ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.5

Utilizing Understanding ISC Mathematics Class 12 Solutions Chapter 10 Probability Ex 10.5 as a study aid can enhance exam preparation.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.5

Question 1.
Bag I contains 3 red and 4 black balls while another bag II contains 5 red and 6 black © balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from bag II. (NCERT)
Answer:
Let the events be
E₁ : bag I is selected ;
E₂ : bag II is selected;
A : ball drawn is red
P (E₁) = P (E₂) = \(\frac{1}{2}\) and P (A/E₁) = \(\frac{3}{7}\); P (A/E₂) = \(\frac{5}{11}\)
we want to find P(E₂/A).

Using Baye’s theorem, we have

Question 1 (Old).
There are two bags I and II. Bag I contains 2 white and 4 red balls and bag II contains 5 white and 3 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 II.
Answer:
Let E₁, E₂ and A be the events defined as follows :
E₁ : bag I is selected;
E₂ : bag II is selected
A : red ball is drawn from selected bag
Then P (E₁) = \(\frac{1}{2}\) =P(E₂)
P (A/E₁) = prob. of drawing a red ball from bag I = \(\)
P (A/E₂) = prob. of drawing a red ball from bag II = \(\)
We want to find P(E₂/A).
Then by Baye’s Theorem, we have

Question 2.
Suppose 5% of men and 0-25% of women have grey hairs. 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 E₁, E₂ and A are as follows :
E₁ : male is selected
E₂ : female is selected
A : A grey haired person is selected
Then P (E₁) = P (E₂) = \(\frac{1}{2}\)
P (A/E₁) = prob. that a grey haired person is male = \(\frac{5}{100}\)
P (A/E₂) = prob. that a grey haired person is women = \(\frac{0.25}{100}=\frac{25}{100 \times 100}=\frac{1}{400}\)
We want to find P (E₂/A)
Then by Baye’s Theorem, we have
P(E₁/A) = \(\frac{P\left(A / E_1\right) P\left(E_1\right)}{P\left(A / E_1\right) P\left(E_1\right)+P\left(A / E_2\right) P\left(E_2\right)}\)
= \(\frac{\frac{5}{100} \times \frac{1}{2}}{\frac{5}{100} \times \frac{1}{2}+\frac{1}{400} \times \frac{1}{2}}\)
= \(\frac{5}{5+\frac{1}{4}}=\frac{20}{21}\)

Question 3.
Two groups are competing for the positions on Board of Directors of a corporation. The probabilities that the first group and the second group will win are 0-6 and 0-4 respectively. Further, if the first group wins, the probability of introducing a new product is 0-7 and the corresponding probability is 0-3 if the second group wins. Find the probability that the new product introduced was by second group.
Answer:
Consider the following events :
E₁ = First group wins,
E₂ = Second group wins,

A = New product is introduced Given,
P(E₁) = 0.6,
P(E₂) = 0.4,
P(A/E₁) = 0.7,
P(A/E₂) = 0.3

Then by using Baye’s Theorem, we have
Required probability = P (E₂/A)
= \(\frac{P\left(E_2\right) P\left(A / E_2\right)}{P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) P\left(A / E_2\right)}\)
= \(\frac{0.4 \times 0.3}{0.6 \times 0.7+0.4 \times 0.3}=\frac{12}{54}=\frac{2}{9}\)
Thus, required probability is \(\frac{2}{9}\).

Question 3(Old).
A box contains 2 gold and 3 silver coins. Another box contains 3 gold and 3 silver coins. A v box is chosen at random and a coin is selected from it. If the selected coin is a gold coin, then find the probability that it is drawn from the second box.
Answer:
Let E₁, E₂ and A be the events defined as follows :
E₁ : purse-I is selected ;
E₂ : purse-II is selected ;
A : selected coin is gold coin
Then P (E₁) = P(E₂)
P (A/E₁) = prob. of chosing a gold coin from purse-I
P (A/E₂) = prob. of drawing a gold coin from purse-II
We want to find P (E₂/A)
Then by Baye’s Theorem, we have

Question 4.
Of the students in a school, it is known that 30% have 100% attendance and 70% students are irregular. Previous years result report that 70% of all students who have 100% attendance attain A grade and 10% irregular students attain A grade. At the end of the year, one student is chosen at random and he was found to have on A grade. What is the probability that the student has 100% attendance.
Answer:
Let us define the events as follows :
E₁ : event that the selected student is having 100% attendance
E₂ : event that the student is irregular.
A : event that the selected student getting grade A
Clearly E₁ and E₂ are mutually exclusive and exhaustive events.
Then P (E₁) = 30% = \(\frac{30}{100}\);
P (E₂) = 70% = \(\frac{70}{100}\)
P (A/E₁) = 70% = \(\frac{70}{100}\)
P (A/E₂) = 10% = \(\frac{10}{100}\)
required probability = P (E₁/A)

Question 5.
In a certain college, 4% of boys and 1% of girls are taller than 1.75 metres. Furthermore, 60% of the students in the college are girls. A student is selected at random from the college and is found to be taller than 1.75 metres. Find the probability that the selected student is a girl.
Answer:
Consier the following events as follows :
E₁ = The selected student is a girl
E₂ = The selected student is not a girl
A = The selected student is taller than 1.75 metres
Then P(E₁) = 60% = \(\frac{60}{100}\) = 0.6 and P(E₂) = 1 – P(E₁) = 1 – 0.6 = 0.4

P (A/E₁) = Probability that the student is taller than 1.75 meters given that the student is a girl
P(A/E₁) = \(\frac{1}{100}\) = 0.01
and P (A/E₂) = Probability that the student is taller than 1-75 meters given that the student is not a girl
P(A/E₂) = \(\frac{4}{100}\) = 0.04
Now, we want to find the probability that the selected student is girl, it is given that it is latter than 1.75 metres.
Required probability = P(E₁/A) = \(\frac{P\left(E_1\right) P\left(A / E_1\right)}{P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) P\left(A / E_2\right)}\)
= \(\frac{0.6 \times 0.01}{0.6 \times 0.01+0.4 \times 0.04}=\frac{\frac{6}{1000}}{\frac{22}{1000}}\)
= \(\frac{3}{11}\)

Question 6.
A company has two plants to manufacture scooters. Plant I manufactures 80% of the scooters and plant II manufactures 20%. At plant I, 85 out of 100 scooters are rated as of Q standard quality and at plant II, 65 out of 100 scooters are rated as of standard quality. A scooter is chosen at random and is found to be of standard quality. What is the probability that it has come from plant I ? (ISC 2006)
Answer:
Let E₁, E₂ and A be the events defined as follows :
E₁ : scooter is manufactured by plant I
E₂ : scooter is manufactured by plant II
A : scooter chosen is of standard quality
Then P(E₁)= \(\frac{80}{100}=\frac{4}{5}\);
p(E₂)= \(\frac{20}{100}=\frac{1}{5}\)
P (A/E₁) = prob. of standard quality scooter manufactured by plant I = \(\frac{85}{100}=\frac{17}{20}\)
P (A/E₂) = prob. of standard quality scooter manufactured by plant II = \(\frac{65}{100}=\frac{13}{20}\)
We want to find P(E₁)/A)
P(E₁)/A) = \(\frac{P\left(A / E_1\right) P\left(E_1\right)}{P\left(A / E_1\right) P\left(E_1\right)+P\left(A / E_2\right) P\left(E_2\right)}\)
= \(\frac{\frac{17}{20} \times \frac{4}{5}}{\frac{17}{20} \times \frac{4}{5}+\frac{13}{20} \times \frac{1}{5}}\)
= \(\frac{\frac{68}{100}}{\frac{68}{100}+\frac{13}{100}}=\frac{68}{68+13}=\frac{68}{81}\)

Question 7.
In a class, having 60% boys, 5% of the boys and 10% of the girls have an I.Q. of more than 150. A student is selected at random and found to have an I.Q. of more than 150. ( Find the probability that the selected student is a boy.
Answer:
Consider the following events :
E₁ = Selected student is boy
E₂ = Selected student is girl
A = A student with IQ more that 150 is selected
Then P (E₁) = \(\frac{60}{100}\) and P (E₂) = \(\frac{40}{100}\)
P (A/E₁) = P (Selected boy has IQ more than 150) = \(\frac{5}{100}\)
and P(A/E₂) = P (Selected girl has IQ more than 150) = \(\frac{10}{100}\)
We want to find the probability of selected student with IQ more than 150 is a boy P(E₁/A)
Then by baye’s theorem, we have
P(E₁/A) = \(\frac{P\left(E_1\right) P\left(A / E_1\right)}{P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) P\left(A / E_2\right)}\)
= \(\frac{\frac{60}{100} \times \frac{5}{100}}{\frac{60}{100} \times \frac{5}{100}+\frac{40}{100} \times \frac{10}{100}}\)
= \(\frac{300}{300+400}=\frac{300}{700}=\frac{3}{7}\)
required probability = \(\frac{3}{7}\)

Question 7(Old).
It is known that 60% students in a college reside in hostel while remaining 40% are day scholars. 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade. If one randomly chosen student has A grade, then what is the probability that he lives in the hostel ? (NCERT)
Answer:
Consider the following events as follows :
E₁ = The chosen student is a hosteller
E₂ = The chosen student is not a hosteller.
A = The student has an A grade in their annual examination
Then P (E₁) = 60% = \(\frac{60}{100}\) = 0.6
P(E₂) =40% = \(\frac{40}{100}\) = 0.4
∴ P(A/E₁) = Probability that the student has an A grade given that the student is a hosteller.
P(A./E₁) = \(\frac{30}{100}\) = 0.3
and P (A/E₂) = Probability that the student has an A grade given that the student is not a hosteller
P (A/E₂) = \(\frac{20}{100}\) = 0.2
Then by Baye’s Theorem, we have Required probability = P (E₁/A)
= \(\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)}\)
= \(\frac{0.6 \times 0.3}{0.6 \times 0.3+0.4 \times 0.2}=\frac{18}{26}=\frac{9}{13}\)

Question 8.
(i) Often it is taken that a truthful person commands more respect in society. A man is known to speak truth 4 out of 5 times. He throws a die and reports that it is a six. Find the probability that it is actually a six. Do you also agree that the value truthfulness leads to more respect in the society ?
(ii) A man is known to speak truth 3 out of 5 times. He throws a die and reports that it is a number greater than 4. Find the probability that it is actually a number greater than 4.
Answer:
(i) Let E₁, E₂ and A be the events defined as follows :
E₁ : die shows six i.e. six has occurred
E₂ : die does not show six i.e. six has not occured
A : A man reports that it is a six.
ThenP(E₁) = \(\frac{1}{6}\); P (E₂) = \(\frac{5}{6}\)
P(A/E₁) = probability that man reports that six occurs and six has occured
= probability that man speaks rruth = \(\frac{4}{5}\)
P (A/E₂) = probability that man reports that six occurs given that six has not occured
= Man tell a lie = 1 – \(\frac{4}{4+5}=\frac{4}{9}\)
We want to find P (E₁/A)
Then by Baye’s Theorem, we have
P(E₁/A) = \(\frac{P\left(A / E_1\right) P\left(E_1\right)}{P\left(A / E_1\right) P\left(E_1\right)+P\left(A / E_2\right) P\left(E_2\right)}\)
= \(\frac{\frac{4}{5} \times \frac{1}{6}}{\frac{4}{5} \times \frac{1}{6}+\frac{1}{5} \times \frac{5}{6}}\)
= \(\frac{4}{4+5}=\frac{4}{9}\)
Yes, truthfulness leads to more respect in the society.

(ii) Let E₁, E₂ and A be the event defined as follows :
E₁ : The die shows a number > 4
E₂ : The die did not shows a number > 4
A : Man reports that the die shows a number > 4.
Then P (E₁) = \(\frac{2}{6}=\frac{1}{3}\); P (E₂) = \(\frac{4}{6}=\frac{2}{3}\)
.-. P(A/E₁) = probability that man reports that die shows a number > 4 given that die show a number > 4
= prob. that man speaks truth = \(\frac{3}{5}\)
P(A/E₂) = prob. that man reports that die shows a number > 4 given that die does not shows a number > 4
= prob. that man tell a lie = 1 – \(\frac{3}{5}=\frac{2}{5}\)
We want to evaluate P (E₁/A)
Then by Baye’s Theorem, we have
P(E₁/A) = \(\frac{\mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right) \mathrm{P}\left(\mathrm{E}_1\right)}{\sum_{i=1}^2 \mathrm{P}\left(\mathrm{A} / \mathrm{E}_i\right) \mathrm{P}\left(\mathrm{E}_i\right)}\)
= \(\frac{\frac{3}{5} \times \frac{1}{3}}{\frac{3}{5} \times \frac{1}{3}+\frac{2}{5} \times \frac{2}{3}}=\frac{3}{3+4}=\frac{3}{7}\)

Question 9.
There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls, and 4 white and 1 black ball, respectively. There is an equal opportunity of each urn being ^ chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that ball drawn was from the second urn. (NCERT Exemplar)
Answer:
Given Urn I contains 2 white and 3 black balls
Urn II contains 3 white and 2 black balls
Urn III contains 4 white and 1 black balls
Consider the following events :
E₁ = urn I is selected ;
E₂ = urn II is selected
E₃ = urn III is selected ;
A = a white ball is drawn 1 1 1
Then P (E₁) = \(\frac{1}{2}\) ; P (E₂) = \(\frac{1}{2}\) and P (E₃) = \(\frac{1}{2}\) [Since there are 3 urns]
Thus, P (A|E₁) = P [Drawing 1 white ball from urn I]
= \(\frac{2}{5}\)

P(A/E₂) = P [Drawing 1 white ball from urn II]
= \(\frac{3}{5}\)

P(A/E₃) = P [Drawing one white ball from urn III]
= \(\frac{4}{5}\)

We want to find
P (Drawn one white ball from urn II) = P(E₂/A)
Then by Baye’s theorem, we have
required probability = P(E₂/A)

Question 10.
A bag contains 5 red and 4 black balla, a second bag contains 3 red and 6 black balls. One of the two bags is selected at random and two balls are drawn at random (without replacement) both of which are found to be red. Find the probability that the balls are drawn from the second bag.
Answer:
Let us defined the events are as follows:
E₁ :bag I is selected
E₂ : bag II is selected
F : Two red balls are drawn without replacement from selected bag.
To find P(E₂/E).

Question 11.
Urn I has 2 white and 3 black balls, urn II has 4 white and 1 black ball and urn III has 3 white and 4 black balls. An urn is selected at random and a ball is drawn at random.
(i) What is the probability of drawing a white ball ?
(ii) If the ball drawn is white, then what is the probability that urn I was selected ?
Answer:
Let us define the event E₁, E₂, E₃ and A are as follows :
E₁ : urn I is selected
E₂ : urn II is selected
E₃ : urn III is selected
A : drawing a white ball from selected urn
ThenP(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)
.’. P (A/E₁) = prob. of drawing a white ball from um-I
= \(\frac{2}{5}\)

P (A/E₂) = probability of drawing a white ball from urn II
= \(\frac{4}{5}\)

P (A/E₃) = probability of drawing a white ball from urn III
= \(\frac{3}{7}\)

(i) P (A) = P (E₁) P (A/E₁) + P (E₂) P (A/E₂) + P (E₃) . P (A/E₃)
= \(\frac{2}{5} \times \frac{1}{3}+\frac{4}{5} \times \frac{1}{3}+\frac{3}{7} \times \frac{1}{3}\)
= \(\frac{2}{15}+\frac{4}{15}+\frac{1}{7}=\frac{2}{5}+\frac{1}{7}=\frac{19}{35}\)

(ii) We want to find P (E₁/A)
Then by Baye’s Theorem, we have

Question 12.
A fair die is rolled. If face 1 turns up, a ball is drawn from bag A. If face 2 or 3 turn up, a ball is drawn from bag B. If face 4 or 5 or 6 turn up, a ball is drawn from bag C. Bag A contains 3 red and 2 white balls, bag B contains 3 red and 4 white balls and bag C contains 4 red and 5 white balls. The die is roiled, a bag is picked up and a ball is drawn. If the ball drawnis red, what is the probability that it is drawn from bag B ? (ISC 2017)
Answer:
Given Bag A contains 3 red and 2 white balls
Bag B contains 3 red and 4 white balls
Bag C contains 4 red and 5 white balls
Let us define the events E₁, E₂, E₃ and E are as follows :
E₁ : event that bag A is selected
E₂ : event that bag B is selected
E₃ : event that bag C is selected
A : ball drawn is red
Then E₁ and E₂, E₃ are mutually exclusive and exhaustive events. Then
P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)
P (A/E₁) = P (prob. of getting a red ball from bag A)
= P (prob. of getting a red ball when bag A is chosen and a die shows 1)
= \(\frac{1}{6} \times \frac{3}{5}=\frac{1}{10}\)

P (A/E₂) = P (prob. of getting a red ball from bag B is chosen)
= \(\frac{2}{6} \times \frac{3}{7}=\frac{1}{7}\)

P (A/E₃) = P (prob. of getting a red ball when bag C is chosen and a die shows 4, 5 or 6)
= \(\frac{3}{6} \times \frac{4}{9}=\frac{2}{9}\)
We want to find P (E₂/A)
Then by Baye’s Theorem, we have

Question 13.
There are two bags, I and II. Bag I contains 3 red and 5 black balls and bag II contains 4 red and 3 black balls. One ball is transferred randomly from bag 1 to bag II and then a ball is drawn randomly from bag II. If the ball so drawn is found to be black in colour, then find the probability that the transferred ball is also black.
Answer:
Let us define the events are as follows :
E₁ : red ball is transferred from bag I to bag II
E₂ : black ball is transferred from bag I to bag II
E : one black ball has been drawn from bag II

To find P (E₂/E):
Here Bag-1 contains 3 red and 5 black balls
P (E₁) = \(\frac{3}{8}\); P(E₂) = \(\frac{5}{8}\)

When E₁ has occurred i.e. one red ball is transferred from bag I to bag II then bag II contains 5 red balls and 3 black balls.
∴ P(E/E₁) = \(\frac{3}{5+3}=\frac{3}{8}\)
When E₂ has occurred i.e. one black ball is transferred from bag I to bag II then bag II contains 4 red balls and 4 black ball.
∴ P(E/E₂) = \(\frac{4}{4+4}=\frac{4}{8}\)

Thus by baye’s theorem, we have
P(E₂/E) = \(\frac{P\left(E / E_2\right) \cdot P\left(E_2\right)}{P\left(E / E_1\right) \cdot P\left(E_1\right)+P\left(E / E_2\right) P\left(E_2\right)}\)
= \(\frac{\frac{4}{8} \times \frac{5}{8}}{\frac{3}{8} \times \frac{3}{8}+\frac{4}{8} \times \frac{5}{8}}=\frac{20}{29}\)

Question 14.
Bag I contaIns 5 red and 4 white balls and Bag II contains 3 red and 3 white balls. Two balls are transfened from Bag I to Bag II and then one ball is drawn from the Bag II. If the ball drawn from the Bag Ills red, then find the probability that one red ball and one white ball are transferred from Bag I to Bag II.
Answer:
Given bag-I contains 5 red and 4 white balls and bag-II contains 3 red and 3 white balls Let us define the events are as follows :
E₁ : transferring two red balls from bag I to bag-II
E₂ : transferring one red and one white balls from bag I to bag II
E₃ : transferring two white balls from bag I to II.
E : red ball has been drawn from bag-II To find P(E₂/E)

Here the events E₁, E₂ and E₃ are mutually exclusive.
When E₁ has occurred i.e. 2 red balls are transferred from I to II so bag-II contains 5 red and 3 white balls.
P(E/E₁) = \(\frac{5}{8}\)

When E₂ has occurred ¡e. in that case, bag II contains 4 red and 4 white balls.
Then P(E/E₂) = \(\frac{4}{8}=\frac{1}{2}\)

When E₃ has occurred, ¡n that case, bag-II contains 3 red balls and 5 white balls.
.‘. P(E/E₃) = \(\frac{3}{8}\)
Thus by Baye’s theorem ; we have

Question 15.
Each of three identical jewellery boxes has two drawers. In each drawer of the first box z there is a gold watch. In each drawer of the second box there is a silver watch. In one drawer of the third box there is a gold watch while in the other there is a silver watch. If we select a box at random, open one of the drawers and find it to contain a silver watch, then what is the probability that the other drawer has the gold watch ?
Answer:
Let the events E₁, E₂, E₃ and A are defined as follows :
E₁ : box-I is selected;
E₂ : box-II is chosen;
E₃ : box III is chosen
A : A silver watch in one of the drawer.
Then P (E₁) = P (E₂) = P (E₃) = \(\frac{1}{3}\)
∴ P (A/E₁) = prob. of drawing a silver watch from box I = \(\frac{0}{2}\) = 0
P (A/E₂) = prob. of drawing a silver watch from box II = \(\frac{2}{2}\) = 1
P (A/E₃) = prob. of drawing a silver watch from box III = \(\frac{1}{2}\)
We want to find the prob. that the other watch in the chosen box is of gold
i.e. we want to find P (E₃/A)
Then by Baye’s Theorem

Question 16.
Given three identical boxes A, B and C, box A contains 2 gold and 1 silver coin, box B contains 1 gold and 2 silver coins and box C contaIns 3 silver coins. A person chooses a box at random and takes out a coin. If the coil drawn Is of silver, find the probability that it has been drawn from the box which has the remaining two coins also of silver. (ISC 2019)
Answer:
Given box A contains 2 gold and 1 silver coin box B contains I gold and 2 silver coins and box C contains 3 silver coins
Let us define the events E₁, E₂, E₃ are as follows:
E₁ : box A is chosen
E₂: box B is chosen
E₃ : box C is chosen

E : drawing coin is of silver
∴ P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)

P(E/E₁) = Prob. of drawing a silver coin when box A has already chosen = \(\frac{1}{3}\)
P(E/E₂) = prob. of drawing a silver coin when box B has already chosen = \(\frac{2}{3}\)
P(E/E₃) = prob. of drawing a silver coin when box C has already chosen = \(\frac{3}{3}\) = 1

we want to find P (E₃/E)
Then by Baye’s Theorem, we have

Question 17.
In a bolt factory, three machines A, B and C manufactures 25%, 35% and 40% of the total production respectively. Of ‘ ‘’Htheir respective outputs, 5%, 4% and 2% are defective. A bolt is drawn at random from the total production and it is found to be defective. Find the probability that it was manufactured by machine C. (ISC 2014)
Answer:
Let us define the events E₁, E₂, E₃ and E are as follows :
E₁ : Bolt produced by Machine A
E₂ : Bolt produced by Machine B
E₃ : Bolt produced by Machine C.
E : bolt drawn is found to be defective
∴ P(E₁) = \(\frac{25}{100}\)
P(E₂) = \(\frac{35}{100}\)
P(E₃) = \(\frac{40}{100}\)

P(E/E₁) = \(\frac{5}{100}\)
P(E/E₂) = \(\frac{4}{100}\)
P(E/E₃) = \(\frac{2}{100}\)
Then by Bay’s theorem, we have

Question 18.
A factory has three machines A, B and C producing 1500,2500 and 3000 bulbs per day respectively. Machine A produces 1.5% defective bulbs, machine B produces 2% defective bulbs and machine C produces 2.5% defective bulbs. At the end of the day, a bulb is drawn at random and it is found to be defective. What is the probability that this defective bulb has been produced by machine B ? (ISC 2010)
Answer:
Let us define the events E₁, E₂, E₃ and A are as follows :
E₁ : bulb produced by machine A
E₂ : bulb produced by machine B
E₃ : bulb produced by machine C
A : A defective bulb is selected

P (A/E₁) = prob. of defective bulb produced by machine A
= 1.5% = \(\frac{1 \cdot 5}{100}=\frac{15}{1000}=\frac{3}{200}\)

P(A/E₂) = probability of getting defective bulb produced by machine B
= \(\frac{2}{100}=\frac{1}{50}\)

P (A/E₃) = prob. of getting defective bulb produced by machine C
= \(\frac{2 \cdot 5}{100}=\frac{25}{1000}=\frac{1}{40}\)

Then we want to find P (E₂/A)
Thus by baye’s Theorem, we have

Question 19.
An insurance company insured 4000 doctors, 8000 teachers and 12000 engineers. The probability 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 person dies before the age of 58 years, find the probability that he is a doctor. (ISC 2009)
Answer:
Let us define the events E₁, E₂, E₃ and A as follows :
E₁ : Insured person is doctor
E₂ : Insured person is teacher
E₃ : Insured person in engineer.
A : One of the insured person dies before the age of 58 years.

P (E₁) = \(\frac{4000}{4000+8000+12000}=\frac{4000}{24000}=\frac{1}{6}\)

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 (A/E₁) = prob. that the insured person who is dies before the age of 58 years be a doctor = 0.01
P (A/E₂) = prob. that the insured person dies before the age of 58 years be a teacher = 0.03
P (A/E₃) = prob. that the insured person dies before the ageof 58 years be an engineer = 0.05
We want to find P (E₂/A)
Then by Baye’s Theorem, we have

Question 20.
In an automobile factory, certain parts are to be fixed into the chassis in a section before it moves to the next section. On a given day, one of the three persons A, B and C carries out the 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 time than the allotted 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 allotted time, what is the probability that A has done the task ? (ISC 2016)
Answer:
Let E₁, E₂, E₃ and A be the events define as follows :
E₁ : person A carries out the task
E₂ : person B carries out the task
E₃ : person C carries out the task.
A : Take taken by one of the selected person is more than the allotted time
Then P (E₁) = \(\frac{45}{100}\)
P (E₂) = \(\frac{35}{100}\)
P(E₃) = \(\frac{20}{100}\)

P (A/E₁) = Probability that A will take more time than the allotted time = \(\frac{1}{6}\)
P (A/E₂) = Probability that B will take more time than the allotted time – \(\frac{1}{10}\)
P (A/E₃) = Probability that C will take more time than the allotted time = \(\frac{1}{20}\)

We want to find P (E₁/A)
Then by Baye’s Theorem, we have P(E₁/A)

Question 21.
There are three coins. One is a two-headed coin, another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed. If it shows heads, then what is the probability that it was the two-headed coin ?
Answer:
Let us consider the following events
E₁ = Chosen coin is two headed ;
E₂ : Chosen coin is biased
E₃ : Chosen coin is unbiased
P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)
Let A be the event that the coin shows heads.
A two-headed coin will always show heads.
P (A/E₁) = P (coin showing heads, given that it is a two-headed coin) = 1
given probability of heads coming up, given that it is a biased coin = 75%
P(A/E₂) = \(\frac{75}{100}=\frac{3}{4}\)
Since the third coin is unbiased-, the probability that heads comes is always \(\frac{1}{2}\)
P(A/E₃) = \(\frac{1}{2}\)
We want to find the probability that the coin is two-headed, given that it shows heads, is given by P(E₁/A).
Then by using Baye’s theorem, we get

Question 21(Old).
Three persons A, B and C apply for a job of Manager in a Private Company. Chances of ^ their selection (A, B and C) are in the ratio 1:2:4. The probabilities that A, B and C can 0 introduce changes to improve profits of the company are 0-8, 0-5 and 0-3 respectively. If the change does not take place, find the probability that it is due to the appointment of C.
Answer:
Let us consider the following events :
E₁ = person A is selected as an manager
E₂ = person B is selected as an manager
E₃ = person C is selected as an manager
D = changes do not take place
Then P(E₁) = \(\frac{1}{1+2+4}=\frac{1}{7}\) P(E₃) = \(\frac{2}{7}\); P(E₃) = \(\frac{4}{7}\)
P (D/E₁) = 1 – 0.8 = 0.2 ;
P (D/E₂) = 1 – 0.5 = 0.5
and P (D/E₃) = 1 – 0.3 = 0.7
Now we want to find the probability that change do not take place due to appointment of C.
required probability

Question 22.
By examining the chest X-ray, the probability that TB Is detected when a person is actually suffering from it Is 099. The probability of an healthy person diagnosed to have TB is 0001. In a certain city, 1 ¡n 1000 people suffers from TB. A person Is selected at random and is diagnosed to have TB. What is the probability that he/she actually has TB? (NCERT Exemplar)
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 TB.
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. 099
and P(A/E₂) = probability that person diagnosed to have TB. and he is not actually having T.B. = 000 1
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

Thus required probability = \(\frac{110}{221}\)

Question 23.
A doctor is to visit a patient. From the past experience, ¡t is known that the probability that he will come by public transport, scooter, taxi or personal car are respectively \(\frac{1}{10}, \frac{1}{5}, \frac{3}{10}\) and \(\frac{2}{5}\). The probabilities that he will be late are \(\frac{1}{4}, \frac{1}{3}\) or \(\frac{1}{12}\) if he comes by public transport, scooter or taxi respectively, but if he comes by personal car he will not be late. When he arrives, he is late. What is the probability that he came by scooter ?
Do you think, we should use public transport and why ? (Value Based)
Answer:
Let us define the events are as follows:
E₁ : doctor goes by public transport
E₂ : doctor goes by scooter
E₃ : doctor goes by taxi ‘
E₄ : doctor goes by personal car
E : doctor reaches late
P (E₁) = \(\frac{1}{10}\)
P (E₂) = \(\frac{1}{5}\)
P(E₃) = \(\frac{3}{10}\);
P (E₄) = \(\frac{2}{5}\)
and P (E/E₁) = \(\frac{1}{4}\)
P (E/E₂) = \(\frac{1}{3}\)
P(E/E₃) = \(\frac{1}{12}\);
P (E/E₄) = prob. that the doctor will be late if he goes by personal car = 0
By Baye’s theorem, we have

We should use public transport because it is safe less expensive and helps in saving petrol/ diesel.

Question 24.
Suppose we have four boxes A, B, C, D containing coloured marbles as given below:

One of the boxes has been selected at random and a marble is drawn from it. If the marble is red, then what are the probabilities that it was drawn from boxes A, B, C, D respectively ? (NCERT)
Answer:
Consider the following events :
E₁ = Box A is selected
E₂ = Box B is selected
E₃ = Box C is selected
E₄ = Box D is selected
A = Marble drawn is of red colour
Then P (E₁) = P (E₂) = P (E₃) = P (E₄) = \(\frac{1}{4}\)
Also P(A/E₁)= \(\frac{1}{10}\)
P(A/E₂) = \(\frac{6}{10}\);
P (A/E₃) = \(\frac{8}{10}\);
P (A/E₄) = \(\frac{0}{10}\) = 0
We want to find the probability drawn marble comes from box A, B and C it is given that it is of red colour.
That is to find P(E₁/A), P(E₂/A) andP(E₃/A).
Then by Baye’s theorem, we have

Question 25.
A coin is tossed. If it turns up heads, 2 bails will be drawn from urn A, otherwise 2 balls s will be drawn from urn 8. Urn A contains 3 black and 5 white balls. Urn B contains 7 black balls and 1 white ball. In both cases, selections are to be made with replacement. What is the probability that urn A is used given that both the balls drawn are black ?
Answer:
Given urn A contains 3 black and 5 white balls and urn B contains 7 black and 1 white ball
Let us define the events E₁, E₂ and A are as follows :
E₁ : urn A is chosen
E₂ : urn B is chosen
A : both drawn balls are black.
Then P (E₁) = \(\frac{1}{2}\) = P (E₂)
Total no. of balls in urn A = 3 + 5 = 8
∴ P (A/E₁) = prob. that both drawn black balls from urn A = \(\frac{3}{8} \times \frac{3}{8}=\frac{9}{64}\)
P (A/E₂) = prob. that both black balls from urn B = \(\frac{7}{8} \times \frac{7}{8}=\frac{49}{64}\)
We want to find P (E₁/A)
Then by Baye’s Theorem ; we have

Question 26.
Redo the above problem assuming that the coin is biased and turns up heads 2 times out of 3.
Answer:
Let us define the events E₁, E₂ and A as follows :
E₁ : urn A is chosen
E₂ : urn B is chosen
A : both drawn balls are black
Then P(E₁) = \(\frac{2}{3}\)
P(E₂) = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
∴ P (A/E₁) = prob. that both drawn black balls from urn A with replacement = \(\frac{3}{8} \times \frac{3}{8}=\frac{9}{64}\)
P (A/E₂) prob. that both drawn black balls from urn B with replacement = \(\frac{7}{8} \times \frac{7}{8}=\frac{49}{64}\)

We want to find P (E₁/A)
Then by Baye’s Theorem, we have

Question 27.
There are three urns. Urn I contains 1 white, 2 black and 3 red balls. Urn II contains 2 white, 1 black and 1 red balls. Urn III contains 4 white, 5 black and 3 red balls. One urnis chosen at random and 2 balls drawn without replacement. They happen to be white and red. What is the probability that they are from urn I, II or III ?
Answer:
Given, urn I contain 1 white, 2 black and 3 red balls urn II contain 2 white, 1 black and 1 red ball urn III contain 4 white, 5 black and 3 red balls.
Let us define the events E₁, E₂, E₃ and A are as follows :
E₁ : urn I is chosen
E₂ : urn II is chosen .
E₃ : urn III is chosen
A : one white and one red ball drawn from one of the selected urn without replacement
Then P (E₁) = \(\frac{1}{3}\) = P (E₂) = P (E₃)

P (A/E₁) = Prob. of drawing one white and one-red ball from urn I without replacement
= \(\frac{1}{6} \times \frac{3}{5}=\frac{1}{10}\)

P (A/E₂) = prob. of drawing one white and one red ball from urn II without replacement
= \(\frac{2}{4} \times \frac{1}{3}=\frac{1}{6}\)

P (A/E₃) = probability of drawing one white and one red ball from urn III without replacement
= \(\frac{4}{12} \times \frac{3}{11}=\frac{1}{11}\)

We want to find P (E₁/A), P (E₂/A) and P (E₃/A)
Then by Baye’s Theorem, we have

Question 28.
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn randomly one-by-one without replacement and are found to be both kings. Find
the probability of the lost card being a king.
Answer:
Let us define the events are as follows:
E₁: lost card is a king card
E₂: lost card is not of king card.
E : Two cards drawn are all king cards.

ThusP(E₁) = \(\frac{4}{52}=\frac{1}{13}\)
P(E₂) = \(\frac{4}{52}=\frac{1}{13}\)

When E₁ has occurred ¡e. lost card is of king card.
Then prob. of drawing 2 cards of kings from remaining pack = \(\frac{{ }^3 C_2}{{ }^{51} C_2}\)
= \(\frac{3 \times 2}{51 \times 50}=\frac{1}{17 \times 25}\) = P(E/E₁)

When E₂ has occurred i.e. lost card is not of king card. Then P (E/E₂) = \(\frac{{ }^4 C_2}{{ }^{51} C_2}\)
= \(\frac{6 \times 2}{51 \times 50}=\frac{2}{17 \times 25}\)

Then by Baye’s Theorem,

Question 29.
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 hearts. Find the probability of the missing card to be a heart.
Answer:
Let the events are :
E₁ : Missing card is a heart
E₂ : Missing card is a spade
E₃ : Missing card is a club
E₄ : Missing card is a diamond and
A : Two drawn cards are heart