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13. Probability
(Conditional Prob. Independent Events and Bayes Theorem)
QUESTION 1 : The events and
are such that
Find:
SOLUTION:
P(A|B)=
/P(B)
= P(A|B) x P(B) = 0.3 x 0.6 = .18
⇒
=P(A) + P(B) 0.2 + 0.6 – 0.18 = 0.62
is the complement of
So: .
QUESTION 2 : The events and are such that:
Determine the value of
in each of the cases when:
(a) and
(b) and
(c)
are mutually exclusive. .
are independent.
SOLUTION:
and
(a)
are mutually exclusive ⇒
=0
=P(A) + P(B)
⇒P(B) = 0.7 – 0.2 = 0.5
(b) If and are independent: .
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Substitute into: .
...............
..............
(c)
.
So we have: .
Substitute into: .
...............
..............
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QUESTION3 :
Suppose two balls are removed from a bag containing three
green and two red balls and that the first ball is not replaced
before the second is removed. What is the probability that
the balls are the same colour?
The addition law for mutually exclusive events can then be used.
(b)
(c)
(d)
(e)
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QUESTION 4:
Two balls are removed from a bag containing three green and two red balls. The first
ball is not replaced before the second is removed. If the second ball is red, what is the
probability that the first ball was green?
SOLUTION:
A = {first ball green} and B = {second ball red}
QUESTION 5: Consider a company with 60 employees (45 males and 15 females).
A
There are three types of employee:
•
•
•
A full time - 10 employees
B part time - 20 employees
C casual - 30 employees
B
C
M
7
15
23
F
3
5
7
If the employee chosen is full time, what is the probability that the employee is a
female?
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SOLUTION:
As event A has occurred, the sample space is restricted to the 10 outcomes of the
event {employee is full time}. Of these outcomes, there are 3 in the event
{employee is female and full time}.
(f)
QUESTION 6: Two cards are drawn from a deck of 52 cards without replacement.
(a) What is the probability of drawing a face card on the first draw and an ace on
the second draw? (b) What is the probability of getting a face card and and ace
from the two draws?
SOLUTION:
(Here,order is not important)
Face cards = 4 x 3 (jack, queen, king) = 12
A) first draw = 12/52 = 3/13
second draw = 4/51
3/13 x 4/51 = 4/221
B) same because you want both a face and an ace. No matter what order you draw
in, you get the same prob once you multiply the two probs.
EX:
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first draw to get an ACE = 4/52 = 1/13
second draw to get a FACE = 12/51 = 4/17
1/13 x 4/17 = 4/221
QUESTION7The probability that student entering a university will graduate is 0.4.
Find the probability that out of 3 students of the university:
(i) none will graduate,
(ii) only one will graduate,
(iii) all will graduate.
SOLUTION:
Let X be the number of students who graduate.
P = Prob. that a student graduates = 0.4
_ q =1 - p =1 - 0.4 = 0.6 and n = 3
Here: ( q + p ) 3 = (0.6 + 0.4) 3
(g)
QUESTION 8: A company has two plants to manufacture motorcycles. 70% motor
cycles are manufactured at the first plant, while 30% are manufactured at
the second plant. At the first plant, 80% motorcycles are rated of the
standard quality while at the second plant, 90% are rated of standard
quality. A motorcycle, randomly picked up, is found to be of standard
quality. Find the probability that it has come out from the second plant.
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SOLUTION: Let A, B and E be the events that a motorcycle is manufactured at
plant I, atplant II and the motor cycle chosen is of standard quality respectively.
Then
(h)
Now by Bayes’ theorem,
(i)
QUESTION 9: Two dice are rolled once. Find the probability that
(i) the numbers on two dice are different
(ii) the total of numbers on the two dice is at least 4.
SOLUTION: Total number of cases = 6 x 6 = 36
(i) When the numbers on the two dice are different,The number of favorable cases
= 6 x 5 = 30.
(ii) Let X = total of the numbers on the two dice
P(X ³ 4) =1- [P(X = 2) + P(X = 3)]
total of the numbers on the dice can not be 0 or 1
(There is only one favourable case for X = 2, namely (1, 1) and two
favourable cases for X = 3, namely (1, 2) and (2, 1))
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QUESTION 10: A card is drawn at random from a well shuffled pack of 52 cards.
Find the probability that it is neither an ace nor a king.
SOLUTION:
Let A be the event of getting an ace card
B be the event of getting a king card.
Total cards = 52
\ P (neither an ace nor a king) = 1– P (A or B)
=1 – [P(A) + P(B) - P(AÇB)]
QUESTION 11: An urn contains 7 red and 4 blue balls. Two balls are drawn at
random with replacement. Finds the probability of getting (a) 2 red balls (b)
2 blue balls (c) one red and one blue ball.
SOLUTION:
Red balls = 7
Blue balls = 4
Total balls = 7+ 4 =11
(a) P (2 red balls) = P (RR)
(j)
(b)P (2 blue balls) = P (BB)
(c) P (one red and one blue) = P (BR) + P (RB)
QUESTION 12: Probability of solving specific problem independently by A and B
are1/2 and1/3respectively. If both try to solve the problem independently, find the
probability that the problem is solved .
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SOLUTION:
Let E1 be the event that A solves the problem, then P(E1) = ½,
so P(E1/)=1- ½=1/2
And E2 be the event that B solves the problem, then P(E2)= 1/3=1-1/3=2/3
P(E2/)= 1-1/3=2/3
P(that the problem is solved) =
P( that atleast one of the two solves the problem)
=1 – P(Niether of the two solves the problem)
=1- P(E1/∩E2/) =1- P(E1/) ∩P(E2/)
= 1 - ½ x 2/3
(k)
=1- 1/3 = 2/3
QUESTION 13: A die is tossed thrice. Find the probability of getting an odd
number at least once.
SOLUTION:
(l)
When a die is tossed thrice, the sample space is {1,2,3,4,5,6} each time
S = {( x, y, z ) : x, y, z ∈ {1, 2, 3, 4,5, 6}}
S contains 6 × 6 × 6 = 216 cases
Let E : 'an odd number appears atleast once'
∴ E ' : 'an odd number appears none of the times'
(m) i.e E ' : 'an even number appears all three times'
E ' = {( x, y, z ) : x, y, z ∈ {2, 4, 6}}
E ' contains 3 × 3 × 3 = 27 cases
Now, P(E) = 1 - P(E ' )
= 1−
27
1 7
= 1− =
216
8 8
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QUESTION 14: In a factory which manufactures bulbs, machines X, Y and Z
manufacture respectively 1000,2000, 3000 of the bulbs. Of their outputs,1%, 1.5%
and 2 % are respectively defective bulbs. A bulb is drawn at random from the
product and is found to be defective. What is the probability that it is manufactured
by the machine X ?
SOLUTION:
Let events B1, B2, B3 be the following :
B1 : the bulb is manufactured by machine X
B2 : the bulb is manufactured by machine Y
B3 : the bulb is manufactured by machine Z
Clearly, B1, B2, B3
are mutually exclusive and exhaustive events and hence, they represent a partition
of the sample space.
Let the event E be the event ‘the bulb is defective’.
The event E occurs with B1 or with B2 or with B3
Given that,
P(B1) = 1000/(1000+2000+3000)= 1/6,
P (B2) = 2000/(1000+2000+3000= 1/3 and
P(B3) = 3000/(1000+2000+3000= 1/2
Again P(E|B1 ) = Probability that the bulb drawn is defective given that it is manufactured by machine A = 1% = 1/ 100
Similarly, P(E| B2) = 1.5%=15/ 100 = 3/ 200 P(E| B3) 2%== 2/100
Hence, by Bayes' Theorem, we have
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P ( B1 E ) =
P( B1 )( PE B1 )
P( B1 )( PE B1 ) + P( B2 )( PE B2 ) + P ( B3 )( PE B3 )
1 1
×
6 100
1 1 1 3
1 2
×
+ ×
+ ×
6 100 3 200 2 100
1
6
=
1 1
+ +1
6 2
1
1
=
=
1 + 3 + 6 10
=
QUESTION 15: . What is the probability of picking a spade from a normal pack of
cards and rolling an odd number on a die?
SOLUTION:
the probability of both occurring is
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