Question 1:
The contents of urns I, II, III are as follows:
Urn I : \displaystyle 1 white, \displaystyle 2 black and \displaystyle 3 red balls
Urn II : \displaystyle 2 white, \displaystyle 1 black and \displaystyle 1 red balls
Urn III : \displaystyle 4 white, \displaystyle 5 black and \displaystyle 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?
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1,E_2 \text{ and } E_3 \text{ denote the events of selecting Urn I, Urn II and Urn III, respectively}
\displaystyle \text{Let } A \text{ be the event that the two balls drawn are white and red}
\displaystyle \therefore P(E_1)=\frac{1}{3}
\displaystyle P(E_2)=\frac{1}{3}
\displaystyle P(E_3)=\frac{1}{3}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{\binom{1}{1}\binom{3}{1}}{\binom{6}{2}}=\frac{3}{15}=\frac{1}{5}
\displaystyle P(A\mid E_2)=\frac{\binom{2}{1}\binom{1}{1}}{\binom{4}{2}}=\frac{2}{6}=\frac{1}{3}
\displaystyle P(A\mid E_3)=\frac{\binom{4}{1}\binom{3}{1}}{\binom{12}{2}}=\frac{12}{66}=\frac{2}{11}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{\frac{1}{5}}{\frac{1}{5}+\frac{1}{3}+\frac{2}{11}}
\displaystyle =\frac{\frac{33}{165}}{\frac{33+55+30}{165}}=\frac{33}{118}
\displaystyle \text{Required probability}=P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{\frac{1}{3}}{\frac{1}{5}+\frac{1}{3}+\frac{2}{11}}
\displaystyle =\frac{\frac{55}{165}}{\frac{33+55+30}{165}}=\frac{55}{118}
\displaystyle \text{Required probability}=P(E_3\mid A)=\frac{P(E_3)P(A\mid E_3)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{\frac{2}{11}}{\frac{1}{5}+\frac{1}{3}+\frac{2}{11}}
\displaystyle =\frac{\frac{30}{165}}{\frac{33+55+30}{165}}=\frac{30}{118}

Question 2:
A bag A contains \displaystyle 2 white and \displaystyle 3 red balls and a bag B contains \displaystyle 4 white and \displaystyle 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.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1 \text{ and } E_2 \text{ denote the events that the ball is red, bag A is chosen and bag B is} \\ \text{chosen, respectively}
\displaystyle \therefore P(E_1)=\frac{1}{2}
\displaystyle P(E_2)=\frac{1}{2}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{3}{5}
\displaystyle P(A\mid E_2)=\frac{5}{9}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{1}{2}\times\frac{5}{9}}{\frac{1}{2}\times\frac{3}{5}+\frac{1}{2}\times\frac{5}{9}}
\displaystyle =\frac{25}{52}

Question 3:
Three urns contain \displaystyle 2 white and \displaystyle 3 black balls; \displaystyle 3 white and \displaystyle 2 black balls and \displaystyle 4 white and \displaystyle 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.
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1,E_2 \text{ and } E_3 \text{ denote the events of selecting Urn I, Urn II and Urn III, respectively}
\displaystyle \text{Let } A \text{ be the event that the ball drawn is white}
\displaystyle \therefore P(E_1)=\frac{1}{3}
\displaystyle P(E_2)=\frac{1}{3}
\displaystyle P(E_3)=\frac{1}{3}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{2}{5}
\displaystyle P(A\mid E_2)=\frac{3}{5}
\displaystyle P(A\mid E_3)=\frac{4}{5}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{2}{2+3+4}=\frac{2}{9}

Question 4:
The contents of three urns are as follows:
Urn \displaystyle 1:7 white, \displaystyle 3 black balls, Urn \displaystyle 2:4 white, \displaystyle 6 black balls, and Urn \displaystyle 3:2 white, \displaystyle 8 black balls.
One of these urns is chosen at random with probabilities \displaystyle 0.20,0.60 and \displaystyle 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 \displaystyle 3?
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1,E_2 \text{ and } E_3 \text{ denote the events of selecting Urn I, Urn II and Urn III, respectively}
\displaystyle \text{Let } A \text{ be the event that the two balls drawn are white}
\displaystyle \therefore P(E_1)=\frac{20}{100}
\displaystyle P(E_2)=\frac{60}{100}
\displaystyle P(E_3)=\frac{20}{100}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{\binom{7}{2}}{\binom{10}{2}}=\frac{21}{45}
\displaystyle P(A\mid E_2)=\frac{\binom{4}{2}}{\binom{10}{2}}=\frac{6}{45}
\displaystyle P(A\mid E_3)=\frac{\binom{2}{2}}{\binom{10}{2}}=\frac{1}{45}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_3\mid A)=\frac{P(E_3)P(A\mid E_3)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{1}{21+18+1}=\frac{1}{40}

Question 5:
Suppose a girl throws a die. If she gets \displaystyle 1 or \displaystyle 2, she tosses a coin three times and notes the number of tails. If she gets \displaystyle 3,4,5 or \displaystyle 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw \displaystyle 3,4,5 or \displaystyle 6 with the die?
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1 \text{ be the event that the outcome on the die is }1\text{ or }2 \text{ and } E_2 \text{ be the} \\ \text{event that the outcome on the die is }3,4,5 \text{ or }6
\displaystyle P(E_1)=\frac{2}{6}=\frac{1}{3} \text{ and } P(E_2)=\frac{4}{6}=\frac{2}{3}
\displaystyle \text{Let } A \text{ be the event of getting exactly one tail}
\displaystyle P(A\mid E_1)=\text{Probability of getting exactly one tail by tossing the coin three times if she gets }1 \\ \text{ or }2=\frac{3}{8}
\displaystyle P(A\mid E_2)=\text{Probability of getting exactly one tail in a single throw of a coin if she gets }3,4,5 \\ \text{ or }6=\frac{1}{2}
\displaystyle \text{As the probability that the girl threw }3,4,5 \text{ or }6 \text{ with the die, if she obtained exactly one} \\ \text{tail, is given by } P(E_2\mid A)
\displaystyle \text{So, by using Bayes' theorem, we get}
\displaystyle P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{2}{3}\times\frac{1}{2}}{\frac{1}{3}\times\frac{3}{8}+\frac{2}{3}\times\frac{1}{2}}
\displaystyle =\frac{\frac{2}{6}}{\frac{1}{8}+\frac{1}{3}}
\displaystyle =\frac{\frac{2}{6}}{\frac{11}{24}}
\displaystyle =\frac{24\times2}{11\times6}
\displaystyle =\frac{8}{11}
\displaystyle \text{So, the probability that she threw }3,4,5 \text{ or }6 \text{ with the die if she obtained exactly one tail is } \frac{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 \displaystyle 0.6 and \displaystyle 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is \displaystyle 0.7 and the corresponding probability is \displaystyle 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1 \text{ and } E_2 \text{ denote the events that the first group and the second group win the competition,} \\ \text{respectively}
\displaystyle \text{Let } A \text{ be the event of introducing a new product}
\displaystyle P(E_1)=\text{Probability that the first group wins the competition}=0.6
\displaystyle P(E_2)=\text{Probability that the second group wins the competition}=0.4
\displaystyle P(A\mid E_1)=\text{Probability of introducing a new product if the first group wins}=0.7
\displaystyle P(A\mid E_2)=\text{Probability of introducing a new product if the second group wins}=0.3
\displaystyle \text{The probability that the new product is introduced by the second group is given by } P(E_2\mid A)
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{0.4\times0.3}{0.6\times0.7+0.4\times0.3}
\displaystyle =\frac{0.12}{0.54}=\frac{2}{9}

Question 7:
Suppose \displaystyle 5 men out of \displaystyle 100 and \displaystyle 25 women out of \displaystyle 1000 are good orators. An orator is chosen at random. Find the probability that a male person is selected. Assume that there are equal number of men and women.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1 \text{ and } E_2 \text{ denote the events that the person is a good orator, is a man and is a woman,} \\ \text{respectively}
\displaystyle \therefore P(E_1)=\frac{1}{2}
\displaystyle P(E_2)=\frac{1}{2}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{5}{100}
\displaystyle P(A\mid E_2)=\frac{25}{1000}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{1}{2}\times\frac{5}{100}}{\frac{1}{2}\times\frac{5}{100}+\frac{1}{2}\times\frac{25}{1000}}
\displaystyle =\frac{1}{1+\frac{1}{2}}=\frac{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?
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1 \text{ and } E_2 \text{ denote the events that the two consecutive letters are visible,} \\ \text{the letter has come from LONDON and the letter has come from CLIFTON, respectively}
\displaystyle \therefore P(E_1)=\frac{1}{2}
\displaystyle P(E_2)=\frac{1}{2}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{2}{5}
\displaystyle P(A\mid E_2)=\frac{1}{6}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{(i) Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{1}{2}\times\frac{2}{5}}{\frac{1}{2}\times\frac{2}{5}+\frac{1}{2}\times\frac{1}{6}}
\displaystyle =\frac{\frac{2}{5}}{\frac{2}{5}+\frac{1}{6}}=\frac{12}{17}
\displaystyle \text{(ii) Required probability}=P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{1}{2}\times\frac{1}{6}}{\frac{1}{2}\times\frac{2}{5}+\frac{1}{2}\times\frac{1}{6}}
\displaystyle =\frac{\frac{1}{6}}{\frac{2}{5}+\frac{1}{6}}=\frac{5}{17}

Question 9:
In a class, \displaystyle 5\% of the boys and \displaystyle 10\% of the girls have an IQ of more than \displaystyle 150. In this class, \displaystyle 60\% of the students are boys. If a student is selected at random and found to have an IQ of more than \displaystyle 150, find the probability that the student is a boy.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1 \text{ and } E_2 \text{ denote the events that the IQ is more than }150,\text{ the selected} \\ \text{student is a boy and the selected student is a girl, respectively}
\displaystyle \therefore P(E_1)=\frac{60}{100}
\displaystyle P(E_2)=\frac{40}{100}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{5}{100}
\displaystyle P(A\mid E_2)=\frac{10}{100}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{60}{100}\times\frac{5}{100}}{\frac{60}{100}\times\frac{5}{100}+\frac{40}{100}\times\frac{10}{100}}
\displaystyle =\frac{300}{300+400}=\frac{300}{700}=\frac{3}{7}

Question 10:
A factory has three machines X, Y and Z producing \displaystyle 1000,2000 and \displaystyle 3000 bolts per day respectively. The machine X produces \displaystyle 1\% defective bolts, Y produces \displaystyle 1.5\% and Z produces \displaystyle 2\% defective bolts. At the end of a 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?
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1,E_2 \text{ and } E_3 \text{ denote the events that machine X produces bolts, machine} \\ \text{Y produces bolts and machine Z produces bolts, respectively}
\displaystyle \text{Let } A \text{ be the event that the bolt is defective}
\displaystyle \text{Total number of bolts}=1000+2000+3000=6000
\displaystyle P(E_1)=\frac{1000}{6000}=\frac{1}{6}
\displaystyle P(E_2)=\frac{2000}{6000}=\frac{1}{3}
\displaystyle P(E_3)=\frac{3000}{6000}=\frac{1}{2}
\displaystyle \text{The probability that the defective bolt is produced by machine X is given by }P(E_1\mid A)
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=1\%=\frac{1}{100}
\displaystyle P(A\mid E_2)=1.5\%=\frac{15}{1000}
\displaystyle P(A\mid E_3)=2\%=\frac{2}{100}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{2}+1}=\frac{\frac{1}{6}}{\frac{1+3+6}{6}}
\displaystyle =\frac{1}{10}

Question 11:
An insurance company insured \displaystyle 3000 scooters, \displaystyle 4000 cars and \displaystyle 5000 trucks. The probabilities of the accident involving a scooter, a car and a truck are \displaystyle 0.02,0.03 and \displaystyle 0.04 respectively. One of the insured vehicles meet with an accident. Find the probability that it is a
(i) scooter
(ii) car
(iii) truck.
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1,E_2 \text{ and } E_3 \text{ denote the events that the vehicle is a scooter, a car and a truck, respectively}
\displaystyle \text{Let } A \text{ be the event that the vehicle meets with an accident}
\displaystyle \text{It is given that there are }3000\text{ scooters, }4000\text{ cars and }5000\text{ trucks}
\displaystyle \text{Total number of vehicles}=3000+4000+5000=12000
\displaystyle P(E_1)=\frac{3000}{12000}=\frac{1}{4}
\displaystyle P(E_2)=\frac{4000}{12000}=\frac{1}{3}
\displaystyle P(E_3)=\frac{5000}{12000}=\frac{5}{12}
\displaystyle \text{The probability that the vehicle which meets with an accident is a scooter is given by }P(E_1\mid A)
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=0.02=\frac{2}{100}
\displaystyle P(A\mid E_2)=0.03=\frac{3}{100}
\displaystyle P(A\mid E_3)=0.04=\frac{4}{100}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{(i) Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{\frac{1}{2}}{\frac{1}{2}+1+\frac{5}{3}}=\frac{\frac{1}{2}}{\frac{3+6+10}{6}}=\frac{3}{19}
\displaystyle \text{(ii) Required probability}=P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{1}{\frac{3+6+10}{6}}=\frac{6}{19}
\displaystyle \text{(iii) Required probability}=P(E_3\mid A)=\frac{P(E_3)P(A\mid E_3)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{\frac{5}{3}}{\frac{3+6+10}{6}}=\frac{10}{19}

Question 12:
Suppose we have four boxes A, B, C, D containing coloured marbles as given below:
Marble Colour
Box A : Red \displaystyle 1, White \displaystyle 6, Black \displaystyle 3
Box B : Red \displaystyle 6, White \displaystyle 2, Black \displaystyle 2
Box C : Red \displaystyle 8, White \displaystyle 1, Black \displaystyle 1
Box D : Red \displaystyle 0, White \displaystyle 6, Black \displaystyle 4
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? box C?
\displaystyle \text{Answer:}
\displaystyle \text{Let } R \text{ be the event of drawing the red marble}
\displaystyle \text{Let } E_A,E_B \text{ and } E_C \text{ denote the events of selecting box A, box B and box C, respectively}
\displaystyle \text{Total number of marbles}=40
\displaystyle \text{Number of red marbles}=15
\displaystyle \therefore P(R)=\frac{15}{40}=\frac{3}{8}
\displaystyle \text{Probability of drawing a red marble from box A is given by }P(E_A\mid R)
\displaystyle \therefore P(E_A\mid R)=\frac{P(E_A\cap R)}{P(R)}=\frac{\frac{1}{40}}{\frac{3}{8}}=\frac{1}{15}
\displaystyle \text{Probability of drawing a red marble from box B is given by }P(E_B\mid R)
\displaystyle \therefore P(E_B\mid R)=\frac{P(E_B\cap R)}{P(R)}=\frac{\frac{6}{40}}{\frac{3}{8}}=\frac{2}{5}
\displaystyle \text{Probability of drawing a red marble from box C is given by }P(E_C\mid R)
\displaystyle \therefore P(E_C\mid R)=\frac{P(E_C\cap R)}{P(R)}=\frac{\frac{8}{40}}{\frac{3}{8}}=\frac{8}{15}

Question 13:
A manufacturer has three machine operators A, B and C. The first operator A produces \displaystyle 1\% defective items, whereas the other two operators B and C produce \displaystyle 5\% and \displaystyle 7\% defective items respectively. A is on the job for \displaystyle 50\% of the time, B on the job for \displaystyle 30\% of the time and C on the job for \displaystyle 20\% of the time. A defective item is produced. What is the probability that it was produced by A?
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1,E_2 \text{ and } E_3 \text{ be the time taken by machine operators A, B and C, respectively}
\displaystyle \text{Let } X \text{ be the event of producing defective items}
\displaystyle \therefore P(E_1)=50
\displaystyle P(E_2)=30
\displaystyle P(E_3)=20
\displaystyle \text{Now,}
\displaystyle P(X\mid E_1)=1
\displaystyle P(X\mid E_2)=5
\displaystyle P(X\mid E_3)=7
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid X)=\frac{P(E_1)P(X\mid E_1)}{P(E_1)P(X\mid E_1)+P(E_2)P(X\mid E_2)+P(E_3)P(X\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{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, \displaystyle 50\% are manufactured on machine A, \displaystyle 30\% on B and \displaystyle 20\% on C. \displaystyle 2\% of the items produced on A and \displaystyle 2\% of items produced on B are defective and \displaystyle 3\% of these produced on C are defective. All the items 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?
\displaystyle \text{Answer:}
\displaystyle \text{Let } E \text{ be the event of getting a defective item}
\displaystyle \text{We have,}
\displaystyle P(A)=50
\displaystyle P(E\mid A)=2
\displaystyle \text{Now,}
\displaystyle P(\text{the defective item drawn was mfg. on machine A})=\frac{P(A)\times P(E\mid A)}{P(A)\times P(E\mid A)+P(B)\times P(E\mid B)+P(C)\times P(E\mid C)}
\displaystyle =\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}}
\displaystyle =\frac{\frac{1}{100}}{\frac{1}{100}+\frac{3}{500}+\frac{3}{500}}
\displaystyle =\frac{\frac{1}{100}}{\frac{5+3+3}{500}}
\displaystyle =\frac{\frac{1}{100}}{\frac{11}{500}}
\displaystyle =\frac{500}{100\times11}
\displaystyle \text{So, the probability that the defective item drawn was manufactured on machine A is }\frac{5}{11}

Question 15:
There are three coins. One is two-headed coin (having head on both faces), another is biased coin that comes up heads \displaystyle 75\% of the times and third is also a biased coin that comes up tail \displaystyle 40\% of the times. 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?
\displaystyle \text{Answer:}
\displaystyle \text{Let :}
\displaystyle A \text{ be the event of choosing a two-headed coin}
\displaystyle B \text{ be the event of choosing a biased coin that comes up head }75\%
\displaystyle C \text{ be the event of choosing a biased coin that comes up tail }40\%
\displaystyle E \text{ be the event of getting a head}
\displaystyle \text{Now,}
\displaystyle P(A)=P(B)=P(C)=\frac{1}{3} \text{ and }
\displaystyle P(E\mid A)=1,\;P(E\mid B)=\frac{3}{4},\;P(E\mid C)=\frac{3}{5}
\displaystyle \text{So, using Bayes' theorem, we get}
\displaystyle P(\text{the head shown was of a two-headed coin})=P(A\mid E)
\displaystyle =\frac{P(A)\times P(E\mid A)}{P(A)\times P(E\mid A)+P(B)\times P(E\mid B)+P(C)\times P(E\mid C)}
\displaystyle =\frac{\frac{1}{3}\times1}{\frac{1}{3}\times1+\frac{1}{3}\times\frac{3}{4}+\frac{1}{3}\times\frac{3}{5}}
\displaystyle =\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{4}+\frac{1}{5}}
\displaystyle =\frac{\frac{1}{3}}{\frac{20+15+12}{60}}
\displaystyle =\frac{\frac{1}{3}}{\frac{47}{60}}
\displaystyle =\frac{60}{3\times47}
\displaystyle =\frac{20}{47}
\displaystyle \text{So, the probability that the head shown was of a two-headed coin is }\frac{20}{47}
\displaystyle \text{Disclaimer: The answer given in the book is incorrect. The same has been corrected here}

Question 16:
In a factory, machine A produces \displaystyle 30\% of the total output, machine B produces \displaystyle 25\% and the machine C produces the remaining output. If defective items produced by machines A, B and C are \displaystyle 1\%,1.2\%,2\% respectively. Three machines working together produce \displaystyle 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?
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1,E_2 \text{ and } E_3 \text{ denote the events that the item is defective, machine A} \\ \text{is chosen, machine B is chosen and machine C is chosen, respectively}
\displaystyle \therefore P(E_1)=\frac{30}{100}
\displaystyle P(E_2)=\frac{25}{100}
\displaystyle P(E_3)=\frac{45}{100}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{1}{100}
\displaystyle P(A\mid E_2)=\frac{1.2}{100}
\displaystyle P(A\mid E_3)=\frac{2}{100}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\frac{\frac{30}{100}\times\frac{1}{100}}{\frac{30}{100}\times\frac{1}{100}+\frac{25}{100}\times\frac{1.2}{100}+\frac{45}{100}\times\frac{2}{100}}
\displaystyle =\frac{30}{30+30+90}=\frac{30}{150}=0.2

Question 17:
A company has two plants to manufacture bicycles. The first plant manufactures \displaystyle 60\% of the bicycles and the second plant \displaystyle 40\%. Out of that \displaystyle 80\% of the bicycles are rated of standard quality at the first plant and \displaystyle 90\% of standard quality at the second plant. A bicycle is picked up at random and found to be standard quality. Find the probability that it comes from the second plant.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1 \text{ and } E_2 \text{ denote the events that the cycle is of standard quality, plant I} \\ \text{is chosen and plant II is chosen, respectively}
\displaystyle \therefore P(E_1)=\frac{60}{100}
\displaystyle P(E_2)=\frac{40}{100}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{80}{100}
\displaystyle P(A\mid E_2)=\frac{90}{100}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{40}{100}\times\frac{90}{100}}{\frac{60}{100}\times\frac{80}{100}+\frac{40}{100}\times\frac{90}{100}}
\displaystyle =\frac{36}{48+36}=\frac{36}{84}=\frac{3}{7}

Question 18:
Three urns A, B and C contain \displaystyle 6 red and \displaystyle 4 white; \displaystyle 2 red and \displaystyle 6 white; and \displaystyle 1 red and \displaystyle 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.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1,E_2 \text{ and } E_3 \text{ denote the events that the ball is red, bag A is chosen, bag B is} \\ \text{chosen and bag C is chosen, respectively}
\displaystyle \therefore P(E_1)=\frac{1}{3}
\displaystyle P(E_2)=\frac{1}{3}
\displaystyle P(E_3)=\frac{1}{3}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{6}{10}=\frac{3}{5}
\displaystyle P(A\mid E_2)=\frac{2}{8}=\frac{1}{4}
\displaystyle P(A\mid E_3)=\frac{1}{6}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{\frac{3}{5}}{\frac{3}{5}+\frac{1}{4}+\frac{1}{6}}=\frac{36}{61}

Question 19:
In a group of \displaystyle 400 people, \displaystyle 160 are smokers and non-vegetarian, \displaystyle 100 are smokers and vegetarian and the remaining are non-smokers and vegetarian. The probabilities of getting a special chest disease are \displaystyle 35\%,20\% and \displaystyle 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?
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1,E_2 \text{ and } E_3 \text{ denote the events that the person suffers from the disease, is} \\ \text{ a smoker and a non-vegetarian, is a smoker and a vegetarian and the person} \\ \text{is a non-smoker and a vegetarian, respectively}
\displaystyle \therefore P(E_1)=\frac{160}{400}
\displaystyle P(E_2)=\frac{100}{400}
\displaystyle P(E_3)=\frac{140}{400}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{35}{100}
\displaystyle P(A\mid E_2)=\frac{20}{100}
\displaystyle P(A\mid E_3)=\frac{10}{100}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{560}{560+200+140}=\frac{560}{900}=\frac{28}{45}

Question 20:
A factory has three machines A, B and C, which produce \displaystyle 100,200 and \displaystyle 300 items of a particular type daily. The machines produce \displaystyle 2\%,3\% and \displaystyle 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.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1,E_2 \text{ and } E_3 \text{ denote the events that the item is defective,} \\ \text{machine A is chosen, machine B is chosen and machine C is chosen, respectively}
\displaystyle \therefore P(E_1)=\frac{100}{600}
\displaystyle P(E_2)=\frac{200}{600}
\displaystyle P(E_3)=\frac{300}{600}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{2}{100}
\displaystyle P(A\mid E_2)=\frac{3}{100}
\displaystyle P(A\mid E_3)=\frac{5}{100}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{2}{2+6+15}=\frac{2}{23}

Question 21:
A bag contains \displaystyle 1 white and \displaystyle 6 red balls, and a second bag contains \displaystyle 4 white and \displaystyle 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 colour. Find the probability that the drawn ball was from the first bag.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1 \text{ and } E_2 \text{ denote the events that the ball is white, bag I is chosen and bag II is chosen,} \\ \text{respectively}
\displaystyle \therefore P(E_1)=\frac{1}{2}
\displaystyle P(E_2)=\frac{1}{2}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{1}{7}
\displaystyle P(A\mid E_2)=\frac{4}{7}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{1}{2}\times\frac{1}{7}}{\frac{1}{2}\times\frac{1}{7}+\frac{1}{2}\times\frac{4}{7}}
\displaystyle =\frac{1}{1+4}=\frac{1}{5}

Question 22:
In a certain college, \displaystyle 4\% of boys and \displaystyle 1\% of girls are taller than \displaystyle 1.75 metres. Further more, \displaystyle 60\% of the students in the colleges are girls. A student selected at random from the college is found to be taller than \displaystyle 1.75 metres. Find the probability that the selected students is girl.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1 \text{ and } E_2 \text{ denote the events that the height of the student is more than } \\ 1.75\text{ m, selected student is a girl and selected student is a boy, respectively}
\displaystyle \therefore P(E_1)=\frac{60}{100}
\displaystyle P(E_2)=\frac{40}{100}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{1}{100}
\displaystyle P(A\mid E_2)=\frac{4}{100}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{60}{100}\times\frac{1}{100}}{\frac{60}{100}\times\frac{1}{100}+\frac{40}{100}\times\frac{4}{100}}
\displaystyle =\frac{6}{6+16}=\frac{6}{22}=\frac{3}{11}

Question 23:
For \displaystyle A,B and \displaystyle C the chances of being selected as the manager of a firm are in the ratio \displaystyle 4:1:2 respectively. The respective probabilities for them to introduce a radical change in marketing strategy are \displaystyle 0.3,0.8 and \displaystyle 0.5. If the change does take place, find the probability that it is due to the appointment of \displaystyle B or \displaystyle C.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1,E_2 \text{ and } E_3 \text{ denote the events that the change takes place, A is selected,} \\ \text{B is selected and C is selected, respectively}
\displaystyle \therefore P(E_1)=\frac{4}{7}
\displaystyle P(E_2)=\frac{1}{7}
\displaystyle P(E_3)=\frac{2}{7}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=0.3
\displaystyle P(A\mid E_2)=0.8
\displaystyle P(A\mid E_3)=0.5
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\frac{\frac{4}{7}\times0.3}{\frac{4}{7}\times0.3+\frac{1}{7}\times0.8+\frac{2}{7}\times0.5}
\displaystyle =\frac{1.2}{1.2+0.8+1}=\frac{1.2}{3}=\frac{12}{30}=\frac{2}{5}
\displaystyle \therefore \text{Required probability}=1-P(E_1\mid A)=1-\frac{2}{5}=\frac{3}{5}

Question 24:
Three persons A, B and C apply for a job of Manager in a private company. Chances of their selections (A, B and C) are in the ratio \displaystyle 1:2:4. The probabilities that A, B and C can introduce changes to improve profits of the company are \displaystyle 0.8,0.5 and \displaystyle 0.3 respectively. If the changes do not take place, find the probability that it is due to the appointment of C.
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1,E_2 \text{ and } E_3 \text{ be the events denoting the selection of A, B and C as} \\ \text{managers, respectively}
\displaystyle P(E_1)=\text{Probability of selection of A}=\frac{1}{7}
\displaystyle P(E_2)=\text{Probability of selection of B}=\frac{2}{7}
\displaystyle P(E_3)=\text{Probability of selection of C}=\frac{4}{7}
\displaystyle \text{Let } A \text{ be the event denoting the change not taking place}
\displaystyle P(A\mid E_1)=\text{Probability that A does not introduce change}=0.2
\displaystyle P(A\mid E_2)=\text{Probability that B does not introduce change}=0.5
\displaystyle P(A\mid E_3)=\text{Probability that C does not introduce change}=0.7
\displaystyle \therefore \text{Required probability}=P(E_3\mid A)
\displaystyle \text{By Bayes' theorem, we have}
\displaystyle P(E_3\mid A)=\frac{P(E_3)P(A\mid E_3)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\frac{\frac{4}{7}\times0.7}{\frac{1}{7}\times0.2+\frac{2}{7}\times0.5+\frac{4}{7}\times0.7}
\displaystyle =\frac{2.8}{0.2+1+2.8}
\displaystyle =\frac{2.8}{4}=0.7

Question 25:
An insurance company insured \displaystyle 2000 scooters and \displaystyle 3000 motorcycles. The probability of an accident involving a scooter is \displaystyle 0.01 and that of a motorcycle is \displaystyle 0.02. An insured vehicle met with an accident. Find the probability that the accidented vehicle was a motorcycle.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1 \text{ and } E_2 \text{ denote the events that the vehicle meets the accident, is a} \\ \text{scooter and is a motorcycle, respectively}
\displaystyle \therefore P(E_1)=\frac{2000}{5000}=0.4
\displaystyle P(E_2)=\frac{3000}{5000}=0.6
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=0.01
\displaystyle P(A\mid E_2)=0.02
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_2\mid A)=\frac{P(E_2)P(A\mid E_2)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{0.6\times0.02}{0.6\times0.02+0.4\times0.01}
\displaystyle =\frac{0.012}{0.012+0.004}=\frac{0.012}{0.016}=\frac{3}{4}

Question 26:
Of the students in a college, it is known that \displaystyle 60\% reside in a hostel and \displaystyle 40\% do not reside in hostel. Previous year results report that \displaystyle 30\% of students residing in hostel attain A grade and \displaystyle 20\% of ones not residing in hostel attain A grade in their annual examination. At the end of the year, one students is chosen at random from the college and he has an A grade. What is the probability that the selected student is a hosteler?
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1 \text{ and } E_2 \text{ denote the events that the selected student attains grade A,} \\ \text{resides in a hostel and does not reside in a hostel, respectively}
\displaystyle \therefore P(E_1)=\frac{60}{100}
\displaystyle P(E_2)=\frac{40}{100}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{30}{100}
\displaystyle P(A\mid E_2)=\frac{20}{100}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{60}{100}\times\frac{30}{100}}{\frac{60}{100}\times\frac{30}{100}+\frac{40}{100}\times\frac{20}{100}}
\displaystyle =\frac{18}{18+8}=\frac{9}{13}

Question 27:
There are three coins. One is two headed coin, another is a biased coin that comes up heads \displaystyle 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?
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1,E_2 \text{ and } E_3 \text{ denote the events of choosing a two-headed coin,} \\ \text{a biased coin and an unbiased coin, respectively}
\displaystyle \text{Let } A \text{ be the event that the coin shows heads}
\displaystyle \therefore P(E_1)=\frac{1}{3}
\displaystyle P(E_2)=\frac{1}{3}
\displaystyle P(E_3)=\frac{1}{3}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=1
\displaystyle P(A\mid E_2)=75\%=\frac{3}{4}
\displaystyle P(A\mid E_3)=\frac{1}{2}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}
\displaystyle =\frac{\frac{1}{3}\times1}{\frac{1}{3}\times1+\frac{1}{3}\times\frac{3}{4}+\frac{1}{3}\times\frac{1}{2}}
\displaystyle =\frac{4}{9}

Question 28:
Assume that the chances of a patient having a heart attack is \displaystyle 40\%. It is also assumed that meditation and yoga course reduces the risk of heart attack by \displaystyle 30\% and prescription of certain drug reduces its chances by \displaystyle 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?
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1 \text{ and } E_2 \text{ denote the events that the selected person had a heart attack,} \\ \text{did yoga and meditation and followed the drug prescriptions, respectively}
\displaystyle \therefore P(E_1)=\frac{1}{2}
\displaystyle P(E_2)=\frac{1}{2}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=0.40\times0.70=0.28
\displaystyle P(A\mid E_2)=0.40\times0.75=0.30
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{1}{2}\times0.28}{\frac{1}{2}\times0.28+\frac{1}{2}\times0.30}
\displaystyle =\frac{14}{29}

Question 29:
Coloured balls are distributed in four boxes as shown in the following table:
\displaystyle \begin{array}{|c|c|c|c|c|} \hline \text{Box} & \text{Black} & \text{White} & \text{Red} & \text{Blue} \\ \hline \text{I} & 3 & 4 & 5 & 6 \\ \hline \text{II} & 2 & 2 & 2 & 2 \\ \hline \text{III} & 1 & 2 & 3 & 1 \\ \hline \text{IV} & 4 & 3 & 1 & 5 \\ \hline \end{array}
A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1,E_2,E_3 \text{ and } E_4 \text{ denote the events that the ball is black, box I} \\ \text{is selected, box II is selected, box III is selected and box IV is selected, respectively}
\displaystyle \therefore P(E_1)=\frac{1}{4}
\displaystyle P(E_2)=\frac{1}{4}
\displaystyle P(E_3)=\frac{1}{4}
\displaystyle P(E_4)=\frac{1}{4}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{3}{18}
\displaystyle P(A\mid E_2)=\frac{2}{8}
\displaystyle P(A\mid E_3)=\frac{1}{7}
\displaystyle P(A\mid E_4)=\frac{4}{13}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_3\mid A)=\frac{P(E_3)P(A\mid E_3)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)+P(E_4)P(A\mid E_4)}
\displaystyle =\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}}
\displaystyle =\frac{\frac{1}{7}}{\frac{1}{6}+\frac{1}{4}+\frac{1}{7}+\frac{4}{13}}
\displaystyle =\frac{156}{947}

Question 30:
If a machine is correctly set up it produces \displaystyle 90\% acceptable items. If it is incorrectly set up it produces only \displaystyle 40\% acceptable items. Past experience shows that \displaystyle 80\% of the setups are correctly done. If after a certain set up, the machine produces \displaystyle 2 acceptable items, find the probability that the machine is correctly set up.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A \text{ be the event that the machine produces two acceptable items}
\displaystyle \text{Also, let } E_1 \text{ represent the event that the machine is correctly set up and }\\ E_2 \text{ represent the event that the machine is incorrectly set up}
\displaystyle \therefore P(E_1)=0.8
\displaystyle P(E_2)=0.2
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=0.9\times0.9=0.81
\displaystyle P(A\mid E_2)=0.40\times0.40=0.16
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{0.8\times0.81}{0.8\times0.81+0.2\times0.16}
\displaystyle =\frac{81}{85}

Question 31:
Bag A contains \displaystyle 3 red and \displaystyle 5 black balls, while bag B contains \displaystyle 4 red and \displaystyle 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 bag A to bag B.
\displaystyle \text{Answer:}
\displaystyle \text{It is given that bag A contains 3 red and 5 black balls and bag B contains 4 red and 4 black balls}
\displaystyle \text{Let } E_1,E_2,E_3 \text{ and } A \text{ be the events as defined below:}
\displaystyle E_1:\text{ Two red balls are transferred from bag A to bag B}
\displaystyle E_2:\text{ One red ball and one black ball is transferred from bag A to bag B}
\displaystyle E_3:\text{ Two black balls are transferred from bag A to bag B}
\displaystyle A:\text{ Ball drawn from bag B is red}
\displaystyle \text{So,}
\displaystyle P(E_1)=\frac{\binom{3}{2}}{\binom{8}{2}}=\frac{3}{28}
\displaystyle P(E_2)=\frac{\binom{3}{1}\binom{5}{1}}{\binom{8}{2}}=\frac{15}{28}
\displaystyle P(E_3)=\frac{\binom{5}{2}}{\binom{8}{2}}=\frac{10}{28}
\displaystyle \text{Also,}
\displaystyle P(A\mid E_1)=\frac{6}{10}
\displaystyle P(A\mid E_2)=\frac{5}{10}
\displaystyle P(A\mid E_3)=\frac{4}{10}
\displaystyle \therefore \text{Required probability}
\displaystyle =\text{Probability that two red balls were transferred from A to B given that the ball drawn from bag B is red}
\displaystyle =P(E_1\mid A)
\displaystyle =\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)}\;[\text{Using Bayes' Theorem}]
\displaystyle =\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}}
\displaystyle =\frac{18}{18+75+40}
\displaystyle =\frac{18}{133}

Question 32:
By examining the chest X-ray, probability that T.B is detected when a person is actually suffering is \displaystyle 0.99. The probability that the doctor diagnoses incorrectly that a person has T.B on the basis of X-ray is \displaystyle 0.001. In a certain city \displaystyle 1 in \displaystyle 1000 persons suffers from T.B. A person is selected at random is diagnosed to have T.B. What is the chance that he actually has T.B.?
\displaystyle \text{Answer:}
\displaystyle \textbf{Let } T \textbf{ = person actually has T.B., and } D \textbf{ = diagnosed (positive) from X-ray.}
\displaystyle P(D\mid T)=0.99,\quad P(D\mid T^c)=0.001,\quad P(T)=\frac{1}{1000}=0.001,\quad P(T^c)=0.999.
\displaystyle P(T\mid D)=\frac{P(T),P(D\mid T)}{P(T)P(D\mid T)+P(T^c)P(D\mid T^c)}
\displaystyle =\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.001}
\displaystyle =\frac{0.00099}{0.00099+0.000999}
\displaystyle =\frac{0.00099}{0.001989}=\frac{990}{1989}=\frac{110}{221}
\displaystyle \therefore\ P(T\mid D)=\frac{110}{221}\approx 0.498.

Question 33:
A test for detection of a particular disease is not fool proof. The test will correctly detect the disease \displaystyle 90\% of the time, but will incorrectly detect the disease \displaystyle 1\% of the time. For a large given population of which an estimated \displaystyle 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 have the disease?
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1 \text{ and } E_2 \text{ denote the events that the person suffers from the disease,} \\ \text{the test detects the disease correctly and the test does not detect the disease correctly, respectively}
\displaystyle \therefore P(E_1)=\frac{90}{100}
\displaystyle P(E_2)=\frac{1}{100}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{2}{1000}
\displaystyle P(A\mid E_2)=\frac{998}{1000}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{90}{100}\times\frac{2}{1000}}{\frac{90}{100}\times\frac{2}{1000}+\frac{1}{100}\times\frac{998}{1000}}
\displaystyle =\frac{180}{180+998}=\frac{180}{1178}=\frac{90}{589}

Question 34:
Let \displaystyle d_1,d_2,d_3 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 \displaystyle 5000 patients: \displaystyle 1800 had disease \displaystyle d_1, \displaystyle 2100 has disease \displaystyle d_2 and the others had disease \displaystyle d_3. \displaystyle 1500 patients with disease \displaystyle d_1, \displaystyle 1200 patients with disease \displaystyle d_2 and \displaystyle 900 patients with disease \displaystyle d_3 showed the symptom. Which of the diseases is the patient most likely to have?
\displaystyle \text{Answer:}
\displaystyle \text{Events }E_1,E_2,E_3 \text{ and } S \text{ be the events defined as follows:}
\displaystyle E_1:\text{ The patient had disease }d_1
\displaystyle E_2:\text{ The patient had disease }d_2
\displaystyle E_3:\text{ The patient had disease }d_3
\displaystyle S:\text{ The patient showed the symptom}
\displaystyle E_1,E_2 \text{ and } E_3 \text{ are mutually exclusive and exhaustive events}
\displaystyle P(E_1)=\frac{1800}{5000}=\frac{18}{50}
\displaystyle P(E_2)=\frac{2100}{5000}=\frac{21}{50}
\displaystyle P(E_3)=\frac{1100}{5000}=\frac{11}{50}
\displaystyle \text{Now,}
\displaystyle P(S\mid E_1)=\text{Probability that the patient showed symptom given that patient} \\ \text{had disease }d_1=\frac{1500}{5000}=\frac{15}{50}
\displaystyle P(S\mid E_2)=\text{Probability that the patient showed symptom given that patient} \\ \text{had disease }d_2=\frac{1200}{5000}=\frac{12}{50}
\displaystyle P(S\mid E_3)=\text{Probability that the patient showed symptom given that patient} \\ \text{had disease }d_3=\frac{900}{5000}=\frac{9}{50}
\displaystyle \text{Using Bayes' theorem, we have probability that patient had disease }d_1\text{ such that symptom showed }=P(E_1\mid S)
\displaystyle P(E_1\mid S)=\frac{P(E_1)P(S\mid E_1)}{P(E_1)P(S\mid E_1)+P(E_2)P(S\mid E_2)+P(E_3)P(S\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{270}{621}
\displaystyle \text{Probability that patient had disease }d_2\text{ such that symptom showed }=P(E_2\mid S)
\displaystyle P(E_2\mid S)=\frac{P(E_2)P(S\mid E_2)}{P(E_1)P(S\mid E_1)+P(E_2)P(S\mid E_2)+P(E_3)P(S\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{252}{621}
\displaystyle \text{Probability that patient had disease }d_3\text{ such that symptom showed }=P(E_3\mid S)
\displaystyle P(E_3\mid S)=\frac{P(E_3)P(S\mid E_3)}{P(E_1)P(S\mid E_1)+P(E_2)P(S\mid E_2)+P(E_3)P(S\mid E_3)}
\displaystyle =\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}}
\displaystyle =\frac{99}{621}
\displaystyle \text{Thus, the patient is most likely to have the disease }d_1

Question 35:
A is known to speak truth \displaystyle 3 times out of \displaystyle 5 times. He throws a die and reports that it is one. Find the probability that it is actually one.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A, E_1 \text{ and } E_2 \text{ denote the events that the man reports the appearance of } 1 \\ \text{ on throwing a die, } 1 \text{ occurs and } 1 \text{ does not occur, respectively.}
\displaystyle \therefore P(E_1)=\frac{1}{6}
\displaystyle P(E_2)=\frac{5}{6}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{3}{5}
\displaystyle P(A\mid E_2)=\frac{2}{5}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{1}{6}\times\frac{3}{5}}{\frac{1}{6}\times\frac{3}{5}+\frac{5}{6}\times\frac{2}{5}}
\displaystyle =\frac{3}{3+10}
\displaystyle =\frac{3}{13}

Question 36:
A speaks the truth \displaystyle 8 times out of \displaystyle 10 times. A die is tossed. He reports that it was \displaystyle 5. What is the probability that it was actually \displaystyle 5?
\displaystyle \text{Answer:}
\displaystyle \text{Let } A \text{ denote the event that the man reports that } 5 \text{ occurs and } E \text{ the event that } 5 \text{ actually turns up.}
\displaystyle \therefore P(E)=\frac{1}{6}
\displaystyle P(\overline{E})=1-\frac{1}{6}=\frac{5}{6}
\displaystyle \text{Also,}
\displaystyle P(A\mid E)=\text{Probability that the man reports that }5\text{ occurs given that }5\text{ actually turns up}
\displaystyle =\text{Probability of the man speaking the truth}
\displaystyle =\frac{8}{10}=\frac{4}{5}
\displaystyle P(A\mid \overline{E})=\text{Probability that the man reports that }5\text{ occurs given that }5\text{ does not turn up}
\displaystyle =\text{Probability of the man not speaking the truth}
\displaystyle =1-\frac{4}{5}=\frac{1}{5}
\displaystyle \therefore \text{Required probability}=P(E\mid A)
\displaystyle =\frac{P(E)P(A\mid E)}{P(E)P(A\mid E)+P(\overline{E})P(A\mid \overline{E})}
\displaystyle =\frac{\frac{1}{6}\times\frac{4}{5}}{\frac{1}{6}\times\frac{4}{5}+\frac{5}{6}\times\frac{1}{5}}
\displaystyle =\frac{4}{9}

Question 37:
In answering a question on a multiple choice test a student either knows the answer or guesses. Let \displaystyle 3/4 be the probability that he knows the answer and \displaystyle 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability \displaystyle 1/4 . What is the probability that a student knows the answer given that he answered it correctly?
\displaystyle \text{Answer:}
\displaystyle \text{Let } A, E_1 \text{ and } E_2 \text{ denote the events that the answer is correct, the student} \\ \text{knows the answer and the student guesses the answer, respectively.}
\displaystyle \therefore P(E_1)=\frac{3}{4}
\displaystyle P(E_2)=\frac{1}{4}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=1
\displaystyle P(A\mid E_2)=\frac{1}{4}
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{\frac{3}{4}\times 1}{\frac{3}{4}\times 1+\frac{1}{4}\times\frac{1}{4}}
\displaystyle =\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}
\displaystyle =\frac{3}{3+\frac{1}{4}}
\displaystyle =\frac{12}{13}

Question 38:
A laboratory blood test is \displaystyle 99\% effective in detecting a certain disease when its infection is present. However, the test also yields a false positive result for \displaystyle 0.5\% of the healthy person tested (i.e. if a healthy person is tested, then, with probability \displaystyle 0.005, the test will imply he has the disease). If \displaystyle 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?
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1 \text{ and } E_2 \text{ denote the events that a person has a disease and a person has no disease,} \\ \text{respectively.}
\displaystyle \text{ }E_1 \text{ and } E_2 \text{ are complementary to each other.}
\displaystyle \therefore P(E_1)+P(E_2)=1
\displaystyle \Rightarrow P(E_2)=1-P(E_1)=1-0.001=0.999
\displaystyle \text{Let } A \text{ denote the event that the blood test result is positive.}
\displaystyle \therefore P(E_1)=0.001
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=0.99
\displaystyle P(A\mid E_2)=0.005
\displaystyle \text{Using Bayes' theorem, we get}
\displaystyle \text{Required probability}=P(E_1\mid A)=\frac{P(E_1)P(A\mid E_1)}{P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)}
\displaystyle =\frac{0.001\times0.99}{0.001\times0.99+0.999\times0.005}
\displaystyle =\frac{990}{5985}
\displaystyle =\frac{22}{133}

Question 39:
There are three categories of students in a class of \displaystyle 60 students:
A : Very hardworking; B : Regular but not so hardworking; C : Careless and irregular.
\displaystyle 10 students are in category A, \displaystyle 30 in category B and 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 \displaystyle 0.002, of category B it is \displaystyle 0.02 and of category C, this probability is \displaystyle 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 of category C.
\displaystyle \text{Answer:}
\displaystyle \text{Let } E \text{ denote the event that the student could not get good marks in the examination.}
\displaystyle \text{Also, } A:\text{ the event that the student is very hardworking}
\displaystyle B:\text{ the event that the student is regular but not so hardworking}
\displaystyle C:\text{ the event that the student is careless and irregular}
\displaystyle \therefore P(A)=\frac{10}{60},\;P(B)=\frac{30}{60}\text{ and }P(C)=\frac{20}{60}
\displaystyle \text{Also,}
\displaystyle P(E\mid A)=\text{Probability that the student of category }A\text{ could not get good marks in} \\ \text{the examination}=0.002
\displaystyle P(E\mid B)=\text{Probability that the student of category }B\text{ could not get good marks in the} \\ \text{examination}=0.02
\displaystyle P(E\mid C)=\text{Probability that the student of category }C\text{ could not get good marks in} \\ \text{the examination}=0.2
\displaystyle \therefore \text{Required probability}=P(C\mid E)
\displaystyle =\frac{P(C)P(E\mid C)}{P(A)P(E\mid A)+P(B)P(E\mid B)+P(C)P(E\mid C)}
\displaystyle =\frac{\frac{20}{60}\times0.2}{\frac{10}{60}\times0.002+\frac{30}{60}\times0.02+\frac{20}{60}\times0.2}
\displaystyle =\frac{4}{4.62}
\displaystyle =\frac{400}{462}
\displaystyle =\frac{200}{231}

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