RD Sharma Class 12 Ex 31.7 Solutions Chapter 31 Probability

Here we provide RD Sharma Class 12 Ex 31.7 Solutions Chapter 31 Probability for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 31.7 Solutions Chapter 31 Probability book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter31
Exercise31.7
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 31.7 Solutions Chapter 31 Probability

Question 1. The contents of urns I, II, III are as follows:

Urn I: 1 white, 2 black, and 3 red balls

Urn II: 2 white, 1 black, and 1 red ball

Urn III: 4 white, 5 black, and 3 red balls

One urn is chosen at random and two balls are drawn. They happen to be white and red. What is the probability that they come from Urns I, II, III?

Solution:

Let us assume that E1, E2 and E3 be the events of selecting Urn I, Urn II and Urn III. And also A be the event that the two balls drawn are white and red.

So, P(E1) = 1/3

P(E2) = 1/3

P(E3) = 1/3

Now,

P(A/E1) = \frac{{}^1 C_1 \times^3 C_1}{{}^6 C_2}   = 3/15 = 1/5

P(A/E2) = \frac{{}^2 C_1 \times^1 C_1}{{}^4 C_2}   = 2/6 = 1/3

P(A/E3) = \frac{{}^4 C_1 \times^3 C_1}{{}^{12} C_2}   = 12/66 = 2/11

By using Bayes’ theorem, the required probability is,

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) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{1}{3} \times \frac{1}{5}}{\frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{11}}

\frac{\frac{1}{5}}{\frac{1}{5} + \frac{1}{3} + \frac{2}{11}}

\frac{\frac{1}{5}}{\frac{33 + 55 + 30}{165}}

= 33/118

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) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{1}{3} \times \frac{1}{3}}{\frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{11}}

\frac{\frac{1}{3}}{\frac{1}{5} + \frac{1}{3} + \frac{2}{11}}

\frac{\frac{1}{3}}{\frac{33 + 55 + 30}{165}}

= 55/118

P(E3/A) = \frac{P\left( E_3 \right)P\left( A/ E_3 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{1}{3} \times \frac{2}{11}}{\frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{11}}

\frac{\frac{2}{11}}{\frac{1}{5} + \frac{1}{3} + \frac{2}{11}}

\frac{\frac{2}{11}}{\frac{33 + 55 + 30}{165}}

= 30/118

Question 2. A bag A contains 2 white and 3 red balls and a 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.

Solution:

Let us assume that the A, E1 and E2 be the events that the ball is red, bag A is chosen and bag B is chosen.

So, P(E1) = 1/2

P(E2) = 1/2

Now,

P(A/E1) = 3/5

P(A/E2) = 5/9

By using Bayes’ theorem, the required probability is,

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{\frac{1}{2} \times \frac{5}{9}}{\frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{5}{9}}

= 25/52

Question 3. Three urns contain 2 white and 3 black balls; 3 white and 2 black balls and 4 white and 1 black ball respectively. One ball is drawn from an urn chosen at random and it was found to be white. Find the probability that it was drawn from the first urn.

Solution:

Let us consider E1, E2 and E3 be the events of selecting Urn I, Urn II and Urn III.

Also, A be the event that the ball drawn is white.

So, P(E1) = 1/3

P(E2) = 1/3

P(E3) = 1/3

Now,

P(A/E1) = 2/5

P(A/E2) = 3/5

P(A/E3) = 4/5

By using Bayes’ theorem, the required probability is,

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) +P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{1}{3} \times \frac{2}{5}}{\frac{1}{3} \times \frac{2}{5} + \frac{1}{3} \times \frac{3}{5} + \frac{1}{3} \times \frac{4}{5}}

\frac{2}{2 + 3 + 4}

= 2/9

Question 4. The contents of the three urns are as follows: 

Urn 1: 7 white, 3 black balls, Urn 2: 4 white, 6 black balls, and Urn 3: 2 white, 8 black balls. One of these urns is chosen at random with probabilities 0.20, 0.60, and 0.20 respectively. From the chosen urn two balls are drawn at random without replacement. If both these balls are white, what is the probability that these came from urn 3?

Solution:

Let us assume that E1, E2 and E3 be the events of selecting Urn I, Urn II and Urn III. Also, A be the event that the two balls drawn are white.

So, P(E1) = 20/100

P(E2) = 60/100

P(E3) = 20/100

Now,

P(A/E1) = \frac{{}^7 C_2}{{}^{10} C_2}    = 21/45

P(A/E2) = \frac{{}^4 C_2}{{}^{10} C_2}    = 6/45

P(A/E3) = \frac{{}^2 C_2}{{}^{10} C_2}    = 1/45

By using Bayes’ theorem, the required probability is,

P(E3/A) = \frac{P\left( E_3 \right)P\left( A/ E_3 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{20}{100} \times \frac{1}{45}}{\frac{20}{100} \times \frac{21}{45} + \frac{60}{100} \times \frac{6}{45} + \frac{20}{100} \times \frac{1}{45}}

\frac{1}{21 + 18 + 1}

= 1/40

Question 5. Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4, 5, or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, then what is the probability that she threw 3, 4, 5, or 6 with the die?       

Solution:

Let us consider E1 be the event that the outcome on the die is 1 or 2 and E2 be the event that outcome on the die is 3, 4, 5 or 6. 

Now, P(E1) = 2/6 = 1/3 and P(E2) = 4/6 = 2/3

Also, let us assume A be the event of getting exactly one ‘tail’.

So, P(A/E1) = 3/8

P(A/E2) = 1/2

Now, P(E2/A) is the probability that the girl threw 3, 4, 5 or 6 with the die, if she obtained exactly one tail

So, by using Bayes’ theorem, the required probability is,

P(E2/A) = \frac{P\left( E_2 \right) \times P\left( A/E_2 \right)}{P\left( E_1 \right) \times P\left( A/ E_1 \right) + P\left( E_2 \right) \times P\left( A/ E_2 \right)}

\frac{\left( \frac{2}{3} \times \frac{1}{2} \right)}{\left( \frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2} \right)}

\frac{ \frac{2}{6}}{ \frac{1}{8} + \frac{1}{3}}

\frac{24 \times 2}{11 \times 6}

= 8/11

Question 6. Two groups are competing for the positions of the Board of Directors of a Corporation. The probabilities that the first and the second groups 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 the second group.

Solution:

Let us consider E1 and E2 be the events that the first group and the second group win the competition. Also, A be the event of introducing a new product.

So, P(E1) = 0.6

P(E2) = 0.4

P(A/E1) = 0.7

P(A/E2) = 0.3

Now, P(E2/A) is the probability that the new product is introduced by the second group

So, by using Bayes’ theorem, the required probability is,

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}

= 0.12/0.54

= 2/9

Question 7. Suppose 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 selected. Assume that there are equal numbers of men and women.

Solution:

Let us assume that A, E1 and E2 be the events that the person is a good orator, is a man and is a woman.

So, P(E1) = 1/2

P(E2) = 1/2

Now,

P(A/E1) = 5/100

P(A/E2) = 25/1000

So, using Bayes’ theorem, the required probability is,

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{1}{2} \times \frac{5}{100}}{\frac{1}{2} \times \frac{5}{100} + \frac{1}{2} \times \frac{25}{1000}}

\frac{1}{1 + \frac{1}{2}}

= 2/3

Question 8. A letter is known to have come either from LONDON or CLIFTON. On the envelope just two consecutive letters ON are visible. What is the probability that the letter has come from

(i) LONDON (ii) CLIFTON?

Solution:

Let us consider A, E1 and E2 be the events that the two consecutive letters are visible and the letter has come from LONDON and CLIFTON.

So, P(E1) = 1/2

P(E2) = 1/2

Now,

P(A/E1) = 2/5

P(A/E2) = 1/6

So, using Bayes’ theorem, the required probability is,

(i) 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{1}{2} \times \frac{2}{5}}{\frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{1}{6}}

\frac{\frac{2}{5}}{\frac{2}{5} + \frac{1}{6}}

\frac{\frac{2}{5}}{\frac{17}{30}}

= 12/17

(ii) P(E2/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{1}{2} \times \frac{1}{6}}{\frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{1}{6}}

\frac{\frac{1}{6}}{\frac{2}{5} + \frac{1}{6}}

\frac{\frac{1}{6}}{\frac{17}{30}}

= 5/17

Question 9. In a class, 5% of the boys and 10% of the girls have an IQ of more than 150. In this class, 60% of the students are boys. If a student is selected at random and found to have an IQ of more than 150, find the probability that the student is a boy.

Solution:

Let us consider A, E1 and Ebe the events that the IQ is more than 150, the selected student is a boy and the selected student is a girl.

So, P(E1) = 60/100

P(E2) = 40/100

Now,

P(A/E1) = 5/100

P(A/E2) = 10/100

So, using Bayes’ theorem, the required probability is,

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}

= 300/700

= 3/7

Question 10. A factory has three machines, X, Y, and Z producing 1000, 2000, and 3000 bolts per day respectively. The machine X produces 1% defective bolts, Y produces 1.5% and Z produces 2% defective bolts. At the end of the day, a bolt is drawn at random and is found to be defective. What is the probability that this defective bolt has been produced by machine X?

Solution:

Let us assume that E1, E2 and E3 be the events that machine X produces bolts, machine Y produces bolts and machine Z produces bolts. Also, A be the event that the bolt is defective.

So, the total number of bolts = 1000 + 2000 + 3000 = 6000

P(E1) = 1000/6000 = 1/6

P(E2) = 2000/6000 = 1/3

P(E3) = 3000/6000 = 1/2

Now, P(E1/A) is the probability that the defective bolt is produced by machine X

So, 

P(A/E1) = 1% = 1/100

P(A/E2) = 1.5% = 15/1000

P(A/E3) = 2% = 2/100

So, using Bayes’ theorem, the required probability is,

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) +  P\left( E_2 \right)P\left( A/ E_2 \right)}

\frac{\frac{1}{6} \times \frac{1}{100}}{\frac{1}{6} \times \frac{1}{100} + \frac{1}{3} \times \frac{15}{1000} + \frac{1}{2} \times \frac{2}{100}}

\frac{\frac{1}{6}}{\frac{1}{6} + \frac{1}{2} + 1}

\frac{\frac{1}{6}}{\frac{1 + 3 + 6}{6}}

= 1/10

Question 11. An insurance company insured 3000 scooters, 4000 cars, and 5000 trucks. The probabilities of the accident involving a scooter, a car, and a truck are 0.02, 0.03, and 0.04 respectively. One of the insured vehicles meets with an accident. Find the probability that it is an (i) scooter (ii) car (iii) truck. 

Solution:

Let us assume that E1, E2 and E3 be the events that the vehicle is a scooter, a car and a truck. Also, A be the event that the vehicle meets with an accident.

From the question, insurance company insured 3000 scooters, 4000 cars and 5000 trucks.

So, the total number of vehicles = 3000 + 4000 + 5000 = 12000

P(E1) = 3000/12000 = 1/4

P(E2) = 4000/12000 = 1/3

P(E3) = 5000/12000 = 5/12

Now, P(E1/A) is the probability that the vehicle, which meets with an accident, is a scooter 

Now,

P(A/E1) = 0.02 = 2/100

P(A/E2) = 0.03 = 3/100

P(A/E3) = 0.04 = 4/100

So, using Bayes’ theorem, the required probability is,

(i) 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) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{1}{4} \times \frac{2}{100}}{\frac{1}{4} \times \frac{2}{100} + \frac{1}{3} \times \frac{3}{100} + \frac{5}{12} \times \frac{4}{100}}

\frac{\frac{1}{2}}{\frac{1}{2} + 1 + \frac{5}{3}}

\frac{\frac{1}{2}}{\frac{3 + 6 + 10}{6}}

= 3/19

(ii) 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) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{1}{3} \times \frac{3}{100}}{\frac{1}{4} \times \frac{2}{100} + \frac{1}{3} \times \frac{3}{100} + \frac{5}{12} \times \frac{4}{100}}

\frac{1}{\frac{1}{2} + 1 + \frac{5}{3}}

\frac{1}{\frac{3 + 6 + 10}{6}}

= 6/19

(iii) P(E3/A) = \frac{P\left( E_3 \right)P\left( A/ E_3 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{5}{12} \times \frac{4}{100}}{\frac{1}{4} \times \frac{2}{100} + \frac{1}{3} \times \frac{3}{100} + \frac{5}{12} \times \frac{4}{100}}

\frac{\frac{5}{3}}{\frac{1}{2} + 1 + \frac{5}{3}}

\frac{\frac{5}{3}}{\frac{3 + 6 + 10}{6}}

= 10/19

Question 12. Suppose we have four boxes A, B, C, D containing colored marbles as given below:

Box        Colour
 RedWhiteBlack
A163
B622
C811
D064

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A, box B, and box C?

Solution:

Let us assume that R be the event of drawing the red marble and EA, EB and EC be the events of selecting box A, box B and box C.

According to the question, the total number of marbles are 40 and the number of red marbles are 15.

So, P(R) = 15/40 = 3/8

P(EA/R) = The probability of drawing a red marble from box A.

P(EA/R) = \frac{P\left( E_A \cap R \right)}{P\left( R \right)}

\frac{\frac{1}{40}}{\frac{3}{8}}

= 1/15

P(EB/R) = The probability of drawing a red marble from box B.

P(EB/R \right) = \frac{P\left( E_B \cap R \right)}{P\left( R \right)}

\frac{\frac{6}{40}}{\frac{3}{8}}

= 2/5

P(EC/R) = The probability of drawing a red marble from box C.

P(EC/R) = \frac{P\left( E_C \cap R \right)}{P\left( R \right)}

\frac{\frac{8}{40}}{\frac{3}{8}}

= 8/15

Question 13. A manufacturer has three machine operators, A, B, and C. The first operator, A produces 1% defective items, whereas the other two operators, B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job for 30% of the time, and C on the job for 20% of the time. A defective item is produced. What is the probability that it was produced by A?

Solution:

Let us consider E1, E2 and E3 be the time taken by machine operators A, B, and C. Also, X be the event of producing defective items.

P(E1) = 50 % = 1/2

P(E2) = 30 % = 3/10

P(E3) = 20 % = 1/5

Now,

P(X/E1) = 1 % = 1/100

P(X/E2) = 5 % = 5/100

P(X/E3) = 7 % = 7/100

So, using Bayes’ theorem, the required probability is,

P(E1/X) = \frac{P\left( E_1 \right)P\left( X/ E_1 \right)}{P\left( E_1 \right)P\left( X/ E_1 \right) + P\left( E_2 \right)P\left( X/ E_2 \right) + P\left( E_3 \right)P\left( X/ E_3 \right)}

\frac{\frac{1}{2} \times \frac{1}{100}}{\frac{1}{2} \times \frac{1}{100} + \frac{3}{10} \times \frac{5}{100} + \frac{1}{5} \times \frac{7}{100}}

= 5/34

Question 14. An item is manufactured by three machines A, B, and C. Out of the total number of items manufactured during a specified period, 50% are manufactured on machine A, 30% on Band 20% on C. 2% of the items produced on A and 2% of items produced on B are defective and 3% of these produced on C are defective. All the items are stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?   

Solution:

Let us assume that E be the event of getting a defective item.

So, we have

P(A) = 50 % = 1/2

P(B) = 30 % = 3/10

P(C) = 20 % = 1/5

Also we have,

P(E/A) = 2 % = 1/50

P(E/B) = 2 % = 1/50

P(E/C) = 3 % = 3/100

Now,

P(the defective item drawn was manufactured on machine A) = \frac{P\left( A \right) \times P\left( E/A \right)}{P\left( A \right) \times P\left( E/A \right) + P\left( B \right) \times P\left( E/B \right) + P\left( C \right) \times P\left( E/C \right)}

\frac{\frac{1}{2} \times \frac{1}{50}}{\frac{1}{2} \times \frac{1}{50} + \frac{3}{10} \times \frac{1}{50} + \frac{1}{5} \times \frac{3}{100}}

\frac{\left( \frac{1}{100} \right)}{\left( \frac{1}{100} + \frac{3}{500} + \frac{3}{500} \right)}

\frac{\left( \frac{1}{100} \right)}{\left( \frac{5 + 3 + 3}{500} \right)}

\frac{\frac{1}{100}}{\frac{11}{500}}

= 500/1100

= 5/11

Question 15. There are three coins. One is a two-headed coin (having a head on both faces), another is a biased coin that comes up heads 75% of the time and the third is also a biased coin that comes up tail 40% of the time. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?     

Solution:

Let us assume 

A = the event of choosing two-headed coin, 

B = the event of choosing a biased coin that comes up head 75 % of the times 

C = the event of choosing a biased coin that comes up tail 40 % of the times

E = the event of getting a head.

Now,

P(A) = 1/3

P(B) = 1/3

P(B) = 1/3

Also we have,

P(E/A) = 1

P(E/B) = 75 % = 75/100 = 3/4

P(E/C) = 60 % = 60/100 = 3/5

By using Bayes’ theorem, the required probability is 

P(the head shown was of two-headed coin) = P (A/E)

\frac{P\left( A \right) \times P\left( E/A \right)}{P\left( A \right) \times P\left( E/A \right) + P\left( B \right) \times P\left( E/B \right) + P\left( C \right) \times P\left( E/C \right)}

\frac{\left( \frac{1}{3} \times 1 \right)}{\left( \frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{3}{5} \right)}

\frac{\left( \frac{1}{3} \right)}{\left( \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \right)}

\frac{\left( \frac{1}{3} \right)}{\left( \frac{20 + 15 + 12}{60} \right)}

\frac{\left( \frac{1}{3} \right)}{\left( \frac{47}{60} \right)}

= 20/47

Question 16. In a factory, a machine A produces 30% of the total output, machine B produces 25%, and machine C produces the remaining output. If defective items produced by machines A, B, and C are 1%, 1.2%, 2% respectively. Three machines working together produce 10000 items in a day. An item is drawn at random from a day’s output and found to be defective. Find the probability that it was produced by machine B?

Solution:

Let us assume that the events are

A = the item is defective

E1 = machine A is chosen

E2 = machine B  is chosen

E3 = machine C is chosen

So, 

P(E1) = 30/100

P(E2) = 25/100

P(E3) = 45/100

Now,

P(A/E1) = 1/100

P(A/E2) = 1.2/1000

P(A/E3) = 2/100

By using Bayes’ theorem, the required probability is 

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) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{30}{100} \times \frac{1}{100}}{\frac{30}{100} \times + \times \frac{1}{100}\frac{25}{100}\frac{1 . 2}{100} + \frac{45}{100} \times \frac{2}{100}}

\frac{30}{30 + 30 + 90}

= 30/150

= 1/5

= 0.2

Question 17. A company has two plants to manufacture bicycles. The first plant manufactures 60% of the bicycles and the second plant 40%. Out of the 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.

Solution:

Let us assume that the events are

A = the cycle is of standard quality

E1 = plant I is chosen

E2 = plant II is chosen.

So, 

P(E1) = 60/100

P(E2) = 40/100

Now,

P(A/E1) = 80/100

P(A/E2) = 90/100

By using Bayes’ theorem, the required probability is 

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) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{40}{100} \times \frac{90}{100}}{\frac{60}{100} \times \frac{80}{100} + \frac{40}{100} \times \frac{90}{100}}

\frac{36}{48 + 36}

= 36/84

= 3/7

Question 18. Three urns A, B, and C contain 6 red and 4 white; 2 red and 6 white; and 1 red and 5 white balls respectively. An urn is chosen at random and a ball is drawn. If the ball drawn is found to be red, find the probability that the ball was drawn from urn A.

Solution:

Let us assume that the events are

A = the ball is red

E1 = urn A is chosen

E2 = urn B is chosen 

E2 = urn C is chosen.

So, P(E1) = 1/3

P(E2) = 1/3

P(E3) = 1/3

Now,

P(A/E1) = 6/10 = 3/5

P(A/E2) = 2/8 = 1/4

P(A/E3) = 1/6

By using Bayes’ theorem, the required probability is 

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) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{1}{3} \times \frac{3}{5}}{\frac{1}{3} \times \frac{3}{5} + \frac{1}{3} \times \frac{1}{4} + \frac{1}{3} \times \frac{1}{6}}

\frac{\frac{3}{5}}{\frac{3}{5} + \frac{1}{4} + \frac{1}{6}}

= 36/61

Question 19. In a group of 400 people, 160 are smokers and non-vegetarian, 100 are smokers and vegetarian and the remaining are non-smokers and vegetarian. The probabilities of getting a special chest disease are 35%, 20%, and 10% respectively. A person is chosen from the group at random and is found to be suffering from the disease. What is the probability that the selected person is a smoker and non-vegetarian?

Solution:

Let us assume that the events are

A = the person suffers from the disease

E= a smoker and a non-vegetarian

E= a smoker and a vegetarian

E3 = a non-smoker and a vegetarian

So, P(E1) = 160/400

P(E2) = 100/400

P(E3) = 140/400

Now,

P(A/E1) = 35/100

P(A/E2) = 20/100

P(A/E3) = 10/100

By using Bayes’ theorem, the required probability is 

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) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{160}{400} \times \frac{35}{100}}{\frac{160}{400} \times \frac{35}{100} + \frac{100}{400} \times \frac{20}{100} + \frac{140}{400} \times \frac{10}{100}}

\frac{560}{560 + 200 + 140}

= 560/900

= 28/45

Question 20. A factory has three machines A, B, and C, which produce 100, 200, and 300 items of a particular type daily. The machines produce 2%, 3% and 5% defective items respectively. One day when the production was over, an item was picked up randomly and it was found to be defective. Find the probability that it was produced by machine A.

Solution:

Let us assume that the events are

A = the item is defective

E= machine A is chosen

E2 = machine B is chosen 

E3 = machine C is chosen

So, P(E1) = 100/600

P(E2) = 200/600

P(E3) = 300/600

Now,

P(A/E1) = 2/100

P(A/E2) = 3/100

P(A/E3) = 5/100

By using Bayes’ theorem, the required probability is 

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) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{100}{600} \times \frac{2}{100}}{\frac{100}{600} \times \frac{2}{100} + \frac{200}{600} \times \frac{3}{100} + \frac{300}{600} \times \frac{5}{100}}

\frac{2}{2 + 6 + 15}

= 2/23

Question 21. A bag contains 1 white and 6 red balls, and a second bag contains 4 white and 3 red balls. One of the bags is picked up at random and a ball is randomly drawn from it and is found to be white in color. Find the probability that the drawn ball was from the first bag.

Solution:

Let us assume the events are 

A = the ball is white

E1 = bag I is chosen

E2 = bag II is chosen

So, P(E1) = 1/2

P(E2) = 1/2

Now,

P(A/E1) = 1/7

P(A/E2) = 4/7

By using Bayes’ theorem, the required probability is

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{1}{2} \times \frac{1}{7}}{\frac{1}{2} \times \frac{1}{7} + \frac{1}{2} \times \frac{4}{7}}

\frac{1}{1 + 4}

= 1/5

Question 22. In a certain college, 4% of boys and 1% of girls are taller than 1.75 meters. Furthermore, 60% of the students in the colleges are girls. A student selected at random from the college is found to be taller than 1.75 meters. Find the probability that the selected student is a girl.

Solution:

Let us assume the events are 

A = The height of the student is more than 1.75 m

E1 = The selected student is a girl 

E2 = The selected student is a boy

So, 

P(E1) = 60/100

P(E2) = 40/100

Now,

P(A/E1) = 1/100

P(A/E1) = 4/100

By using Bayes’ theorem, the required probability is

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{1}{100}}{\frac{60}{100} \times \frac{1}{100} + \frac{40}{100} \times \frac{4}{100}}

\frac{6}{6 + 16}

= 6/22

= 3/11

Question 23. For A, B, and C the chances of being selected as the manager of a firm are in the ratio 4:1:2 respectively. The respective probabilities for them to introduce a radical change in marketing strategy are 0.3, 0.8, and 0.5. If the change does take place, find the probability that it is due to the appointment of B or C.

Solution:

Let us assume the events are 

A = The change takes place

E= A is selected

E2 = B is selected 

E3 = C is selected

So, 

P(E1) = 4/7

P(E2) = 1/7

P(E3) = 2/7

Now,

P(A/E1) = 0.3

P(A/E2) = 0.8

P(A/E3) = 0.5

By using Bayes’ theorem, the required probability is

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) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{4}{7} \times 0 . 3}{\frac{4}{7} \times 0 . 3 + \frac{1}{7} \times 0 . 8 + \frac{2}{7} \times 0 . 5}

\frac{1 . 2}{1 . 2 + 0 . 8 + 1}

= 1.2/3

= 2/5

1 – P(E1/A) = 1 – 2/5 = 3/5

Question 24. 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 introduce changes to improve the 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.

Solution:

Let us assume the events are 

E= The selection of A as manager

E= The selection of B as manager

E= The selection of C as manager

So, 

P(E1) = The probability of selection of A = 1/7

P(E2) = The probability of selection of B = 2/7

P(E3) = The probability of selection of C = 4/7

Let us assume that A be the event representing the change not taking place.

P(A/E1) = Probability that A does not introduce change = 0.2

P(A/E2) = Probability that B does not introduce change = 0.5

P(A/E3) = Probability that C does not introduce change = 0.7

So,  the required probability = P(A/E3)

By using Bayes’ theorem, the required probability is

P(A/E3) = \frac{P\left( E_3 \right)P\left( A/ E_3 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)}

\frac{\frac{4}{7} \times 0 . 7}{\frac{1}{7} \times 0 . 2 + \frac{2}{7} \times 0 . 5 + \frac{4}{7} \times 0 . 7}

\frac{2 . 8}{0 . 2 + 1 + 2 . 8}

= 2.8/4

= 0.7

Question 25. An insurance company insured 2000 scooters and 3000 motorcycles. The probability of an accident involving a scooter is 0.01 and that of a motorcycle is 0.02. An insured vehicle met with an accident. Find the probability that the accident vehicle was a motorcycle.

Solution:

Let us assume the events are 

A = The vehicle meets the accident

E1 = is a scooter 

E2 = is a motorcycle

So, 

P(E1) = 2000/5000 = 0.4

P(E2) = 3000/5000 = 0.6

Now,

P(A/E1) = 0.01

P(A/E2) = 0.02

By using Bayes’ theorem, the required probability is

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 . 6 \times 0 . 02}{0 . 6 \times 0 . 02 + 0 . 4 \times 0 . 01}

\frac{0 . 012}{0 . 012 + 0 . 004}

= 0.012/0.016

= 3/4

Question 26. Of the students in a college, it is known that 60% reside in a hostel and 40% do not reside in a hostel. Previous year results report that 30% of students residing in hostel attain A grade and 20% of ones not residing in hostel attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade. What is the probability that the selected student is a hosteler?

Solution:

Let us assume the events are 

A = The selected student attains grade A

E1 = resides in a hostel 

E2 = does not reside in a hostel

So, P(E1) = 60/100

P(E2) = 40/100

Now,

P(A/E1) = 30/100

P(A/E2) = 20/100

By using Bayes’ theorem, the required probability is

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{30}{100}}{\frac{60}{100} \times \frac{30}{100} + \frac{40}{100} \times \frac{20}{100}}

\frac{18}{18 + 8}

= 9/13

Question 27. 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, it shows heads. What is the probability that it was the two-headed coin?
Solution:

Let us assume that the events are
E= choosing a two-headed coin
E= choosing a biased coin 
E3 = choosing an unbiased coin
A = the coin shows heads.
So, 
P(E1) = 1/3
P(E2) = 1/3
P(E3) = 1/3
Now,
P(A/E1) = 1
P(A/E2) = 75% = 3/4
P(A/E3) = 1/2
By using Bayes’ theorem, the required probability is 
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) + P\left( E_3 \right)P\left( A/ E_3 \right)}
\frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{2}}
= 4/9
Question 28. Assume that the chances of a patient having a heart attack is 40%. It is also assumed that meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options and patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
Solution:
Let us assume that the events are
E= follow the course of yoga and meditation
E2 = follow the drug prescriptions
A = the selected person had heart attacak
So, 
P(E1) = 1/2
P(E2) = 1/2
Now,
P(A/E1) = 0.4 x 0.70 = 0.28
P(A/E2) = 0.4 x 0.75 = 0.30
By using Bayes’ theorem, the required probability is 
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{1}{2} \times 0.28}{\frac{1}{2} \times 0.28 + \frac{1}{2} \times 0.30}
= 14/29
Question 29. Colored balls are distributed in four boxes as shown in the following table:
Box
Colour
 
Black
White
Red
Blue
I
3
4
5
6
II
2
2
2
2
III
1
2
3
1
IV
4
3
1
5
A box is selected at random and then a ball is randomly drawn from the selected box. The color of the ball is black, what is the probability that the ball drawn is from box III.
Solution:
Let us assume that the events are
A = the ball is black
E= box I selected
E= box II selected
E3 = box III is selected 
E4 = box IV is selected
So, 
P(E1) = 1/4
P(E2) = 1/4
P(E3) = 1/4
P(E4) = 1/4
Now,
P(A/E1) = 3/18
P(A/E2) = 2/8
P(A/E3) = 1/7
P(A/E4) = 4/13
By using Bayes’ theorem, the required probability is 
P(E3/A) = \frac{P\left( E_3 \right)P\left( A/ E_3 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)+P\left( E_3 \right)P\left( A/ E_3 \right)+P\left( E_4 \right)P\left( A/ E_4 \right)}
\frac{\frac{1}{4} \times \frac{1}{7}}{\frac{1}{4} \times \frac{3}{18} + \frac{1}{4} \times \frac{2}{8} + \frac{1}{4} \times \frac{1}{7} + \frac{1}{4} \times \frac{4}{13}}
\frac{\frac{1}{7}}{\frac{1}{6} + \frac{1}{4} + \frac{1}{7} + \frac{4}{13}}
= 156/947
Question 30. If a machine is correctly set up it produces 90% acceptable items. If it is incorrectly set up it produces only 40% acceptable item. Past experience shows that 80% of the setups are correctly done. If after a certain setup, the machine produces 2 acceptable items, find the probability that the machine is correctly set up.
Solution:
Let us assume that the events are
A = the machine produces two acceptable items.
E1 = the machine is correctly set up 
E2 = the machine is incorrectly set up
So, 
P(E1) = 0.8
P(E2) = 0.2
Now,
P(A/E1) = 0.9 (0.9) = 0.81
P(A/E2) = 0.40 (0.40) = 0.16
By using Bayes’ theorem, the required probability is 
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 . 8 \times 0 . 81}{0 . 8 \times 0 . 81 + 0 . 2 \times 0 . 16}
= 81/85
Question 31. Bag A contains 3 red and 5 black balls, while bag B contains 4 red and 4 black balls. Two balls are transferred at random from bag A to bag B and then a ball is drawn from bag B at random. If the ball drawn from bag B is found to be red find the probability that two red balls were transferred from A to B.
Solution:
According to the question, bag A contains 3 red and 5 black balls and bag B contains 4 red and 4 black balls.
Let us assume that the events are
E1 = Two red balls are transferred from bag A to bag B.
E2 = One red ball and one black ball is transferred from bag A to bag B.
E3 = Two black balls are transferred from bag A to bag B.
A = Ball drawn from bag B is red.
So,
P(E1) = \frac{{^3}{}{C}_2}{{^8}{}{C}_2}   = 3/28
P(E2) = \frac{{^3}{}{C}_1 \times {^5}{}{C}_1}{^{8}{}{C}_2}   = 15/28
P(E3) = \frac{^{5}{}{C}_2}{^{8}{}{C}_2}   = 10/28
Also,
P(A/E1) = 6/10
P(A/E2) = 5/10
P(A/E3) = 4/10
P(E1/A) is the required probability, that two red balls were transferred from A to B given that the ball drawn from bag B is red  
So, by using Bayes’ theorem, the required probability is 
\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)+P\left( E_3 \right)P\left( A/ E_3 \right)}
\frac{\frac{3}{28} \times \frac{6}{10}}{\frac{3}{28} \times \frac{6}{10} + \frac{15}{28} \times \frac{5}{10} + \frac{10}{28} \times \frac{4}{10}}
\frac{18}{18 + 75 + 40}
= 18/133
Question 32. Let d1, d2, d3 be three mutually exclusive diseases. Let S be the set of observable symptoms of these diseases. A doctor has the following information from a random sample of 5000 patients: 1800 had disease d1, 2100 has disease d2 and the others had disease d3. 1500 patients with disease d1, 1200 patients with disease d2 and 900 patients with disease d3 showed the symptoms. Which of the diseases is the patient most likely to have?
Solution:
Let us assume that the events are
A = the patient shows symptoms S
E= has disease d1
E2 = has disease​ d2
E3 = has disease​ d3
So, 
P(E1) = 1800/5000
P(E2) = 2100/5000
P(E3) = 1100/5000
Now,
P(A/E1) =1500/1800
P(A/E2) = 1200/2100
P(A/E3) = 900/1100
By using Bayes’ theorem, the required probabilities are
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) + P\left( E_3 \right)P\left( A/ E_3 \right)}
\frac{\frac{1800}{5000} \times \frac{1500}{1800}}{\frac{1800}{5000} \times \frac{1500}{1800} + \frac{2100}{5000} \times \frac{1200}{2100} + \frac{1100}{5000} \times \frac{900}{1100}}
\frac{15}{15 + 12 + 9}
= 15/36
= 5/12
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) + P\left( E_3 \right)P\left( A/ E_3 \right)}
\frac{\frac{2100}{5000} \times \frac{1200}{2100}}{\frac{1800}{5000} \times \frac{1500}{1800} + \frac{2100}{5000} \times \frac{1200}{2100} + \frac{1100}{5000} \times \frac{900}{1100}}
\frac{12}{15 + 12 + 9}
= 12/36
= 1/3
P(E3/A) = \frac{P\left( E_3 \right)P\left( A/ E_3 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)}
\frac{\frac{1100}{5000} \times \frac{900}{1100}}{\frac{1800}{5000} \times \frac{1500}{1800} + \frac{2100}{5000} \times \frac{1200}{2100} + \frac{1100}{5000} \times \frac{900}{1100}}
\frac{9}{15 + 12 + 9}
= 9/36
= 1/4
As P(E1/A ) is maximum, so it is most likely that the person have ddisease.
Question 33. A test for the detection of a particular disease is not foolproof. The test will correctly detect the disease 90% of the time, but will incorrectly detect the disease 1% of the time. For a large population of which an estimated 0.2% have the disease, a person is selected at random, given the test, and told that he has the disease. What are the chances that the person actually has the disease?
Solution:
Let us assume that the events are
A = the person suffers from the disease
E1 = the test detects the disease correctly 
E2 = the test does not detect the disease correctly
So, 
P(E1) = 90/100
P(E2) = 1/100
Now,
P(A/E1) = 2/1000
P(A/E2) = 998/1000
By using Bayes’ theorem, the required probability is
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{90}{100} \times \frac{2}{1000}}{\frac{90}{100} \times \frac{2}{1000} + \frac{1}{100} \times \frac{998}{1000}}
\frac{180}{180 + 998}
= 180/1178
= 90/589
Question 34. Let d1, d2, d3, d4 be three mutually exclusive diseases. Let S be the set of observable symptoms of these diseases. A doctor has the following information from a random sample of 5000 patients: 1800 had disease d1, 2100 has disease d2, and others had disease d3. 1500 patients with disease d1, 1200 patients with disease d2, and 900 patients with disease d3 showed the symptoms. Which of the diseases is the patient most likely to have?
Solution:
Let us assume that the events are
E= The patient had disease d1
E= The patient had disease d2
E= The patient had disease d3
S = The patient showed the symptom
Also, E1, E2, and E3 are mutually exclusive and exhaustive events.
So, 
P(E1) = 1800/1500 = 18/50
P(E2) = 2100/5000 = 21/50
P(E3) = 1100/5000 = 11/50
Now,  
P(S/E1) = The probability that the patient had disease dand showed symptoms S = 1500/5000 = 15/50
P(S/E2) = The probability that the patient had disease dand showed symptoms S = 1200/5000 = 12/50
P(S/E3) = The probability that the patient had disease dand showed symptoms S = 900/5000 = 9/50
By using Bayes’ theorem,
P(E1/S) is the probability that patient had disease d1 such that symptoms of d1 appears
\frac{P\left( E_1 \right)P\left( \frac{S}{E_1} \right)}{P\left( E_1 \right)P\left( \frac{S}{E_1} \right) + P\left( E_2 \right)P\left( \frac{S}{E_2} \right) + P\left( E_3 \right)P\left( \frac{S}{E_3} \right)}
\frac{\frac{18}{50} \times \frac{15}{50}}{\frac{18}{50} \times \frac{15}{50} + \frac{21}{50} \times \frac{12}{50} + \frac{11}{50} \times \frac{9}{50}}
= 270/621
P(E2/S) is the probability that patient had disease d2 such that symptom of d2 appears
\frac{P\left( E_2 \right)P\left( \frac{S}{E_2} \right)}{P\left( E_1 \right)P\left( \frac{S}{E_1} \right) + P\left( E_2 \right)P\left( \frac{S}{E_2} \right) + P\left( E_3 \right)P\left( \frac{S}{E_3} \right)}
\frac{\frac{21}{50} \times \frac{12}{50}}{\frac{18}{50} \times \frac{15}{50} + \frac{21}{50} \times \frac{12}{50} + \frac{11}{50} \times \frac{9}{50}}
= 252/621
P(E3/S) is the probability that patient had disease dsuch that symptom of d3 appears
\frac{P\left( E_3 \right)P\left( \frac{S}{E_3} \right)}{P\left( E_1 \right)P\left( \frac{S}{E_1} \right) + P\left( E_2 \right)P\left( \frac{S}{E_2} \right) + P\left( E_3 \right)P\left( \frac{S}{E_3} \right)}
\frac{\frac{11}{50} \times \frac{9}{50}}{\frac{18}{50} \times \frac{15}{50} + \frac{21}{50} \times \frac{12}{50} + \frac{11}{50} \times \frac{9}{50}}
= 99/621
Hence, P(E1/A) is the maximum, then the patient is most likely to have the disease d1.
Question 35. A is known to speak the truth 3 times out of 5 times. He throws a die and reports that it is one. Find the probability that it is actually one.
Solution:
Let us assume that the events are
A = the man reports the appearance of 1 on throwing a die
E1 = 1 occurs 
E2 = 1 does not occur
So, 
P(E1) = 1/6
P(E2) = 5/6
Now,
P(A/E1) = 3/5
P(A/E2) = 2/5
By using Bayes’ theorem, the required probability is
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{1}{6} \times \frac{3}{5}}{\frac{1}{6} \times \frac{3}{5} + \frac{5}{6} \times \frac{2}{5}}
\frac{3}{3 + 10}
= 3/13
Question 36. A speaks the truth 8 times out of 10 times. A die is tossed. He reports that it was 5. What is the probability that it was actually 5?
Solution:
Let us assume that A be the event that man reports that 5 occurs and E the event that 5 actually turns up.
So, 
P(E) = 1/6
P(\overline{E})    = 1 – 1/6 = 5/6
Also,  
P(A/E)  = Probability that man reports that 5 occurs given that 5 actually turns up 
= Probability of man speaking the truth 
= 8/10
= 4/5
P\left(A/\overline{E}\right)   = Probability that man reports that 5 occurs given that 5 does not turns up 
= Probability of man not speaking the truth 
=  1 – 4/5
= 1/5
Therefore, the required probability = P(E/A)
\frac{P\left( E \right)P\left( \frac{A}{E} \right)}{P\left( E \right)P\left( \frac{A}{E} \right) + P\left(\overline{ (E) } \right)P\left( \frac{A}{\overline{(E)}} \right)}
\frac{\frac{1}{6} \times \frac{4}{5}}{\frac{1}{6} \times \frac{4}{5} + \frac{5}{6} \times \frac{1}{5}}
= 4/9
Question 37. In answering a question on a multiple-choice test a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with a probability of 1/4. What is the probability that a student knows the answer given that he answered it correctly?
Solution:
Let us assume that the events are
A = the answer is correct
E1 = the student knows the answer
E2 = the student guesses the answer
So, 
P(E1) = 3/4
P(E2) = 1/4
Now, 
P(A/E1) = 1
P(A/E2) = 1/4
By using Bayes’ theorem, the required probability is
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{3}{4} \times 1}{\frac{3}{4} \times 1 + \frac{1}{4} \times \frac{1}{4}}
= 3/3 + 1/4
= 12/13
Question 38. A laboratory blood test is 99% effective in detecting a certain disease when its infection is present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1% of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Solution:
Let us assume that E1 and E2 be the events that a person has a disease and a person has no disease.
Also, E1 and E2 are complimentary to each other.
So, P(E1) + P(E2) = 1
P(E1) = 0.001
=>  P (E2) = 1 − P (E1) = 1 − 0.001 = 0.999
Let us assume that A be the event that the blood test result is positive.
Now,
P(A/E1) = 99% = 0.99
P(A/E2) = 0.5 % = 0.005
By using Bayes’ theorem, the required probability is
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 . 001 \times 0 . 99}{0 . 001 \times 0 . 99 + 0 . 999 \times 0 . 005}
= 990/5985
= 22/133
Question 39. There are three categories of students in a class of 60 students:
A: Very hardworking; B: Regular but not so hardworking; C: Careless and irregular 10 students are in category A, 30 in category B, and the rest in category C. It is found that the probability of students of category A, unable to get good marks in the final year examination is 0.002, of category B it is 0.02 and of category C, this probability is 0.20. A student selected at random was found to be one who could not get good marks in the examination. Find the probability that this student is category C.
Solution:
Let us assume that the events are
E = the student could not get good marks in the examination.
A = student is very hardworking
B = student is regular but not so hardworking
C = student is careless and irregular
Here, we have
P(A) = 10/60
P(B) =  30/60
P(C) = 20/60
Also,
P(E/A)  = Probability that category A student could not get good marks in the examination = 0.002  
P(E/B) = Probability that a category B student could not get good marks in the examination = 0.02
P(E/C)  = Probability that a category C student could not get good marks in the examination = 0.2
So, P(C/E) is the required probability 
\frac{P\left( C \right)P\left( \frac{E}{C} \right)}{P\left( A \right)P\left( \frac{E}{A} \right) + P\left( B \right)P\left( \frac{E}{B} \right) + P\left( C \right)P\left( \frac{E}{C} \right)}
\frac{\frac{20}{60} \times 0 . 2}{\frac{10}{60} \times 0 . 002 + \frac{30}{60} \times 0 . 02 + \frac{20}{60} \times 0 . 2}
= 4/4.62
= 400/462
= 200/231

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Leave a Comment

Your email address will not be published.