RS Aggarwal Solutions for Class 12 Chapter 30 Bayess Theorem and its Applications

Question 1. In a bulb factory, three machines, A, B, C, manufacture 60%, 25% and 15% of the total production respectively. Of their respective outputs, 1%, 2% and 1% are defective. A bulb is drawn at random from the total product, and it is found to be defective. Find the probability that it was manufactured by machine C.
Answer: Let D represent a defective bulb. We want to find P(C|D), which is the probability that the selected defective bulb came from machine C. Using Bayes's theorem:

\( P(C|D) = \frac{P(C) \cdot P(D|C)}{P(A) \cdot P(D|A) + P(B) \cdot P(D|B) + P(C) \cdot P(D|C)} \)

Where: P(A) = \( \frac{60}{100} \), P(B) = \( \frac{25}{100} \), P(C) = \( \frac{15}{100} \), P(D|A) = \( \frac{1}{100} \), P(D|B) = \( \frac{2}{100} \), P(D|C) = \( \frac{1}{100} \)

\( P(C|D) = \frac{15}{60 + 50 + 15} = \frac{15}{125} = \frac{3}{25} \)

Therefore, the probability that the selected defective bulb came from machine C is \( \frac{3}{25} \).
In simple words: Out of all defective bulbs in the factory, machine C is responsible for only a small fraction. We calculate this by weighing how many defective bulbs machine C typically produces against all the defective bulbs from all three machines combined.

Exam Tip: Always verify your final answer by checking that the sum of numerators across all posterior probabilities equals the total denominator - this ensures your calculation is consistent.

Question 2. A company manufactures scooters at two plants, A and B. Plant A produces 80% and plant B produces 20% of the total product. 85% of the scooters produced at plant A and 65% of the scooters produced at plant B are of standard quality. A scooter produced by the company is selected at random, and it is found to be of standard quality. What is the probability that it was manufactured at plant A?
Answer: Let S denote standard quality. We want to find P(A|S), representing the probability that the selected standard scooter originated from plant A. Using Bayes's theorem:

\( P(A|S) = \frac{P(A) \cdot P(S|A)}{P(A) \cdot P(S|A) + P(B) \cdot P(S|B)} \)

Where: P(A) = \( \frac{80}{100} \), P(B) = \( \frac{20}{100} \), P(S|A) = \( \frac{85}{100} \), P(S|B) = \( \frac{65}{100} \)

\( P(A|S) = \frac{(80)(85)}{(80)(85) + (20)(65)} = \frac{6800}{6800 + 1300} = \frac{6800}{8100} = \frac{68}{81} \)

Therefore, the probability that the selected standard scooter came from plant A is \( \frac{68}{81} \).
In simple words: Since plant A makes more scooters and also produces a higher quality rate, a randomly selected good scooter is much more likely to have originated from plant A than from plant B.

Exam Tip: Remember to multiply the probability of picking each plant by the probability of getting a standard scooter from that plant before applying Bayes's theorem.

Question 3. In a certain college, 4% of boys and 1% of girls are taller than 1.75 meters. Furthermore, 60% of the students are girls. If a student is selected at random and is taller than 1.75 meters, what is the probability that the selected student is a girl?
Answer: Let T denote students taller than 1.75 meters, B denote boys, and G denote girls. We want to find P(G|T), the probability that a randomly selected tall student is a girl. Using Bayes's theorem:

\( P(G|T) = \frac{P(G) \cdot P(T|G)}{P(G) \cdot P(T|G) + P(B) \cdot P(T|B)} \)

Where: P(G) = \( \frac{60}{100} \), P(B) = \( \frac{40}{100} \), P(T|G) = \( \frac{1}{100} \), P(T|B) = \( \frac{4}{100} \)

\( P(G|T) = \frac{(\frac{60}{100})(\frac{1}{100})}{(\frac{60}{100})(\frac{1}{100}) + (\frac{40}{100})(\frac{4}{100})} = \frac{60}{60 + 160} = \frac{60}{220} = \frac{3}{11} \)

Therefore, the probability that a randomly selected tall student is a girl is \( \frac{3}{11} \).
In simple words: Even though there are more girls in the college, a smaller percentage of girls are tall. So among all tall students, there are actually fewer girls than boys.

Exam Tip: Pay close attention to conditional probabilities - the percentage of girls who are tall differs significantly from the percentage of boys who are tall, which affects the final result.

Question 4. 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.
Answer: Let I denote students having IQ greater than 150, B denote boys, and G denote girls. We want to find P(B|I), the probability that a randomly selected student with IQ greater than 150 is a boy. Using Bayes's theorem:

\( P(B|I) = \frac{P(B) \cdot P(I|B)}{P(B) \cdot P(I|B) + P(G) \cdot P(I|G)} \)

Where: P(B) = \( \frac{60}{100} \), P(G) = \( \frac{40}{100} \), P(I|B) = \( \frac{5}{100} \), P(I|G) = \( \frac{10}{100} \)

\( P(B|I) = \frac{(\frac{60}{100})(\frac{5}{100})}{(\frac{60}{100})(\frac{5}{100}) + (\frac{40}{100})(\frac{10}{100})} = \frac{300}{300 + 400} = \frac{300}{700} = \frac{3}{7} \)

Therefore, the probability that a randomly selected student with IQ greater than 150 is a boy is \( \frac{3}{7} \).
In simple words: Although more boys are in the class overall, a higher percentage of girls have very high IQs. This balances out, so when you find someone with an exceptionally high IQ, that person is more likely to be a girl than a boy.

Exam Tip: Notice that the group with lower initial percentage (girls at 40%) actually has a higher conditional probability of high IQ (10%), which significantly influences the posterior probability.

Question 5. Suppose 5% of men and 0.25% of women have grey hair. A grey-haired person is selected at random. What is the probability of this person being male? Assume that there is an equal number of males and females.
Answer: Let MG denote men having grey hair, WG denote women having grey hair, and G denote having grey hair. We want to find P(MG|G), the probability that a randomly selected grey-haired person is male. Since there are equal numbers of males and females, we set P(M) = P(W) = \( \frac{1}{2} \). Using Bayes's theorem:

\( P(MG|G) = \frac{P(M) \cdot P(G|M)}{P(M) \cdot P(G|M) + P(W) \cdot P(G|W)} \)

Where: P(M) = \( \frac{1}{2} \), P(W) = \( \frac{1}{2} \), P(G|M) = \( \frac{5}{100} \), P(G|W) = \( \frac{0.25}{100} \)

\( P(MG|G) = \frac{(\frac{1}{2})(\frac{5}{100})}{(\frac{1}{2})(\frac{5}{100}) + (\frac{1}{2})(\frac{0.25}{100})} = \frac{5}{5.25} = \frac{5}{5.25} = \frac{20}{21} \)

Therefore, the probability that a randomly selected grey-haired person is male is \( \frac{20}{21} \).
In simple words: Even though men and women exist in equal numbers, men develop grey hair at a much higher rate. So when you see a grey-haired person, it is far more likely to be a man.

Exam Tip: When probabilities of populations are equal, focus on the conditional probabilities to determine which outcome is more likely - the larger conditional probability will dominate the posterior calculation.

Question 6. Two groups are competing for the positions on 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 when the second groups win, the corresponding probability is 0.3. Find the probability that the new product introduced was by the second group.
Answer: Let F denote the first group, S denote the second group, and N denote introducing a new product. We want to find P(S|N), the probability that the second group introduced a new product. Using Bayes's theorem:

\( P(S|N) = \frac{P(S) \cdot P(N|S)}{P(S) \cdot P(N|S) + P(F) \cdot P(N|F)} \)

Where: P(F) = 0.6, P(S) = 0.4, P(N|F) = 0.7, P(N|S) = 0.3

\( P(S|N) = \frac{(0.4)(0.3)}{(0.4)(0.3) + (0.6)(0.7)} = \frac{0.12}{0.12 + 0.42} = \frac{0.12}{0.54} = \frac{2}{9} \)

Therefore, the probability that the second group introduced a new product is \( \frac{2}{9} \).
In simple words: The first group wins more often and also introduces new products more frequently when they do win. So even if a new product appears, it is more likely to come from the first group.

Exam Tip: When applying Bayes's theorem with non-equal prior probabilities, make sure to weight each possibility by both its likelihood of occurring and its conditional probability for the observed event.

Question 7. A bag A contains 1 white and 6 red balls. Another bag contains 4 white and 3 red balls. One of the bags is selected at random, and a ball is drawn from it, which is found to be white. Find the probability that the ball is drawn is from bag A.
Answer: Let W denote a white ball, A denote bag A, and B denote bag B. We want to find P(A|W), the probability that the selected white ball came from bag A. Since each bag is equally likely to be selected, P(A) = P(B) = \( \frac{1}{2} \). Using Bayes's theorem:

\( P(A|W) = \frac{P(A) \cdot P(W|A)}{P(A) \cdot P(W|A) + P(B) \cdot P(W|B)} \)

Where: P(A) = \( \frac{1}{2} \), P(B) = \( \frac{1}{2} \), P(W|A) = \( \frac{1}{7} \), P(W|B) = \( \frac{4}{7} \)

\( P(A|W) = \frac{(\frac{1}{2})(\frac{1}{7})}{(\frac{1}{2})(\frac{1}{7}) + (\frac{1}{2})(\frac{4}{7})} = \frac{1}{1 + 4} = \frac{1}{5} \)

Therefore, the probability that the selected white ball came from bag A is \( \frac{1}{5} \).
In simple words: Bag A has very few white balls compared to bag B. So when a white ball is drawn, it is much more likely to have come from bag B, which is better stocked with white balls.

Exam Tip: Always calculate the conditional probability of the observed outcome for each scenario - here, the probability of drawing white from each bag is crucial to determining which bag the ball most likely came from.

Question 8. There are two bags I and II. The bag I contains 3 white and 4 black balls, and bag II contains 5 white and 6 black balls. One ball is drawn at random from one of the bags and is found to be white. Find the probability that it was drawn from the bag I.
Answer: Let W denote a white ball, X denote the first bag, and Y denote the second bag. We want to find P(X|W), the probability that the selected white ball came from the first bag. Since each bag is equally likely to be selected, P(X) = P(Y) = \( \frac{1}{2} \). Using Bayes's theorem:

\( P(X|W) = \frac{P(X) \cdot P(W|X)}{P(X) \cdot P(W|X) + P(Y) \cdot P(W|Y)} \)

Where: P(X) = \( \frac{1}{2} \), P(Y) = \( \frac{1}{2} \), P(W|X) = \( \frac{3}{7} \), P(W|Y) = \( \frac{5}{11} \)

\( P(X|W) = \frac{(\frac{1}{2})(\frac{3}{7})}{(\frac{1}{2})(\frac{3}{7}) + (\frac{1}{2})(\frac{5}{11})} = \frac{\frac{3}{7}}{\frac{3}{7} + \frac{5}{11}} = \frac{\frac{33}{77}}{\frac{33}{77} + \frac{35}{77}} = \frac{33}{68} \)

Therefore, the probability that the selected white ball came from the first bag is \( \frac{33}{68} \).
In simple words: Both bags contain white balls, but in different proportions. The first bag has slightly fewer white balls relative to its total, so when a white ball is drawn, it is somewhat more likely to have come from the second bag.

Exam Tip: When bags have different composition ratios, compare the conditional probabilities carefully - a higher proportion of white in bag II shifts the posterior probability even though both bags are equally likely initially.

Question 9. A box contains 2 gold and 3 silver coins. Another box contains 3 gold and 3 silver coins. A box is chosen at random, and a coin is drawn from it. If the selected coin is a gold coin, find the probability that it was drawn from the second box.
Answer: Let G denote a gold coin, A denote the first box, and B denote the second box. We want to find P(B|G), the probability that the selected gold coin came from the second box. Since each box is equally likely to be chosen, P(A) = P(B) = \( \frac{1}{2} \). Using Bayes's theorem:

\( P(B|G) = \frac{P(B) \cdot P(G|B)}{P(A) \cdot P(G|A) + P(B) \cdot P(G|B)} \)

Where: P(A) = \( \frac{1}{2} \), P(B) = \( \frac{1}{2} \), P(G|A) = \( \frac{2}{5} \), P(G|B) = \( \frac{3}{6} = \frac{1}{2} \)

\( P(B|G) = \frac{(\frac{1}{2})(\frac{1}{2})}{(\frac{1}{2})(\frac{2}{5}) + (\frac{1}{2})(\frac{1}{2})} = \frac{\frac{1}{4}}{\frac{1}{5} + \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{9}{20}} = \frac{5}{9} \)

Therefore, the probability that the selected gold coin came from the second box is \( \frac{5}{9} \).
In simple words: The second box has a higher proportion of gold coins (50% versus 40%). So when a gold coin appears, it is more likely to have originated from the second box.

Exam Tip: Convert all fractions to a common denominator before adding them in the denominator of Bayes's theorem to avoid arithmetic errors.

Question 10. Three urns A, B and C contains 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 balls was drawn from the first urn A.
Answer: Let R denote a red ball, and A, B, C denote the three urns. We want to find P(A|R), the probability that the selected red ball came from urn A. Since each urn is equally likely to be selected, P(A) = P(B) = P(C) = \( \frac{1}{3} \). Using Bayes's theorem:

\( P(A|R) = \frac{P(A) \cdot P(R|A)}{P(A) \cdot P(R|A) + P(B) \cdot P(R|B) + P(C) \cdot P(R|C)} \)

Where: P(A) = \( \frac{1}{3} \), P(B) = \( \frac{1}{3} \), P(C) = \( \frac{1}{3} \), P(R|A) = \( \frac{6}{10} \), P(R|B) = \( \frac{2}{8} \), P(R|C) = \( \frac{1}{6} \)

\( P(A|R) = \frac{(\frac{1}{3})(\frac{6}{10})}{(\frac{1}{3})(\frac{6}{10}) + (\frac{1}{3})(\frac{2}{8}) + (\frac{1}{3})(\frac{1}{6})} = \frac{\frac{6}{10}}{\frac{6}{10} + \frac{2}{8} + \frac{1}{6}} = \frac{\frac{36}{60}}{\frac{36}{60} + \frac{15}{60} + \frac{10}{60}} = \frac{36}{61} \)

Therefore, the probability that the selected red ball came from urn A is \( \frac{36}{61} \).
In simple words: Urn A has the highest proportion of red balls (60%), while urns B and C have much lower proportions. This makes urn A the most likely source when a red ball is drawn.

Exam Tip: When dealing with three or more categories, always include all possibilities in both the numerator and denominator of Bayes's theorem - missing even one term will give an incorrect answer.

Question 11. 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.
Answer: Let W denote a white ball, and A, B, C denote the three urns. We want to find P(A|W), the probability that the selected white ball came from urn A. Since each urn is equally likely to be selected, P(A) = P(B) = P(C) = \( \frac{1}{3} \). Using Bayes's theorem:

\( P(A|W) = \frac{P(A) \cdot P(W|A)}{P(A) \cdot P(W|A) + P(B) \cdot P(W|B) + P(C) \cdot P(W|C)} \)

Where: P(A) = \( \frac{1}{3} \), P(B) = \( \frac{1}{3} \), P(C) = \( \frac{1}{3} \), P(W|A) = \( \frac{2}{5} \), P(W|B) = \( \frac{3}{5} \), P(W|C) = \( \frac{4}{5} \)

\( P(A|W) = \frac{(\frac{1}{3})(\frac{2}{5})}{(\frac{1}{3})(\frac{2}{5}) + (\frac{1}{3})(\frac{3}{5}) + (\frac{1}{3})(\frac{4}{5})} = \frac{\frac{2}{5}}{\frac{2}{5} + \frac{3}{5} + \frac{4}{5}} = \frac{2}{9} \)

Therefore, the probability that the selected white ball came from urn A is \( \frac{2}{9} \).
In simple words: Urn A has the fewest white balls proportionally (only 40% white). Urns B and C have higher proportions of white balls, making them more likely sources for a white ball that is drawn.

Exam Tip: Notice that when the conditional probability for the first urn is the smallest, its posterior probability will also be the smallest - the evidence (drawing white) works against the first urn being the source.

Question 13. Urn A contains 7 white and 3 black balls; urn B contains 4 white and 6 black balls; urn C contains 2 white and 8 black balls. One of these urns is chosen at random with probabilities 0.2, 0.6 and 0.2 respectively. From the chosen urn, two balls are drawn at random without replacement. Both the balls happen to be white. Find the probability that the balls are drawn are from urn C.
Answer: Let WW denote drawing two white balls, and A, B, C denote the three urns. We want to find the probability that two white balls came from urn C. The urns are chosen with non-equal probabilities: P(A) = 0.2, P(B) = 0.6, P(C) = 0.2. Using Bayes's theorem:

Probability of picking 2 white balls from urn A = \( \frac{7 \cdot 6}{10 \cdot 9} = \frac{42}{90} = \frac{7}{15} \)

Probability of picking 2 white balls from urn B = \( \frac{4 \cdot 3}{10 \cdot 9} = \frac{12}{90} = \frac{2}{15} \)

Probability of picking 2 white balls from urn C = \( \frac{2 \cdot 1}{10 \cdot 9} = \frac{2}{90} = \frac{1}{45} \)

\( P(C|WW) = \frac{(0.2)(\frac{1}{45})}{(0.2)(\frac{7}{15}) + (0.6)(\frac{2}{15}) + (0.2)(\frac{1}{45})} = \frac{\frac{1}{40}}{\frac{21}{45} + \frac{6}{45} + \frac{1}{45}} = \frac{1}{40} \)

Therefore, the probability that both selected white balls came from urn C is \( \frac{1}{40} \).
In simple words: Urn C has very few white balls relative to the others. Even though urn C is chosen with some probability, drawing two white balls from it is extremely unlikely, making its posterior probability quite small.

Exam Tip: When urns are chosen with unequal prior probabilities, ensure these probabilities are explicitly multiplied into Bayes's formula - they significantly affect which outcome is most likely.

Question 14. There are 3 bags, each containing 5 white and 3 black balls. Also, there are 2 bags, each containing 2 white and 4 black balls. A white ball is drawn at random. Find the probability that this ball is from a bag of the first group.
Answer: Let A represent the group of 3 bags (each with 5 white and 3 black), and B represent the group of 2 bags (each with 2 white and 4 black). We can combine each group into one equivalent bag: bag A contains 15 white and 9 black balls total, while bag B contains 4 white and 8 black balls total. We want to find P(A|W), the probability that a selected white ball came from group A. The probability of selecting group A is \( \frac{3}{5} \) (3 bags out of 5 total), and the probability of selecting group B is \( \frac{2}{5} \). Using Bayes's theorem:

\( P(A|W) = \frac{P(A) \cdot P(W|A)}{P(A) \cdot P(W|A) + P(B) \cdot P(W|B)} \)

Where: P(A) = \( \frac{3}{5} \), P(B) = \( \frac{2}{5} \), P(W|A) = \( \frac{15}{24} \), P(W|B) = \( \frac{4}{12} \)

\( P(A|W) = \frac{(\frac{3}{5})(\frac{15}{24})}{(\frac{3}{5})(\frac{15}{24}) + (\frac{2}{5})(\frac{4}{12})} = \frac{\frac{45}{120}}{\frac{45}{120} + \frac{8}{60}} = \frac{45}{61} \)

Therefore, the probability that the selected white ball came from the first group is \( \frac{45}{61} \).
In simple words: The first group has bags with a higher proportion of white balls (roughly 63% white) compared to the second group (roughly 33% white). When a white ball is drawn, it is much more likely to have come from the first group.

Exam Tip: For grouped problems, combine all bags in each group into a single equivalent container before applying Bayes's theorem - this simplifies the calculation while maintaining accuracy.

Question 15. There are four boxes, A, B, C and D, containing marbles. A contains 1 red, 6 white and 3 black marbles; B contains 6 red, 2 white and 2 black marbles; C contains 8 red, 1 white and 1 black marbles; and D contains 6 white and 4 black marbles. One of the boxes is 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 the box A?
Answer: Let R denote a red marble, and A, B, C, D denote the four boxes. We want to find P(A|R), the probability that the selected red marble came from box A. Since each box is equally likely to be selected, P(A) = P(B) = P(C) = P(D) = \( \frac{1}{4} \). Using Bayes's theorem:

\( P(A|R) = \frac{P(A) \cdot P(R|A)}{P(A) \cdot P(R|A) + P(B) \cdot P(R|B) + P(C) \cdot P(R|C) + P(D) \cdot P(R|D)} \)

Where: P(A) = \( \frac{1}{4} \), P(B) = \( \frac{1}{4} \), P(C) = \( \frac{1}{4} \), P(D) = \( \frac{1}{4} \), P(R|A) = \( \frac{1}{10} \), P(R|B) = \( \frac{6}{10} \), P(R|C) = \( \frac{8}{10} \), P(R|D) = 0

\( P(A|R) = \frac{(\frac{1}{4})(\frac{1}{10})}{(\frac{1}{4})(\frac{1}{10}) + (\frac{1}{4})(\frac{6}{10}) + (\frac{1}{4})(\frac{8}{10}) + (\frac{1}{4})(0)} = \frac{\frac{1}{10}}{\frac{1}{10} + \frac{6}{10} + \frac{8}{10}} = \frac{1}{15} \)

Therefore, the probability that the selected red marble came from box A is \( \frac{1}{15} \).
In simple words: Box A has very few red marbles (only 1 out of 10), while boxes B and C have many red marbles. When a red marble is drawn, it is far more likely to have come from B or C rather than from A.

Exam Tip: Box D contains no red marbles, so its contribution to the denominator is zero - this eliminates it from consideration immediately, simplifying the problem.

Question 16. A car manufacturing factory has two plants X and Y. Plant X manufactures 70% of the cars, and plant Y manufactures 30%. At plant X, 80% of the cars are rated of standard quality, and at plant Y, 90% are rated of standard quality. A car is picked up at random and is found to be of standard quality. Find the probability that it has come from plant X.
Answer: Let S denote standard quality, X denote plant X, and Y denote plant Y. We want to find P(X|S), the probability that a selected standard quality car came from plant X. Using Bayes's theorem:

\( P(X|S) = \frac{P(X) \cdot P(S|X)}{P(X) \cdot P(S|X) + P(Y) \cdot P(S|Y)} \)

Where: P(X) = \( \frac{70}{100} \), P(Y) = \( \frac{30}{100} \), P(S|X) = \( \frac{80}{100} \), P(S|Y) = \( \frac{90}{100} \)

\( P(X|S) = \frac{(\frac{70}{100})(\frac{80}{100})}{(\frac{70}{100})(\frac{80}{100}) + (\frac{30}{100})(\frac{90}{100})} = \frac{5600}{5600 + 2700} = \frac{5600}{8300} = \frac{56}{83} \)

Therefore, the probability that the selected standard quality car came from plant X is \( \frac{56}{83} \).
In simple words: Even though plant Y produces a higher percentage of quality cars (90% versus 80%), plant X makes far more cars overall. This means most standard quality cars in the market came from plant X, despite its slightly lower quality rate.

Exam Tip: When one source is much larger than another, its posterior probability remains high even if its conditional probability is slightly lower - volume and quality both influence the final result.

Question 17. An insurance company insured 2000 scooters and 3000 motorcycles. The probability of an accident involving a scooter is 0.01, and that of motorcycles is 0.02. An insured vehicle met with an accident. Find the probability that the accidented vehicle was a motorcycle.
Answer: Let M denote a motorcycle, S denote a scooter, and A denote an accident. We want to find P(M|A), the probability that an insured vehicle involved in an accident is a motorcycle. The proportion of scooters is \( \frac{2000}{5000} = \frac{2}{5} \), and the proportion of motorcycles is \( \frac{3000}{5000} = \frac{3}{5} \). Using Bayes's theorem:

\( P(M|A) = \frac{P(M) \cdot P(A|M)}{P(M) \cdot P(A|M) + P(S) \cdot P(A|S)} \)

Where: P(M) = \( \frac{3}{5} \), P(S) = \( \frac{2}{5} \), P(A|M) = 0.02, P(A|S) = 0.01

\( P(M|A) = \frac{(\frac{3}{5})(0.02)}{(\frac{3}{5})(0.02) + (\frac{2}{5})(0.01)} = \frac{0.012}{0.012 + 0.004} = \frac{0.012}{0.016} = \frac{3}{4} \)

Therefore, the probability that the vehicle involved in an accident is a motorcycle is \( \frac{3}{4} \).
In simple words: Motorcycles are both more numerous in the fleet and twice as accident-prone as scooters. This combination makes it very likely that any accident reported was caused by a motorcycle.

Exam Tip: Always convert counts to proportions first - here, using the relative frequencies of scooters and motorcycles as prior probabilities is essential for an accurate calculation.

Question 18. In a bulb factory, machines A, B and C manufactures 60%, 30% and 10% bulbs respectively. Out of these bulbs 1%, 2% and 3% of the bulbs produced respectively by A, B and C are found to be defective. A bulb is picked up at random from the total production and found to be defective. Find the probability that this bulb was produced by machine A.
Answer: Let D denote a defective bulb, and A, B, C denote the three machines. We want to find P(A|D), the probability that a selected defective bulb was produced by machine A. Using Bayes's theorem:

\( P(A|D) = \frac{P(A) \cdot P(D|A)}{P(A) \cdot P(D|A) + P(B) \cdot P(D|B) + P(C) \cdot P(D|C)} \)

Where: P(A) = \( \frac{60}{100} \), P(B) = \( \frac{30}{100} \), P(C) = \( \frac{10}{100} \), P(D|A) = \( \frac{1}{100} \), P(D|B) = \( \frac{2}{100} \), P(D|C) = \( \frac{3}{100} \)

\( P(A|D) = \frac{(\frac{60}{100})(\frac{1}{100})}{(\frac{60}{100})(\frac{1}{100}) + (\frac{30}{100})(\frac{2}{100}) + (\frac{10}{100})(\frac{3}{100})} = \frac{60}{60 + 60 + 30} = \frac{60}{150} = \frac{2}{5} \)

Therefore, the probability that the selected defective bulb was produced by machine A is \( \frac{2}{5} \).
In simple words: Machine A produces the most bulbs overall, but its defect rate is the lowest. Machines B and C have higher defect rates. When a defective bulb is found, the probabilities balance out: machine A's volume is offset by its lower defect rate, while machines B and C contribute more per unit despite lower volumes.

Exam Tip: Always include all three (or more) machines in the denominator, even if one has a very low defect rate - omitting any term will invalidate the entire calculation.