Class: XI
Subject:
√ Math’s
Topic: Probability
No. of Questions: 20
Let Q, R, T be three events. If the probability of occurring exactly one of the event out of Q
and R is 1-y, out of R and T is 1-2y, out of T and Q is 1-y, and that of occurring three events
2
simultaneously is y , then the probability (P) that at least one out of Q, R, T will, occur is
A
ia
Sol:
greater than 1/2
greater than 1
greater than and equal to 1/2
less than and equal to 1
ns
A.
B.
C.
D.
P(Q) + P(R) – 2P (Q
R) = 1 – y
……..(1)
P(R) + P(T) – 2 P(R
T) = 1 – 2y
…….(2)
P(T) + P(Q) – 2 P(T
Q) = 1 – y
……..(3)
IT
1.
………(4)
and P(Q
R
T) = y
Adding (1), (2) and (3), we get
R) – P(R
kI
P(Q) + P(R) + P(T) – P(Q
T) – P(T
Probability that atleast one out of Q, R, T will occur = P(Q
R) – P(R
as
P(Q) + P(R) + P(T) – P(Q
=
R
T)
Q) + P(Q
R
T)
2
+y
2
= y – 2y +
2.
T) – P(T
…….(5)
Q) =
2
= (y – 1) +
>
A,B, C are events such that p(A) = 0.3, p(B) = 0.4, p(C) = 0.8 , P(A ∩ B) = 0.08, P(A
0.28, P(A
A.
B.
C.
D.
B
C) = 0.09. If P(A
B
C) >= 0.75, then P(B
C) =
C) lies in the interval
(- 0.48, 0.23)
(0.23, 0.48)
(0.75, 0.48)
None of these
Page|1
Sol:
B
P(A
B
0.75
0.75
C)
0.75 (Given)
P(A
B
C)
1
P(A) + P(B) + P(C) – P(A
B) – P(B
0.75
0.3 + 0.4 + 0.8 – 0.08 – P(B
After solving we get
0.23
C)
C) – 0.28 + 0.09
B
C)
1
1
0.48
If A, B and C are three mutually exclusive and exhaustive events of an experiment such that
3P(A) = 2P(B) = P(C), then P(A) is equal to
ia
A.
B.
IT
C.
B
kI
D.
Option (2) is the correct answer. Let 3P (A) = 2P(B) = P(C) = p which gives p (A) =
, P(B) =
as
Sol:
C) + P(A
ns
3.
P(B
C) – P(A
and P(C) = p
Since A, B and C are mutually exclusive and exhaustive events, we have P(A) + P(B) + P(C)
=1
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4.
Four digit numbers are formed using the digits 0 to 4 without repetition. The probability that a
number so formed is divisible by 2 is
A.
B.
C.
D.
D
ns
Sol:
A.
B.
C.
kI
Three squares of a chess board are selected at random. The probability of getting 2 squares
of one colour and other of a different colour is
as
5.
=
IT
Required probability =
ia
Total numbers formed = 4.4.3.2 = 96
Numbers divisible by 2 = (4.3.2.1) {zero in the last place} = (3.3.2.2)
{2 or 4 in the last place} = 60
D.
Sol:
A
Option (1) is the correct answer. In a chess board, there are 64 squares of which 32 are white
and 32 are black. Since 2 of one colour and 1 of other can be 2W, 1B, or 1W, 2B, the number
32
66
of ways is ( C2
32C1)
2 and also, the number of ways of choosing any 3 boxes is C3.
Hence, the required probability =
=
Page|3
6.
One mapping (function) is selected at random from all the mappings of the set A = {1, 2, 3,
….., n} into itself. The probability that the mapping selected is one to one is
A.
B.
C.
D. None of these
Sol:
C
A.
B.
kI
Three digit numbers are formed using the digits 0, 2, 4, 6, 8. A number is chosen at random
out of these numbers. What is the probability that this number has the same digits?
as
7.
IT
Hence, the required probability is
ia
ns
Option (3) is the correct answer. Total number of mappings from a set A having n elements
n
onto itself is n .Now, for one to one mapping the first element in A can have any of the n
nd
images in A; the 2 element in A can have any of the remaining (n – 1) images, counting like
this, the nth element in A can have only 1 image.
Therefore, the total number of one to one mapping is n!.
C.
D.
Sol:
D
Option (4) is the correct answer. Since a-3-digit number cannot start with digit 0, so the
hundredth place can have any of the 4 digits. Now, the tens and units place can have all the
5 digits. Therefore, the total possible 3-digit numbers are 4
5
5, i.e., 100.The total
possible 3 digit numbers having all digits same = 4.
Hence, P(3-digit number with same digits) =
=
.
Page|4
8.
A bag contains 7 white and 9 blue balls. If two balls are drawn at random, what is the chance
that one ball is white and the other is blue?
A.
B.
C.
D.
D
Sol:
B
1/7
2/7
3/7
4/7
kI
A.
B.
C.
D.
IT
In a survey of 70 businessmen, it is found that 20 of them own only scooters, 10 own only
bikes and 15 own only cars. 5 businessmen own all the three. Find the probability that a
businessmen selected at random possess only two items.
as
9.
=
ia
Required probability =
ns
Sol:
From the Venn diagram, 20 + 10 + 15 + 5 + (x + y + z) = 70.
Therefore, x + y + z = 20.
So, the required probability =
=
.
Page|5
10.
A bag contains 3 different white, 4 different black and 2 different red balls. Two balls are
chosen at random. What is the probability that (a) one white and one red ball is chosen,
(b) no white ball is chosen, (c) exactly one black ball is chosen?
A.
B.
C.
D.
Sol:
1/6, 5/36, 4/9
5/6, 21/36, 5/9
1/6, 5/12, 5/9
None of these
C
9
Total number of outcomes = C2 = 36
3
2
3
=
ns
(a) Number of favourable outcomes = ( C1) ( C1) = 6; Probability =
3
6
9
(b) Probability that a white ball is chosen = [ C2 + C1 C1] / C2 =
4
=
=
ia
Probability that no white ball is chosen =1 –
5
2
A.
B.
C.
D.
Sol:
A
1/18
13/18
5/18
1/9
kI
The coefficients b and c of the equation x + bx + c = 0 are determined by throwing an ordinary
die. The probability that the equation has equal roots is
as
11.
=
IT
(c) Number of favourable outcomes = ( C1 C1) = 20; Probability =
2
Condition for the equation to have equal roots is b - 4c=0
This is satisfied only for two sets of values (2, 1),(4, 4)
Reqd. prob. = 2/36 = 1/18
Page|6
12.
Six letters are to be placed in six addressed envelopes. If the letters are placed at random into
the envelopes, the probability that all of them are placed in the correct envelopes is
A.
B.
C.
D.
Sol:
1
0
1/6!
5/6!
C
There is only one chance for the reqd. event, so the reqd. prob. =1/6!
The probability that a teacher will give an unannounced test during any class meeting is
. If
a student is absent twice, then the probability that the student will miss at least one test is
ns
13.
ia
A.
IT
B.
C.
D
Prob. that one test is held = 2 x
as
Sol:
kI
D.
x
Prob. that test is held on both days =
=
x
=
Prob. that the student misses at least one test =
Page|7
14.
The probability that a man lives after 10 years is
and that his wife is alive after 10 years is
. The probability that neither of them is alive after 10 years is
A.
B.
C.
P(M) =
P(
, P(W) =
)=1-
=
,P(
) , P(
)=1-
)=
=
,
=
kI
Reqd. prob. = P(
If the letters of the word “MISSISSIPPI” are written down at random in a row, the probability
that no two „S‟ occur together is
A.
as
15.
ia
A
IT
Sol:
ns
D.
B.
C.
D. None of these
Sol:
B
First we will arrange the letters other than S
This can be done in
ways
Page|8
Now, 4 S can be arranged in between at 7 places in 8C4 ways
Hence, favourable outcomes =
Total ways =
Required probability =
ns
In a box there are 2 red, 3 black and 4 white balls. Out of these, three balls are drawn
together. The probability of the balls being drawn of the same colour is
ia
16.
B.
C
Total number of equally likely cases
9
as
Sol:
kI
C.
D. None of these
IT
A.
= C3 =
= 3 x 4 x 7 = 84
3
4
Favorable case = C3 C3 = 1 + 4 = 5
Required probability =
Page|9
17.
The chances of wining of two race horses are 1/3 and 1/6 respectively. What is the probability
that at least one will win, when the horses are running in different races?
A.
B.
C.
D.
Sol:
1/2
1/18
4/9
None of these
C
P (at least one will win) = 1 - P (none wins) = 1 -
=
.
Six guys are waiting for an interview in a conference hall. There are 8 rooms in the office in
which the interview is to be held. The probability that each of them enters a different room
for the interview is
ia
A.
C.
D. None of these
kI
B
IT
B.
Sol:
=1-
ns
18.
×
as
Six people can enter in 8 rooms in P(8, 6) ways. The favourable number of ways, i.e. if they
6
can enter more than one room = 8 . So, the required probability =
19.
The probabilities of the three doctors A, B and C getting success in an operation are 0.5, 0.2
and 0.3 respectively. Find the probability that the operation is not successful.
A.
B.
C.
D.
Sol:
.
0.78
0.64
0.56
0.28
D
Since A, B or C could do the operation independently, these are mutually exclusive events.
Therefore, the required probability is (1 – 0.5) ´ (1 – 0.2) ´ (1 – 0.3) = 0.28.
Page|10
If a card is drawn at random from a packet of 100 cards numbered 1 to 100, the
probability of drawing a number on the card that is a cube is
A.
B.
C.
D.
B
Cubes in between 1 and 100 =1,8,27,64 = 4 in no.
IT
ia
ns
So, required probability is 4/100 = 1/25
kI
Sol:
3/100
1/25
9/100
1/10
as
20.
Page|11