2) =
5. If P(A) =
2
1
, P(A1 Ç A2) =
. Find P A1C ∪ A2C
3
6
(
) and P ( A1C ∩ A2C ) .
1
1
1
, P(B) =
, P(A Ç B) =
. Find P(AC È BC).
4
3
6
C10\N-PROB\CHP1-2
44
Probability, Random Processes and Queueing Theory
6. A problem in statistics is given to three students A, B, C whose chances of solving it are
7.
8.
9.
10.
1 3 1
, , respectively. What is the probability that the problem will be solved if all of them try
2 4 5
independently.
Suppose there are three boxes containing 2 white and 3 black; 3 white and two black, 4 white and
1 black balls respectively. There is equal probability of each box being chosen. One ball is drawn
from a box at random. What is the probability that a ball drawn is white.
A bag contains 8 white and 6 red balls. Find the probability of drawing two balls of the same
colour.
A box contains 4 bad and 6 good tubes. Two are drawn out together. One of them is tested and
found to be good. What is the probability that the other one is also good.
Show that the multiplication theorem P(A Ç B) = P(A/B) P(B) established for two events, may be
generalized to three events as follows.
P(A Ç B Ç C) = P(A/B Ç C) P(B/C) P(C).
11. Prove: If P(A/B) > P(A), then P(B/A) > P(B).
12. Suppose that colored balls are distributed in three indistinguishable boxes as follows:
Box
Red
White
Blue
I
2
3
5
II
4
1
3
III
3
4
3
A box is selected at random from which a ball is selected at random and it is observed to be red
what is the probability that box 3 was selected.
13. A factory has two machines A and B. Past records show that machine A produces 30% of the total
output and machine B the remaining 70%. Machine A produces 5% defective articles and machine B produces 1% defective articles. An item is drawn at random and found to be defective.
What is the probability that it was produced by machine A.
14. A coin is tossed. If it turns up head, two balls will be drawn from an urn A, otherwise two balls
will be drawn from urn B. Urn A contains 3 black and 5 white balls. Urn B contains 7 black and
1 white ball. In both cases selections are to be made with replacements, what is the probability
that urn A is used given that both the balls drawn are black.
ANSWERS
1.
11
12
5. P(AC È BC) =
9.
5
9
C10\N-PROB\CHP1-2
5
6
(
)
3.
7
20
C
C
4. P A1 ∪ A2 =
6.
29
32
7.
12.
3
10
(
)
1
5
C
C
, P A1 ∩ A2 =
3
6
2
9
8.
43
91
13. 0.682
14.
1
.
8