OP Malhotra Class 12 Maths Solutions Chapter 19 Baye’s Theorem Ex 19

Effective OP Malhotra ISC Class 12 Solutions Chapter 19 Baye’s Theorem Ex 19 can help bridge the gap between theory and application.

S Chand Class 12 ICSE Maths Solutions Chapter 19 Baye’s Theorem Ex 19

Question 1.
Bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was-drawn from bag B.
Answer:
Given Bag A contains 2 white and 3 red balls Bag B contains 4 white and 5 red balls Consider the following events as :
E₁ = bag A is chosen
E₂ = bag B is chosen
A = Drawing one red ball
Then P(E₁) = \(\frac{1}{2}\); P(E₂) = \(\frac{1}{2}\)
[Since there are 2 bags]
∴ P (A | E₁)
= P[Drawing one red ball from bag A]
= \(\frac{3}{5}\)
P(A / E₂)
= P[Drawing one red ball from bag B]
= \(\frac{5}{9}\)
We want to find P (Drawing, one red ball is from bag B) = P(E₂ / A)
Then By Baye’s theorem, we have
Required probability = P(E₂/ A)
= \(\frac{P(E_2) P(A / E_2)}{P(E_1) P(A / E_1)+P(E_2) P(A / E_2)}\)
= \(\frac{\frac{1}{2} \times \frac{5}{9}}{\frac{1}{2} \times \frac{3}{5}+\frac{1}{2} \times \frac{5}{9}}\)
= \(\frac{\frac{5}{9}}{\frac{3}{5}+\frac{5}{9}}\)
= \(\frac{\frac{5}{9}}{\frac{27+25}{45}}\)
= \(\frac{5}{9}\) × \(\frac{45}{52}\)
= \(\frac{25}{52}\)

Question 2.
There are two bags I and II, containing 3 red and 4 white balls, and 2 red and 3 white balls respectively. A bag is selected at random and a ball is drawn from it. It is found to be a red ball, find the probability that it is drawn from the first bag.
Answer:
Given bag-I contains 3 red and 4 white balls bag-II contains 2 red and 3 white balls Let us define the events as follows:
E₁ : bag- I is selected
E₂ : bag-II is selected
E : a red ball is drawn
∴ P(E₁) = P(E₂) = \(\frac{1}{2}\)
Clearly E₁ and E₂ are mutually exclusive and exhausitive events.
P(E / E₁) = prob. of drawing red ball from bag-I = \(\frac{3}{7}\)
P(E / E₂) = prob. of drawing a red ball from bag-II = \(\frac{2}{5}\),
we want to find the prob. that ball drawn from first bag when it is found that it is red. i.e. To find P(E₁ / E)
Then by Baye’s Theorem, we have
P(E₁ / E) = \(\frac{P(E / E_1) P(E_1)}{P(E / E_1) P(E_1)+P(E / E_2) P(E_2)}\)
= \(\frac{\frac{3}{7} \times \frac{1}{2}}{\frac{3}{7} \times \frac{1}{2}+\frac{2}{5} \times \frac{1}{2}}\)
= \(\frac{\frac{3}{7}}{\frac{3}{7}+\frac{2}{5}} \)
= \(\frac{\frac{3}{7}}{\frac{15+14}{35}}\)
= \(\frac{3}{7}\) × \(\frac{35}{29}\)
= \(\frac{15}{29}\)

Question 3.
Suppose that 5 men out of 100 and 25 women out of 1000 are good orators. An orator is chosen at random. Find the probability that a male person is chosen. Assume that there are equal number of men and women.
Answer:
Consider the following events :
E₂ = Selected person is male
E₂ = Selected person is female
A = Selected person is an orator
Then P(E₁) = \(\frac{1}{2}\); P(E₂) = \(\frac{1}{2}\)
[Since number of males and females are equal] and P(A | E₁) = P (Selecting person be a male orator)
= \(\frac{5}{100}\) = \(\frac{1}{20}\)
P(A/ E₂) = P(Selecting person be a female orator)
= \(\frac{25}{1000}\) = \(\frac{1}{40}\)
We want to find the probability that the orator chosen be a male = P(E₁/ A)
Then By baye’s theorem, we have
P(E₁/ A) = \(\frac{P(E_1) P(A / E_1)}{P(E_1) P(A / E_1)+P(E_2) P(A / E_2)}\)
= \(\frac{\frac{1}{2} \times \frac{1}{20}}{\frac{1}{2} \times \frac{1}{20}+\frac{1}{2} \times \frac{1}{40}}\)
= \(\frac{1}{1+\frac{1}{2}}\)
= \(\frac{2}{3}\)
∴ Required probability = \(\frac{2}{3}\)

Question 4.
Suppose that 5 % of men and 0.25 % of women have grey hair. A grey haired person is selected at random, what is the probability of this person being male? Assume that there are equal number of males and females.
Answer:
Let us define the events are as follows :
E₁ : person selected is male
E₂ : person selected is female
E : person is having grey hair
Then P(E₁) = P(E₂) = \(\frac{1}{2}\)
Thus E₁ and E₂ are mutually exclusive and exhausitive events.
P(E/ E₁) = prob. that person is having grey hair is male = \(\frac{5}{100}\)
P(E / E₂) = prob. that person is having grey hair is women = 0.25 % = \(\frac{0.25}{100}\)
Now we want to find the prob. that person being male when it is given that he is having grey hair i.e. to find P(E₁ / E).
Thus by Baye’s Theorem, we have
P(E₁|E) = \(\frac{\mathrm{P}(\mathrm{E} \mid \mathrm{E}_1) \mathrm{P}(\mathrm{E}_1)}{\sum_{i=1}^2 \mathrm{P}(\mathrm{E} \mid \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{\frac{5}{100} \times \frac{1}{2}}{\frac{5}{100} \times \frac{1}{2}+\frac{0.25}{100} \times \frac{1}{2}}\)
= \(\frac{\frac{5}{100}}{\frac{5}{100}+\frac{0.25}{100}}\)
= \(\frac{5}{5.25}\)
= \(\frac{500}{525}\)
= \(\frac{20}{21}\)

Question 5.
A company has two plants to manufacture scooters. Plant I manufactures 7 0 % of the scooters and plant II manufactures 30 %. At plant I, 80 % of scooters are rated standard quality and at plant II, 9 0 % of scooters are rated standard quality. A scooter is picked up at random and is found to be of standard quality. What is the chance that it was from plant II ?
Answer:
Let us define the events as follows :
E₁ : scooter is manufactured by plant-I
E₂ : scooter is manufactured by plant-II
Then P(E₁) = 70 % = \(\frac{70}{100}\) = \(\frac{7}{10}\);
P(E₂) = 30 % = \(\frac{30}{100}\) = \(\frac{3}{10}\)
Let E : scooter is of standard quality Thus E₁ and E₂ are mutually exclusive and exhausitive events.
Also P(E | E₁) = prob. of getting scooter of standard quality and coming from plant I
= 80 % = \(\frac{80}{100}\)
P(E | E₂) = prob. of getting scooter of standard quality from plant II = \frac{90}{100}
= \(\frac{9}{10}\)
Now we want to find P(E₂)
Thus by Baye’s Theorem, we have
P(E₂) = \(\frac{P(E | E_2) P(E_2)}{P(E | E_1) P(E_1)+P(E | E_2) P(E_2)}\)
= \(\frac{\frac{9}{10} \times \frac{3}{10}}{\frac{8}{10} \times \frac{7}{10}+\frac{9}{10} \times \frac{3}{10}}\)
= \(\frac{\frac{27}{100}}{\frac{83}{100}}\)
= \(\frac{27}{83}\)

Question 6.
A company has two plants to manufacture bicycles. The first plant manufactures 60 % of the bicycles and the second plant 40 % .80 % of the bicycles are rated of standard quality at the first plant and 90 % of standard quality at the second plant. A bicycle is picked up at random and found to be of standard quality. Find the probability that it comes from the second plant.
Answer:
Let us consider the events :
E₁ = Selecting bicycle manufactured by first plant
E₂ = Selecting bicycle manufactured by second plant
A = selected bicycle is of standard quality Thus, P(E₁) = \(\frac{60}{100}\);
P(E₂) = \(\frac{40}{100}\); and P(A | E₁ = P
(Selecting a bicycle of standard quality from first plant ) = \(\frac{80}{100}\)
P(A / E₂) = P( Selecting bicycle of standard quality from second plant ) = \(\frac{90}{100}\)
We want to find, P (Selected bicycle of standard quality is from second plant)
= P(₂/ A)
Then By Baye’s theorem, we have
= P(E₂/ A)
= \(\frac{P(E_2) P(A / E_2)}{P(E_1) P(A / E_1)+P(E_2) P(A / E_2)}\)
= \(\frac{\frac{40}{100} \times \frac{90}{100}}{\frac{60}{100} \times \frac{80}{100}+\frac{40}{100} \times \frac{90}{100}}\)
= \(\frac{3600}{4800+3600}\)
= \(\frac{3600}{8400}\)
= \(\frac{3}{7}\)
∴ Required probability = \(\frac{3}{7}\).

Question 7.
A has an alarm which will ring at the appointed time with probability 0.9 . If the alarm rings, it will awake him and he will/ each the examination hall in time with probability 0.8 . If the alarm does not ring, A will get up at his own time to reach the
examination hall in time, with probability 0.3. K now ing that the person \boldsymbol{A} reached the hall in time, find the probability that the alarm ring.
Answer:
Let us define the events are as follows :
E₁ : alarm rings
E₂ : alarm donot rings
E : person A reached the hall in time.
Thus P(E₁) = 0.9 ; P(E₂) = 1 – 0.9 = 0.1
∴ E₁ and E₂ are mutually exclusive and exhausitive events.
Given, P(E | E₁) = prob. that person A reached the hall in time when it is given that alarm rings = 0.8
P(E | E₂) = prob. that person A reached the hall in time when it is given that alarm donot rings = 0.3
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{\mathrm{P}(\mathrm{E} \mid \mathrm{E}_1) \mathrm{P}(\mathrm{E}_1)}{\sum_{i=1}^2 \mathrm{P}(\mathrm{E} \mid \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{0.8 \times 0.9}{0.8 \times 0.9+0.3 \times 0.1}\)
= \(\frac{0.72}{0.75}\)
= \(\frac{72}{75}\) = \(\frac{24}{25}\) = 0.96

Question 8.
By examining the chest X-rays, probability that T.B. is detected when a person is actually suffering, is 0.99 . The probability that the doctor diagnoses incorrectly that a person has T.B. on the basis of X-ray is 0.001 . In a certain city 1 in 1000 persons suffers from T.B. A person selected at random is diagnosed to have T.B. What is the chance that he actually has T.B.
Answer:
Consider the following events:
E₁ = The person selected is actually suffering from T.B.
E₂ = The person selected is not suffering from T.B.
A = The selected person diagnosed to have T.B.
Then P(E₁) = \(\frac{1}{1000}\);
P(E₂) = \(\frac{999}{1000}\)
∴ P(A/ E₁) = probability that person diagnosed to have T.B. and he is actually having T.B. = 0.99
and P(A/ E₂) = probability that person diagnosed to have T.B. and he is not actually having T.B. = 0.001
We want to find, probability that person diagnosed to have T.B. is actually having T.B. = P(E₁/ A)
Then by baye’s theorem, we have
P(E₁/ A) = \(\frac{P(E_1) P(A / E_1)}{P(E_1) P(A / E_1)+P(E_2) P(A / E_2)}\)
= \(\frac{\frac{1}{1000} \times 0.99}{\frac{1}{1000} \times 0.99+\frac{999}{1000} \times 0.001}\)
= \(\frac{990}{990+999}\)
= \(\frac{990}{1989}\)
= \(\frac{110}{221}\)
Thus, required probability = \(\frac{110}{221}\).

Question 9.
Shoes are produced by two machines A and B .50 % of the shoes are produced by machine A with an estimate of 10 % of them being defective. On machine B, 20 % of the shoes produced are defective if a shoe taken at random is found to be defective, what is the probability that shoe was produced by machine A I
Answer:
Let us define the events are as follows:
E₁: shoes are produced by machine A
E₂ : shoes are produced by machine B
Then P(E₁) = 50 % = \(\frac{1}{2}\);
P(E₂) = \(\frac{1}{2}\)
∴ E₁ and E₂ are mutually exclusive and exhausitive events.
Let E : shoe is taken to be defective
∴ P(E | E₁) = 10 % = \(\frac{1}{10}\);
∴ P(E | E₂) = 20 %
= \(\frac{20}{100}\)
= \(\frac{2}{10}\)
we want to find P(E₁ | E).
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1)+P(E | E_2) P(E_2)}\)
= \(\frac{\frac{1}{10} \times \frac{1}{2}}{\frac{1}{10} \times \frac{1}{2}+\frac{2}{10} \times \frac{1}{2}}\)\(
= [latex]\frac{\frac{1}{10}}{\frac{1}{10}+\frac{2}{10}}\)
= \(\frac{1}{3}\)

Question 10.
In a large company, 15 % of the employees are graduates (G), and of these, 80 % work in administrative posts (A). Of the non-graduate (NG) employees of the company, 10 % work in administrative posts. Find the probability that an employee of this company selected at random from those working in administrative posts will be a graduate.
Answer:
Let us define the events are as follows:
E₁ : employees are graduate
E₂ : employees are Non-graduate
E : employee of the company working in administrative post
Then P(E₁) = 15 % = \(\frac{15}{100}\)
∴ P(E₂) = 85 % = \(\frac{85}{100}\)
∴ P(E | E₁) = 80 % = \(\frac{80}{100}\);
∴ P(E | E₂) = 10 % = \(\frac{10}{100}\)
We want to find (E₁ | E).
Then by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1)+P(E | E_2) P(E_2)}\)
= \(\frac{\frac{80}{100} \times \frac{15}{100}}{\frac{80}{100} \times \frac{15}{100}+\frac{10}{100} \times \frac{85}{100}}\)
= \(\frac{1200}{1200+850}\)
= \(\frac{1200}{2050}\)
= \(\frac{120}{205}\)
= 0.585

Question 11.
Three urns contain 6 red and 4 black, 4 red and 6 black, and 5 red and 5 black balls respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is red, find the probability that it is drawn from the first urn.
Answer:
Given urn I contains 6 red and 4 black balls urn II contains 4 red and 6 black balls urn III contains 5 red and 5 red balls Let us define the events are as follows:
E₁ : urn-I is chosen
E₂ : urn-II is chosen
E₃ : urn-III is chosen
E : ball drawn is red.
Then P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)
∴ E₁ and E₂, E₃ are mutually exclusive and exhausitive events.
Also P(E | E₁) = prob, of drawing a red ball from urn l = \(\frac{6}{10}\)
P(E | E₂) = prob. of drawing red ball from urn 11 = \(\frac{4}{10}\)
P(E | E₃) = prob. of drawing red ball from urn III = \(\frac{5}{10}\)
We want to find P(E₁ | E).
Thus by Baye’s theorem, we have
P(E₁ | E) = \(\frac{P(E | E_1) P(E_1)}{\sum_{i=1}^3 P(E | E_i) P(E_i)}\)
= \(\frac{\frac{6}{10} \times \frac{1}{3}}{\frac{6}{10} \times \frac{1}{3}+\frac{4}{10} \times \frac{1}{3}+\frac{5}{10} \times \frac{1}{3}}\)
= \(\frac{\frac{6}{10}}{\frac{6}{10}+\frac{4}{10}+\frac{5}{10}}\)
= \(\frac{6}{15}\)
= \(\frac{2}{5}\)

Question 12.
Three mris are given, each containing red and black balls as indicated below:
Urn I: 6 rid and 4 black balls.
Urn II: 2 red and 6 black balls.
Urn III: 1 red and 8 black balls.
An urn is chosen at random and a ball is drawn from the urn. The ball drawn is red. Find the probability that the ball is drawn.either from urn II or from urn III.
Answer:
Given urn I : 6 red and 4 black ball
urn II : 2 red and 6 black ball
urn III : 1 red and 8 black ball
Let us define the events are as follows :
E₁ : urn I is chosen
E₂ : urn II is chosen
E₃ : urn III is chosen
Then P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)
∴ E₁, E₂ and E₃ be mutually exclusive and exhausitive events.
Let E : ball drawn is red.
∴ P(E | E₁) = prob. of drawing a red ball from urn-I = \(\frac{6}{10}\)
P(E | E₂) = prob. of drawing a red ball from urn-II = \(\frac{2}{8}\)
P(E | E₃) = prob. of drawing a red ball from urn-III = \(\frac{1}{9}\)
Thus by Baye’s Theorem,

Question 13.
Suppose that there is a chance for a newly constructed house to collapse whether the design is faulty or not. The chance that the design is faulty is 2 0 %. The chance that the house collapses if the design is faulty is 98 % and otherwise it is 25 %. It is seen that the house collapsed. What is the probability, that it is due to faulty design?
Answer:
Let us define the events are as follows :
E₁ : design of house is faulty
E₂ : design of house is not faulty
E : house is collapsed
Then P(E₁) = 20 % = 0.2;
P(E₂) = 1 – 0.2 = 0.8
∴ E₁ and E₂ are mutually exclusive and exhaustive events.
Also P(E | E₁) = Prob. that house is collapsed when it is given that design is faulty = 98 % = 0.98
P(E} | E₂) = prob. that house is collapsed when it is given that design is not faulty = 25 % = 0.25
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{P(E | E_1) P(E_1)}{P(E | E_1) P(E_1)+P(E | E_2) P(E_2)}\)
= \(\frac{0.98 \times 0.2}{0.98 \times 0.2+0.25 \times 0.8}\)
= \(\frac{0.196}{0.196+0.2}\)
= \(\frac{0.196}{0.396}\)
= \(\frac{196}{396}\)
= \(\frac{49}{99}\)

Question 14.
In an automobile factory, certain parts are to be fixed to the chasis in a section before it moves into another section. On a given day, one of the three persons A, B and C carries out this task. A has 45 % B has 35 % and C has 20 % chance of doing it. The probability that A, B and C will take more than the allotted time are 1 / 6, 1 / 10 and 1 / 20 respectively. If it is found that none of them has taken more time, what is the probability that A has taken more time?
Answer:
Let us define the events are as follows:
E₁ : Task carry out by person A
E₂ : Task carry out by person B
E₃ : Task carry out by person C
E : event of taking more time than the allotted time
Then P(E₁) = \(\frac{45}{100}\);
P(E₂) = \(\frac{35}{100}\);
P(E₃) = \(\frac{20}{100}\)
P(E | E₁) = prob. that A takes more time than the allotted time = \(\frac{1}{6}\)
P(E | E₂) = \(\frac{1}{10}\);
P(E | E₃) = \(\frac{1}{20}\)
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{\mathrm{P}(\mathrm{E} \mid \mathrm{E}_1) \mathrm{P}(\mathrm{E}_1)}{\sum_{i=1}^3 \mathrm{P}(\mathrm{E} \mid \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{\frac{1}{6} \times \frac{45}{100}}{\frac{1}{6} \times \frac{45}{100}+\frac{1}{10} \times \frac{35}{100}+\frac{1}{20} \times \frac{20}{100}}\)
= \(\frac{\frac{45}{600}}{\frac{450+210+60}{6000}}\)
= \(\frac{45}{600} \times \frac{6000}{720}\)
= \(\frac{45}{72}\)
= \(\frac{5}{8}\)

Question 15.
Urn A contains 2 white, 1 black and 3 red balls, urn B contains 3 white, 2 black and 4 red balls and urn C contains 4 white, 3 black and 2 red balls. One urn is chosen at random and 2 balls are drawn at random from the urn. If the chosen balls happen to be red and black, what is the probability that both balls come from urn B?
Answer:
Given urn A contains 2 white, 1 black and 3 red balls
urn B contains 3 white, 2 black and 4 red balls
and urn C contains 4 white, 3 black and 2 red balls
Let us define the events as follows :
E₁ : urn A is chosen
E₂ : urn B is chosen
E₃ : urn C is chosen
E : chosen ball happen to be red and black.
Then P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)
P(E | E₁) = prob. of drawing 1 red and 1
black ball from urn A = \(\frac{{ }^3 \mathrm{C}_1 \times{ }^1 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}\)
= \(\frac{3}{\frac{6 \times 5}{2}}\)
= \(\frac{1}{5}\)
P(E | E₂) = probability of drawing 1 red and 1 black ball from urn B = \(\frac{{ }^2 C_1 \times{ }^4 C_1}{{ }^9 C_2}\)
= \(\frac{2 \times 4}{\frac{9 \times 8}{2}}\)
= \(\frac{16}{72}\)
= \(\frac{2}{9}\)
P(E | E₃) = prob. of drawing 1 red and 1
black ball from urn C = \(\frac{{ }^3 \mathrm{C}_1 \times{ }^2 \mathrm{C}_1}{{ }^9 \mathrm{C}_2}\)
= \(\frac{3 \times 2}{\frac{9 \times 8}{2}}\) = \(\frac{1}{6}\)
We want to find P(E₂ | E)
Thus by Baye’s Theorem, we have
P(E₂ | E) = \(=\frac{\mathrm{P}(\mathrm{E} \mid \mathrm{E}_2) \mathrm{P}(\mathrm{E}_2)}{\sum_{i=1}^3 \mathrm{P}(\mathrm{E} \mid \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{\frac{2}{9} \times \frac{1}{3}}{\frac{1}{5} \times \frac{1}{3}+\frac{2}{9} \times \frac{1}{3}+\frac{1}{6} \times \frac{1}{3}}\)
= \(\frac{\frac{2}{9}}{\frac{1}{5}+\frac{2}{9}+\frac{1}{6}}\)
= \(\frac{\frac{2}{9}}{\frac{18+20+15}{90}}\)
= \(\frac{2}{9} \times \frac{90}{53}\)
= \(\frac{20}{53}\)

Question 16.
Three bags contain balis as shown in the following table :

Bag	Number of
	White balls	Black balls	Red balls
I	1	2	3
II	2	1	1
III	4	3	2

A bag is chosen at random and two balts are drawn. They happen to be white and red. What is the probability that they came from the third bag?
Answer:
Let us define the events are as follows:
E₁ : bag I is chosen
E₂ : bag II is chosen
E₃ : bag III is chosen
E : Two drawn balls are white and red
Since E₁, E₂ and E₃ are mutually exclusive and exhaustive events.
∴ P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)
Also, P(E | E₁) = prob. of drawing one white and one red ball from bag-I
= \(\frac{{ }^1 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}\)
= \(\frac{3}{\frac{6 \times 5}{2}}\)
= \(\frac{1}{5}\)
P(E | E₂) = prob. of drawing one white and one red ball from bag II
= \(\frac{{ }^2 \mathrm{C}_1 \times{ }^1 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}\)
= \(\frac{2}{\frac{4 \times 3}{2}}\) = \(\frac{1}{3}\)
P(E | E₃) = porb. of drawing one white and one red ball from bag III
= \(\frac{{ }^4 C_1 \times{ }^2 C_1}{{ }^9 C_2}\)
= \(\frac{4 \times 2}{\frac{9 \times 8}{2}}\)
= \(\frac{2}{9}\)
We want to find P(E₃ | E)
Thus by Baye’s Theorem, we have
P(E₃ | E) = \(\frac{\mathrm{P}(\mathrm{E} \mid \mathrm{E}_3) \mathrm{P}(\mathrm{E}_3)}{\sum_{i=1}^3 \mathrm{P}(\mathrm{E} \mid \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{\frac{2}{9} \times \frac{1}{3}}{\frac{1}{5} \times \frac{1}{3}+\frac{1}{3} \times \frac{1}{3}+\frac{2}{9} \times \frac{1}{3}}\)
= \(\frac{\frac{2}{9}}{\frac{1}{5}+\frac{1}{3}+\frac{2}{9}}\)
= \(\frac{\frac{2}{9}}{\frac{9+15+10}{45}}\)
= \(\frac{2}{9}\) × \(\frac{45}{34}\)
= \(\frac{5}{17}\)

Question 17.
In a bolt factory, machines A, B and C manufacture 25 %, 35 % and 40 % respectively. Of the total of their output 5,4 , and 2 per cent are defective. A bolt is drawn and found to be defective.
(i) What are the probabilities that it was manufactured by the machines A, B and C ?
(ii) Find the probability that it was manufactured by either machine A or C.
Answer:
(i) Let us define the events are as follows :
E₁ : bolt is manufactured by machine A
E₂ : bolt is manufactured by machine B
E₃ : bolt is manufactured by machine C
Then P(E₁) = 25 % = \(\frac{25}{100}\);
P(E₂) = 35 % = \(\frac{35}{100}\)
and P(E₃) = 40 % = \(\frac{40}{100}\)
Let E : drawing a defective bolt. P(E | E₁) = prob. of drawing a defective bolt from machine A = 5 % = \(\frac{5}{100}\)
P(E | E₂) = prob. of drawing a defective bolt from machine B = 4 % = \(\frac{4}{100}\)P(E | E₃) = prob. of drawing a defective bolt from machine C = 2 % = \(\frac{2}{100}\) We want to find P(E₁ | E), P(E₂ | E) and P(E₃ | E)
Thus by Baye’s Theorem; we have

(ii) ∴ required probability
=P(E₁ | E)+P(E₃| E)
= \(\frac{25}{69}\) + \(\frac{16}{69}\)
= \(\frac{41}{69}\)

Question 18.
A company has two plants to manufacture cars. Plant I manufactures 80 per cent of the cars and plant II manufactures 20 per cent. At plant I, 85 out of 100 cars are rated standard quality or better. At plant II, only 65 out of 100 cars are rated standard or better.
(i) What is the probability that cars selected at random came from plant I and it is known that the car is of standard quality?
(ii) What is the probability that the cars came from plant II if it is known that the car is of standard quality?
Answer:
(i) Let us define the events are as follows :
E₁ : car is manufactured by plant-I
E₂ : car is manufactured by plant-II
∴P(E₁) = \(\frac{80}{100}\);
P(E₂) = \(\frac{20}{100}\)
Thus E₁ and E₂ are mutually exclusive and exhaustive events.
Let E : car is of standard or better quality. Also, P(E | E₁)
= Prob. that a standard quality car came from plant-I = \(\frac{85}{100}\) P(E | E₂)
= prob. that a standard quality car came from plant -II = \(\frac{65}{100}\)
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have
P(E₁ | E)
= \(\frac{\mathrm{P}(\mathrm{E} \mid \mathrm{E}_1) \mathrm{P}(\mathrm{E}_1)}{\sum_{i=1}^2 \mathrm{P}(\mathrm{E} \mid \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)

Question 19.
A factory has three machines X, Y, and Z producing 1000,2000 and 3000 bolts per day respectively. The machines X produces 1 % defective bolts, F produces 1.5 % and Z produces 2 % defective bolts. At the end of a day, a bolt is drawn at random and is found defective. What is the probability that this defective bolt has been produced by the machine X ?
Answer:
Consider the following events :
E₁= Bolt produced by machine X
E₂ = Bolt produced y} machine Y
E₃ = Bolt produced by machine Z
A = A bolt drawn is defective.
∴ P(E₁) = \(\frac{1000}{6000}\) = \(\frac{1}{6}\);
P(E₂) = \(\frac{2000}{6000}\)
= \(\frac{1}{3}\)
and P(E₃) = \(\frac{3000}{6000}\)
= \(\frac{1}{2}\)
Thus P(A | E₁) = probability defective bolt from machine X = \(\frac{1}{100}\)
P(A | E₂)= probability of drawing defective bolt from machine Y = \(\frac{1.5}{100}\)
= \(\frac{3}{200}\)
P(A/E₃) = probability of drawing defective bolt from machine
Z = \(\frac{2}{100}\)
We want to find, probability of defective bolt drawn is produced by machine X
= P(E₁ / A)
P(E₁ / A) = \(\frac{P(E_1) P(A / E_1)}{P(E_1) P(A / E_1)+P(E_2) P(A / E_2)}+P(E_3) P(A / E_3)\)
= \(\frac{\frac{1}{6} \times \frac{1}{100}}{\frac{1}{6} \times \frac{1}{100}+\frac{1}{3} \times \frac{3}{200}+\frac{1}{2} \times \frac{2}{100}}\)
= \(\frac{\frac{1}{600}}{\frac{1}{600}+\frac{3}{600}+\frac{1}{100}}\)
= \(\frac{1}{10}\)
Thus, required probability = \(\frac{1}{10}\).

Question 20.
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Answer:
Let us define the events are as follows :
E₁ : lost card is a diamond card
E₂ : lost card is not of diamond.
E : both drawn cards, are of diamond suit Then P(E₁) = prob. of drawing a diamond
card = \(\frac{13}{52}\) = \(\frac{1}{4}\)
P(E₂) = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
∴P(E | E₁) = prob. of drawing two diamond cards when it is given that lost card is a diamond card
= \(\frac{{ }^{12} C_2}{{ }^{51} C_2}\)
P(E | E₂) = prob. of drawing two diamond cards given that lost card is not of diamond card = \(\frac{{ }^{13} \mathrm{C}_2}{{ }^{51} \mathrm{C}_2}\)
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have

Question 21.
A speaks the truth 2 out of 3 times and B, 4 out of 5 times. They agree in the assertion that from a bag containing 6 balts of different colours a black ball has been drawn. Find the probability that the statements are true.
Answer:
The prob. of drawing a black ball = \(\frac{1}{6}\) prob. of drawing a non-black ball
= 1 – \(\frac{1}{6}\)
= \(\frac{5}{6}\)
Given P( A speaks truth )= \(\frac{2}{3}\)
P(B speaks truth )= \(\frac{4}{5}\)
∴ Prob. that both A and B agree in assertion that it is black ball = \(\frac{1}{6}\) × \(\frac{2}{3}\) × \(\frac{4}{5}\)
= \(\frac{4}{5}\) = P(I)
Then prob. that A} asserts wrongly i.e. a certain ball is black = P(A tell a lie ). P(of drawing a ball from 5 balls other than black)
= (1 – \(\frac{2}{3}\)) × \(\frac{1}{5}\)
= \(\frac{1}{15}\)
The prob. that B} asserts wrongly i.e. a certain ball is black = P(B tell a lie).P (of drawing a ball from 5 balls other from black)
= (1 – \(\frac{4}{5}\)) × \(\frac{1}{5}\)
= \(\frac{1}{25}\)
Thus when a non black ball is drawn and both A and B agree in asserting that it is black
= \(\frac{5}{6}\) × \(\frac{1}{15}\) × \(\frac{1}{25}\)
= \(\frac{1}{450}\) = P (II)
Thus by Baye’s Theorem, required probability that statement is true
= \(\frac{\mathrm{P}(\mathrm{I})}{\mathrm{P}(\mathrm{I})+\mathrm{P}(\mathrm{II})}\)
= \(\frac{\frac{4}{45}}{\frac{4}{45}+\frac{1}{450}}\)
= \(\frac{\frac{4}{45}}{\frac{40+1}{450}}\)
= \(\frac{4}{45}\) × \(\frac{450}{41}\)
= \(\frac{40}{41}\)

Question 22.
In 2004, there will be three candidates for the position of principal C₁, C₂ and C₃. The chances of their selection are in the proportion 4 : 2 : 3 respectively. The probability that C₁, if selected, will introduce co-education in the college is 0.3 . The probabilities of C₂ and C₃ doing the same are respectively 0.5 and 0.8 .
(i) What is the probability that there will be co-education in the college in 2004?
(ii) Also, find the probability that principal C₂ introduces co-education in the college.
Answer:
Let us define the events are as follows :
E₁ : C₁ be selected as a principal
E₂ : C₂ be selected as a principal
E₃ : C₃ be selected as a principal
E : Introduction of co-education in the college

Question 23.
You note that your officer is happy on 60 % of your calls, so you assign a probability of his being happy on your visit as 0.6 or 6 / 10. You have noticed also that if he is happy, he accedes to your request with a probability of 0.4 or 4 / 10 whereas if he is not happy, he accedes to the request with a probability of 0.1 or 1/10. You call one day, and he accedes to your request. What is the probability of his being happy?
Answer:
Let us define the events are as follows :
E₁ : The officer is being happy
E₂ : The officer is be not happy
E : The officer accedes to request
Then P(E₁) = \(\frac{6}{10}\)
and P(E₂) = 1 – \(\frac{6}{10}\)
= \(\frac{4}{10}\)
Also, P(E | E₁) = prob. that he accedes to request when it is given that officer is being happy
= \(\frac{4}{10}\)
P(E | E₁) = probability that he accedes to request when it is given that officer is not happy
= \(\frac{1}{10}\)
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{P(E / E_1) P(E_1)}{P(E / E_1) P(E_1)+P(E / E_1) P(\bar{E}_1)}\)
= \(\frac{\frac{4}{10} \times \frac{6}{10}}{\frac{4}{10} \times \frac{6}{10}+\frac{1}{10} \times \frac{4}{10}}\)
= \(\frac{\frac{24}{100}}{\frac{24}{100}+\frac{4}{100}}\)
= \(\frac{24}{28}\)
= \(\frac{6}{7}\)

Question 24.
The chance that a female worker in a chemical factory will contract an occupational disease is 0.04 and the chance for a male worker 0.06 . Out of 1000 workers in a factory 200 are females. One worker is selected at random and the worker is found to have contracted the disease. What is the probability that the worker is a female?
Answer:
Let us define the events are as follows :
E₁ : worker is a female
E₂ : worker is a male
E : The worker is found to have contracted the disease.
Then
P(E₁) = \(\frac{200}{1000}\) = \(\frac{2}{10}\);
P(E₂) = \(\frac{800}{1000}\) = \(\frac{8}{10}\)
∴ P(E | E₁) = probability that female worker will contract an occupational disease = 0.04
P(E | E₂) = probability that male worker will contract an occupational disease = 0.06
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{P(E / E_1) P(E_1)}{P(E / E_1) P(E_1)+P(E / E_2) P(E_2)}\)
= \(\frac{0.04 \times \frac{2}{10}}{0.04 \times \frac{2}{10}+0.06 \times \frac{8}{10}}\)
= \(\frac{\frac{8}{1000}}{\frac{8}{1000}+\frac{48}{1000}}\)
= \(\frac{8}{56}\)
= \(\frac{1}{7}\)

Question 25.
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident involving a scooter driver, car driver and a truck driver is 0.01,0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver? [Type solved Ex. 20.]
Answer:
Let us define the events are as follows :
E₁ : Insured person is scooter driver
E₂ : Insured person is car driver
E₃ : Insured person is truck driver
E : Insured person meets an accident
Then
P(E₁) = \(\frac{2000}{2000+4000+6000}\)
= \(\frac{2000}{12000}\) = \(\frac{1}{6}\)
P(E₂) = \(\frac{4000}{2000+4000+6000}\)
= \(\frac{4000}{12000}\)
= \(\frac{1}{3}\)
P(E₃) = \(\frac{6000}{2000+4000+6000}\)
= \(\frac{6000}{12000}\) = \(\frac{1}{2}\)
Also P(E | E₁) = probability of an accident involving a scooter driver = 0.01
P(E | E₂) = probability of an accident involving a car driver = 0.03
P(E | E₃) = probability of an accident involving a truck driver = 0.15
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{2000}{2000+4000+6000}\)
= \(\frac{0.01 \times \frac{1}{6}}{0.01 \times \frac{1}{6}+0.03 \times \frac{1}{3}+0.15 \times \frac{1}{2}}\)
= \(\frac{\frac{0.01}{6}}{\frac{0.01+0.06+0.45}{6}}\)
= \(\frac{0.01}{0.52}\) = \(\frac{1}{52}\)

Question 26.
There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75 % of the times and the third is also a biased coin that comes up tails 40 % of the times. One of the three coins is chosen at random and tossed and it shows head. What is the probability that it was the two headed coin?
Answer:
Let us consider the events :
E₁ : first coin is chosen
E₂ : second coin is chosen
E₃ : third coin is chosen
∴ P(E₁) = \(\frac{1}{3}\), P(E₂) = \(\frac{1}{3}\),
P(E₃) = \(\frac{1}{3}\)
Let A denote the event when the toss shows a heads.
It is also given that
P(A / E₁) = 1, P(A / E₂) = 0.75,
P(A / E₂)= 0.6
We want to find P(E₁ / A).
Then by Baye’s theorem, we have
= \(\frac{\frac{1}{3}(1)}{\frac{1}{3}(1)+\frac{1}{3}(0 \cdot 75)+\frac{1}{3}(0 \cdot 60)}\)
= \(\frac{1 / 4}{47 / 60}\)
= \(\frac{20}{47}\)

Question 27.
In a factory which manufactures bolts, machines A, B and C manufacture respectively 30 %, 50 % and 20 % of the bolts of their outputs, 3 %, 4% and 1 % respectively are defective bolts. A bolt is drawn at random from the product and is found to be defective. Find the probability that this is not manufactured by machine B.
Answer:
Let us define the events are as follows :
E₁ : bolt manufactured by machine A
E₂ : bolt manufactured by machine B
E₃ : bolt manufactured by machine C
E : bolt is found to be defective
Then
P(E₁) = 30 % = \(\frac{30}{100}\) = \(\frac{3}{10}\);
P(E₂) = \(\frac{50}{100}\) = \(\frac{5}{10}\);
P(E₃) = 20 % = \(\frac{20}{100}\) = \(\frac{2}{10}\)
Also P(E | E₁) = prob. that bolt is defective and it is given that it is produced by
machine A = \(\frac{3}{100}\)
P(E | E₂) = \(\frac{4}{100}\);
P(E | E₃) = \(\frac{1}{100}\)
Thus by Baye’s Theorem, we have

Examples:

Question 1.
A firm produces steel pipes in three plants A, B and C with daily production of 500, 1000 and 2000 units respectively. It is known that fraction of defective output
produced by the three plants are respectively 0.005,0.008 and 0.0l.A pipe ¡s selected at random from a day’s total production and found to be defective.
What ¡s the probability that ¡t came from the first plant? (ISC 2005, 2000)
Answer:
Let us define the events are as follows:
E₁ : steel pipe is produced by plant A
E₂ : steel pipe is produced by plant B
E₃ : steel pipe is produced by plant C
E : selected pipe ¡s found to be defective
P(E₁) = \(\frac{500}{500+1000+2000}\) = \(\frac{500}{3500}\) = \(\frac{1}{7}\)
P(E₂) = \(\frac{1000}{500+1000+2000}\) = \(\frac{1000}{3500}\) = \(\frac{2}{7}\)
P(E₃) = \(\frac{2000}{500+1000+2000}\) = \(\frac{2000}{3500}\) = \(\frac{4}{7}\)
Also, P(E | E₁) = probability that steel pipe is defective and produced by plant A = 0.005
P(E | E₂) = prob. that steel pipe is defective and it is given that is is produced by plant B = 0.008
P(E | E₃) = prob. tha steel pipe is defective and it is given that it is produced by plant C = 0.01
We want to find P(E₁| E)
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{\mathrm{P}(\mathrm{E} | \mathrm{E}_1) \mathrm{P}(\mathrm{E}_1)}{\sum_{i=1}^3 \mathrm{P}(\mathrm{E}_i | \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{0.005 \times \frac{1}{7}}{0.005 \times \frac{1}{7}+0.008 \times \frac{2}{7}+0.01 \times \frac{4}{7}}\)
= \(\frac{0.005}{0.005+0.016+0.04}\)
= \(\frac{0.005}{0.061}\)
= \(\frac{5}{61}\)

Question 2.
An insurance company insured 6000 scooter drivers, 3000 car drivers and 9000 truck drivers. The probability of an accident involving a scooter, a car or a truck is 0.02,0.06 and 0.3 respectively. One of the injured persons meets with an accident. Find the probability that he is a car driver.
Answer:
Let us define the events are as follows :
E₁ : Insured person is scooter driver
E₂ : Insured person is car driver
E₃ : Insured person is truck driver
E : Insured person meets an accident
Then P(E₁) = \(\frac{6000}{6000+3000+9000}\)
= \(\frac{6}{18}\)
= \(\frac{1}{3}\)
P(E₂) = \(\frac{3000}{6000+3000+9000}\)
= \(\frac{3000}{18000}\)
= \(\frac{1}{6}\)
P(E₃) = \(\frac{9000}{6000+3000+9000}\)
= \(\frac{9000}{18000}\)
= \(\frac{1}{2}\)
Also, P(E | E₁) = prob.
that insured person meets an accident when it is given that insured person is a scooter driver = 0.02
P(E | E₂) = prob. that insured person meets an accident when it is given that insured person is a car driver =0.06 and
P(E | E₃) = 0.6
We want to find P(E₂ | E)
By Baye’s Theorem, we have
∴ P(E₂ | E) = \(\frac{\mathrm{P}(\mathrm{E} | \mathrm{E}_2) \mathrm{P}(\mathrm{E}_2)}{\sum_{i=1}^3 \mathrm{P}(E | \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{0.06 \times \frac{1}{6}}{0.02 \times \frac{1}{3}+0.06 \times \frac{1}{6}+0.3 \times \frac{1}{2}}\)
= \(\frac{\frac{0.06}{6}}{\frac{0.04+0.06+0.9}{6}}\)
= \(\frac{0.06}{1.0}\) = 0.06

Question 3.
A company has two plants which manufacture scooters. Plant I manufactures 80 % of the scooters while plant II manufactures 20 % of the scooters. At plant I, 85 out of 100 scooters are being of standard quality, while at plant II only 65 out of 100 scooters are rated as being of standard quality. If a scooter is of standard quality, what is the probability that it comes from plant I ?
Answer:
Let us define the events are as follows :
E₁ : car is manufactured by plant-I
E₂ : car is manufactured by plant-II
∴ P(E₁) = \(\frac{80}{100}\);
P(E₂) = \(\frac{20}{100}\)
Thus E₁ and E₂ are mutually exclusive and exhaustive events.
Let E : car is of standard or better quality. Also, P(E | E₁) = Prob. that a standard quality car came from plant-I = \(\frac{85}{100}\)
P(E | E₂) prob. that a standard quality car came from plant-II = \(\frac{65}{100}\)
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{\mathrm{P}(\mathrm{E} | \mathrm{E}_1) \mathrm{P}(\mathrm{E}_1)}{\sum_{i=1}^2 \mathrm{P}(\mathrm{E} | \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{\frac{85}{100} \times \frac{80}{100}}{\frac{85}{100} \times \frac{80}{100}+\frac{65}{100} \times \frac{20}{100}}\)
= \(\frac{6800}{6800+1300}\)
= \(\frac{68}{81}\)

Question 4.
An insurance company insured 1500 scooter drivers, 2500 car drivers and 4500 truck drivers. The probability of a scooter, a car and struck meeting with an accident is 0.01,0.02 and 0.04 respectively. If one of the insured persons meets with an accident, find the probability that he is a scooter driver.
Answer:
Let us define the events are as follows:
E₁ : Insured person is a scooter driver
E₂ : Insured person is a car driver
E₃ : Insured person is a truck driver
E : Insured person meets an accident
P(E₁) = \(\frac{1500}{1500+2500+4500}\)
= \(\frac{1500}{8500}\) = \(\frac{3}{17}\)
P(E₂) = \(\frac{2500}{1500+2500+4500}\)
= \(\frac{2500}{8500}\) = \(\frac{5}{17}\)
P(E₃) = \(\frac{4500}{8500}\) = \(\frac{9}{17}\)
Also, P(E | E₁) = prob. that insured person meets an accident when it is given that insured person is scooter driver = 0.01
P(E | E₂) = 0.02 ; P(E | E₃) = 0.04
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{P(E / E_1) P(E_1)}{P(E / E_1) P(E_1)+P(E / E_2) P(E_2)} +P(E / E_3) P(E_3)\)
= \(\frac{0.01 \times \frac{3}{17}}{0.01 \times \frac{3}{17}+0.02 \times \frac{5}{17}+0.04 \times \frac{9}{17}}\)
= \(\frac{0.03}{0.03+0.1+0.36}\)
= \(\frac{0.03}{0.49}\)
= \(\frac{3}{49}\)

Question 5.
A class consists of 50 students out of which there are 10 girls. In the class 2 girls and 5 boys are rank holders in an examination. If a student is selected at random from the class and is found to be a rank holder, what is the probability that the student selected is a girl.
Answer:
Let us define the events are as follows:
E₁ : chosen student is a girl
E₂ : selected student is a boy
E : selected student is found to be rank holder
Then P(E₁) = \(\frac{10}{50}\) = \(\frac{1}{5}\)
and P(E₂) = \(\frac{40}{50}\) = \(\frac{4}{5}\)
Also P(E | E₁) = prob. that student is found to be rank holder when it is given that it is a
girl = \(\frac{2}{10}\) = \(\frac{1}{5}\)
P(E | E₂) = prob. that student is found to be rank holder when it is given that it is a boy
= \(\frac{5}{40}\)
= \(\frac{1}{8}\)
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{P(E / E_1) P(E_1)}{P(E / E_1) P(E_1)+P(E / E_2) P(E_2)}\)
= \(\frac{\frac{1}{5} \times \frac{1}{5}}{\frac{1}{5} \times \frac{1}{5}+\frac{1}{8} \times \frac{4}{5}}\)
= \(\frac{\frac{1}{25}}{\frac{1}{25}+\frac{1}{10}}\)
= \(\frac{\frac{1}{25}}{\frac{2+5}{50}}\)
= \(\frac{50}{7 \times 25}\) = \(\frac{2}{7}\)

Question 6.
An insurance company insured 4000 doctors, 8000 teachers and 12000 engineers. The probabilities of a doctor, a teacher and an engineer dying before the age of 58 years are 0.01,0.03 and 0.05 respectively. If one of the insured persons dies before the age of 58 years, find the probability that he is a doctor.
Answer:
Let us define the events are as follows :
E₁ : Insured person is a doctor
E₂ : Insured person is a teacher
E₃ : Insured person is an engineer
E : Insured person dies before the age of 58 years
Then
P(E₁) = \(\frac{4000}{4000+8000+12000}\)
= \(\frac{4000}{24000}\) = \(\frac{4}{24}\) = \(\frac{1}{6}\)
P(E₂) = \(\frac{8000}{4000+8000+12000}\)
= \(\frac{8000}{24000}\) = \(\frac{1}{3}\)
P(E₃) = \(\frac{12000}{4000+8000+12000}\)
= \(\frac{12000}{24000}\) = \(\frac{1}{2}\)
∴ P(E | E₁) = The probability that the doctor dying before the age of 58 years = 0.01
P(E | E₂) = The prob. that the teacher dying before the age of 58 years =0.03
P(E | E₃) = The prob. that the engineer dying before the age of 58 years = 0.05
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{\mathrm{P}(\mathrm{E} | \mathrm{E}_1) \mathrm{P}(\mathrm{E}_1)}{\sum_{i=1}^3 \mathrm{P}(\mathrm{E} | \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{0.01 \times \frac{1}{6}}{0.01 \times \frac{1}{6}+0.03 \times \frac{1}{3}+0.05 \times \frac{1}{2}}\)
= \(\frac{\frac{0.01}{6}}{\frac{0.01+0.06+0.15}{6}}\)
= \(\frac{0.01}{0.22}\)
= \(\frac{1}{2}\)
= 0.045

Question 7.
A factory has three machines A, B and C producing 1500, 2500 and 3000 bulbs per day respectively. Machine produces 1.5 % defective bulbs, machine B produces 2 % defective bulbs and machine E produces 2.5 % defective bulbs. At the end of the day, a bulb is drawn at random and is found to be defective. What is the probability that this defective bulb has been produced by machine E?
Answer:
Let us define the events are as follows:
E₁ : bulb produced by machine A
E₂ : bulb produced by machine B
E₃ : bulb produced by machine C
E : drawn bulb is found to be defective
Then P(E₁) = \(\frac{1500}{1500+2500+3000}\)
= \(\frac{1500}{7000}\) = \(\frac{15}{70}\) = \(\frac{3}{14}\)
P(E₂) = \(\frac{2500}{1500+2.500+3000}\)
= \(\frac{2500}{7000}\) = \(\frac{5}{14}\)
and
P(E₃) = \(\frac{3000}{1500+2500+3000}\)
= \(\frac{3000}{7000}\) = \(\frac{3}{7}\)
Also, P(E | E₁) = prob. that defective bulb is produced by machine A = 1.5 % = \(\frac{1.5}{100}\)
P(E | E₂) = \(\frac{2}{100}\)
P(E | E₃) = \(\frac{2.5}{100}\)
We want to find P(E₂ | E)
Thus by Baye’s Theorem, we have
P(E₂ | E) = \(\frac{\mathrm{P}(\mathrm{E} | \mathrm{E}_2) \mathrm{P}(\mathrm{E}_2)}{\sum_{i=1}^3 \mathrm{P}(\mathrm{E}_i | \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{\frac{5}{14} \times \frac{2}{100}}{\frac{1.5}{100} \times \frac{3}{14}+\frac{2}{100} \times \frac{5}{14}+\frac{2.5}{100} \times \frac{3}{7}}\)
= \(\frac{\frac{10}{1400}}{\frac{4.5}{1400}+\frac{10}{1400}+\frac{7.5}{700}}\)
= \(\frac{\frac{10}{1400}}{\frac{4.5+10+15}{1400}}\)
= \(\frac{10}{29.5}\)
= \(\frac{100}{295}\)
= \(\frac{20}{59}\)

Question 8.
Bag A contains 2 white, 1 black and 3 red balls, Bag B contains 3 white, 2 black and 4 red balls and Bag C contains 4 white, 3 black and 2 red balls. One bag is chosen at random and 2 balls are drawn at random from that bag. If the randomly drawn balls happen to be red and black, what is the probability that both balls come from Bag B ?
Answer:
Given bag A contains 2 white, 1 black and 3 red balls
bag B contains 3 white, 2 black and 4 red balls
and bag C contains 4 white, 3 black and 2 red balls
Let us define the events as follows :
E₁ : bag A is chosen,
E₂ : bag B is chosen
E₃ : bag C is chosen
E : chosen ball happen to be red and black.
Then P|(E₁) = P|(E₂) = P|(E₃) = \(\frac{1}{3}\)
P|(E | E₁) = prob. of drawing 1 red and 1 black ball from bag A = \(\frac{{ }^3 \mathrm{C}_1 \times{ }^1 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}\)
= \(\frac{3}{\frac{6 \times 5}{2}}\)
= \(\frac{1}{5}\)
P|(E | E₂) = probability of drawing 1 red and 1 black ball from bag B = \(\frac{{ }^2 C_1 \times{ }^4 C_1}{{ }^9 C_2}\)
= \(\frac{2 \times 4}{\frac{9 \times 8}{2}}\)
= \(\frac{16}{72}\)
= \(\frac{2}{9}\)
P(E | E₃) = prob. of drawing 1 red and 1 black ball from bag C = \(\frac{{ }^3 \mathrm{C}_1 \times{ }^2 \mathrm{C}_1}{{ }^9 \mathrm{C}_2}\)
= \(\frac{3 \times 2}{\frac{9 \times 8}{2}}\) = \(\frac{1}{6}\)
We want to find P(E₂ | E)
Thus by Baye’s Theorem, we have
P(E₂ | E) = \(\frac{\mathrm{P}(\mathrm{E} | \mathrm{E}_2) \mathrm{P}(\mathrm{E}_2)}{\sum_{i=1}^3 \mathrm{P}(\mathrm{E} | \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{\frac{2}{9} \times \frac{1}{3}}{\frac{1}{5} \times \frac{1}{3}+\frac{2}{9} \times \frac{1}{3}+\frac{1}{6} \times \frac{1}{3}}\)
= \(\frac{\frac{2}{9}}{\frac{1}{5}+\frac{2}{9}+\frac{1}{6}}\)
= \(\frac{\frac{2}{9}}{\frac{18+20+15}{90}}\)
= \(\frac{2}{9} \times \frac{90}{53}\)
= \(\frac{20}{53}\)

Question 9.
In a class of 75 students, 15 are above average, 45 are average and the rest below average achievers. The probability that an above average achieving student fails is 0.005 that an average achieving student fails is 0.05 and the probability of a below average achieving student failing is 0.15. If a student is known to have passed, what is the probability that he is a below average achiever?
Answer:
Let us define the events are as follows:
E₁ : a student is above average
E₂ : a student is an average student
E₃ : a student is below average
E : a student is known to have passed.
Then P(E₁) = \(\frac{15}{75}\) = \(\frac{1}{5}\);
P(E₂) = \(\frac{45}{75}\)
= \(\frac{3}{5}\);
P(E₃) = \(\frac{15}{75}\) = \(\frac{1}{5}\)
P(E | E₁) = prob. that student is passed when it is given that student is above average
= 1 – 0.005 = 0.995
P(E | E₂) = 1 – 0.05 = 0.95 ;
P(E | E₃) = 1 – 0.15 = 0.85
We want to find P(E₃ | E)
Thus by Baye’s Theorem, we have
P(E₃ | E) = \(\frac{\mathrm{P}(\mathrm{E} | \mathrm{E}_3) \mathrm{P}(\mathrm{E}_3)}{\sum_{i=1}^3 \mathrm{P}(\mathrm{E} | \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{0.85 \times \frac{1}{5}}{0.995 \times \frac{1}{5}+0.95 \times \frac{3}{5}+0.85 \times \frac{1}{5}}\)
= \(\frac{0.85}{0.995+2.85+0.85}\)
= \(\frac{0.85}{4.695}\)
= 0.54

Question 10.
For A, B and C the chances of being selected as the manager of a firm are 4 : 1 : 2, respectively. The probabilities for them to introduce a radical change in the marketing strategy are 0.3,0.8, and 0.5 respectively. If a change takes place, find the probability that it is due to the appointment of B.
Answer:
Let us consider the events as follows :
E₁ = A is appointed as manager of firm
E₂ = B is appointed as manager of firm
E₃ = C is appointed as manager of firm
E = a change does take place
Then P(E₁) = \(\frac{4}{7}\);
P(E₂) = \(\frac{1}{7}\);
and P(E₃) = \(\frac{2}{7}\)
∴ P(E | E₁) = prob. that changes take place by
A = 0.3
P(E / E₂) = probability that changes take place by B = 0.8
and P(E / E₃) = probability that changes take place by C = 0.5
We want to find, the probability changes were taken place by B = P(E₂ / E)
Then By Baye’s theorem, we have
P(E₂ / E) = \(\frac{P(E_2) P(E / E_2)}{P(E_1) P(E / E_1)+P(E_2) P(E / E_2)} +P(E_3) P(E / E_3)\)
= \(\frac{\frac{1}{7} \times \frac{8}{10}}{\frac{4}{7} \times \frac{3}{10}+\frac{1}{7} \times \frac{8}{10}+\frac{2}{7} \times \frac{5}{10}}\)
= \(\frac{\frac{8}{70}}{\frac{30}{70}}\)
= \(\frac{8}{30}\)
= \(\frac{4}{15}\)
Thus, required probability = \(\frac{4}{15}\)

Question 11.
In a bolt factory, three machines A, B and C manufacture 25 %, 35 % and 40 % of the total production respectively. Of their respective outputs, 5 %, 4 % and 2 % are defective. A bolt is E drawn at random from the total production and it is found to be defective. Find the probability that it was manufactured by machine C.
Answer:
Consider the following events:
E₁ = Bolt is manufactured by machine A,
E₂ = Bolt is manufactured by machine B,
E₃ = Bolt is manufactured by machine C, and
E = Bolt is defective
Clearly, P(E₁) = \(\frac{25}{100}\) = \(\frac{1}{4}\);
P(E₂) = \(\frac{35}{100}\) = \(\frac{7}{20}\);
P(E₃) = \(\frac{40}{100}\) = \(\frac{2}{5}\);
Also, P(E / E₁) = \(\frac{5}{100}\);
P(E / E₂) = \(\frac{4}{100}\);
P(E / E₃) = \(\frac{2}{100}\)
We want to find the probability that item was manufactured on machine C as it is given that item is defective.
Thus Required probability = P(E₃/ E)
= \(\frac{P(E_3) P(E / E_3)}{P(E_1) P(E / E_1)+P(E_2) P(E / E_2)} +P(E_3) P(E / E_3)\)
= \(\frac{\frac{2}{5} \times \frac{2}{100}}{\frac{1}{4} \times \frac{5}{100}+\frac{7}{20} \times \frac{4}{100}+\frac{2}{5} \times \frac{2}{100}}\)
= \(\frac{\frac{4}{5}}{\frac{5}{4}+\frac{7}{5}+\frac{4}{5}}\)
= \(\frac{\frac{4}{5}}{\frac{69}{20}}\)
= \(\frac{16}{69}\)

Question 12.
Box I contains two white and three black balls. Box II contains four white and one black balls and box III contains three white and four black balls. A die having three red, two yellow and one green face, is thrown to select the box. If red face turns up, we pick up box I, if a yellow face turns up we pick up box II, otherwise, we pick up box III. Then, we draw a ball from the selected box. If the ball drawn is white, what is the probability that the die had turned up with a red face?
Answer:
Given box-I contains 2 white and three black balls
box-II contains 4 white and one black ball box-III contains 3 white and 4 black balls
Let us define the events are as follows :
E₁ : red face turns up
E₂ : yellow face turns up
E₃ : green face turns up
E : ball drawn is white
Then P(E₁) = \(\frac{3}{6}\) = \(\frac{1}{2}\);
P(E₂) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
and P(E₃) = \(\frac{1}{6}\)
P(E | E₁) = prob. of drawing a white ball when it is given that red face turn up i.e. we pick up box-I = \(\frac{2}{5}\)
P(E | E₂) = prob. of drawing white ball when it is given that yellow face turns up i.e. we pick up box-II = \(\frac{4}{5}\)
P(E | E₃) = prob. of drawing white ball when it is given that green face turn up i.e. we pick up box-III = \(\frac{3}{7}\)
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{\mathrm{P}(\mathrm{E} | \mathrm{E}_1) \mathrm{P}(\mathrm{E}_1)}{\sum_{i=1}^3 \mathrm{P}(\mathrm{E} | \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{\frac{2}{5} \times \frac{1}{2}}{\frac{2}{5} \times \frac{1}{2}+\frac{4}{5} \times \frac{1}{3}+\frac{3}{7} \times \frac{1}{6}}\)
= \(\frac{\frac{1}{5}}{\frac{1}{5}+\frac{4}{15}+\frac{1}{14}}\)
= \(\frac{\frac{1}{5}}{\frac{42+56+15}{210}}\)
= \(\frac{210}{5 \times 113}\)
= \(\frac{42}{113}\)

Question 13.
In an automobile factory, certain parts are to be fixed into the class is in a section before it moves into another section. On a given day, one of the three persons A, B and C carries out this task. A has 45 % chance, B has 35 % chance and C has. 20 % chance of doing the task. The probability that A, B and C will take more than the alloted time is \(\frac{1}{6}\), \(\frac{1}{10}\) and \(\frac{1}{20}\) respectively. If it is found that the time taken is more than the alloted time, what is the probability that A has done the task?
Answer:
Let us define the events are as follows :
E₁ : A carries out the task
E₂ : B carries out the task
E₃ : C carries out the task
E : time taken by the person is more than allotted time
Then P(E₁) = 45 % = \(\frac{45}{100}\);
P(E₂) = \(\frac{35}{100}\);
P(E₃) = \(\frac{20}{100}\)
∴ E₁, E₂ and E₃ are mutually exclusive and exhaustive events
P(E | E₁) = \(\frac{1}{6}\),
P(E | E₂) = \(\frac{1}{10}\),
P(E | E₃) = \(\frac{1}{20}\)
We want to find P(E₁ | E)
Thus by Baye’s Theorem, we have
P(E₁ | E) = \(\frac{\mathrm{P}(\mathrm{E} | \mathrm{E}_1) \mathrm{P}(\mathrm{E}_1)}{\sum_{i=1}^3 \mathrm{P}(\mathrm{E}_1 \mathrm{E}_i) \mathrm{P}(\mathrm{E}_i)}\)
= \(\frac{\frac{1}{6} \times \frac{45}{100}}{\frac{1}{6} \times \frac{45}{100}+\frac{1}{10} \times \frac{35}{100}+\frac{1}{20} \times \frac{20}{100}}\)
= \(\frac{\frac{45}{600}}{\frac{45}{600}+\frac{35}{1000}+\frac{20}{2000}}\)
= \(\frac{\frac{45}{600}}{\frac{450+210+60}{6000}}\)
= \(\frac{45}{600} \times \frac{6000}{720}\)
= \(\frac{45}{72}\)
= \(\frac{5}{8}\)