Bayes Theorem Questions and Answers

Q: A retail shop sells three brands of tea, namely Brand A, Brand B, and Brand C, and each in two varieties, namely regular and premium. The proportion of customers buying the Brands A, B, and C are 40%, 35%, and 25%, respectively. Out of those customers who buy Brand A, 30% buy the premium variety, out of those who buy Brand B, 40% buy the premium variety, while out of those who buy Brand C, 60% buy the premium variety. Given that a randomly selected customer has bought the premium variety of tea, the probability that he/she has bought Brand B equals

The probability is \(\frac{14}{41}\).Let \(Pr\) be the event that the customer buys premium. By the theorem of total probability:\[ \begin{aligned} P(Pr) &= 0.4(0.3) + 0.35(0.4) + 0.25(0.6) \\ &= 0.12 + 0.14 + 0.15 = 0.41 \end{aligned} \]Then by Bayes' theorem:\[ \begin{aligned} P(B \mid Pr) &= \frac{0.35 \times 0.4}{0.41} \\ &= \frac{0.14}{0.41} = \frac{14}{41} \end{aligned} \]About 34%, a little under the 35% share of Brand B overall, because Brand B's premium rate (40%) is close to the average premium rate (41%). The other fractions do not arise from this data.

Q: In 1989 there were three candidates for the position of Principal, Mr. Chatterji, Mr. Ayangar and Dr. Singh, whose chances of getting the appointment are in the proportion 4 : 2 : 3 respectively. The probability that Mr. Chatterji if selected would introduce co-education in the college is 0.3. The probabilities of Mr. Ayangar and Dr. Singh doing the same are respectively 0.5 and 0.8. What is the probability that there was co-education in the college in 1990 ?

The probability is \(\frac{23}{45}\).The chances 4 : 2 : 3 give selection probabilities of 4/9, 2/9 and 3/9. By the theorem of total probability:\[ \begin{aligned} P &= \tfrac{4}{9}(0.3) + \tfrac{2}{9}(0.5) + \tfrac{3}{9}(0.8) \\ &= \frac{1.2 + 1.0 + 2.4}{9} = \frac{4.6}{9} \\ &= \frac{46}{90} = \frac{23}{45} \end{aligned} \]The other fractions come from wrong denominators: 55 does not arise from the 4 + 2 + 3 = 9 split, and 32/45 is not the weighted sum. A quick check: the answer, about 0.51, must lie between the smallest and largest individual probabilities, 0.3 and 0.8.

A retail shop sells three brands of tea, namely Brand A, Brand B, and Brand C, and each in two varieties, namely regular and premium. The proportion of customers buying the Brands A, B, and C are 40%, 35%, and 25%, respectively. Out of those customers who buy Brand A, 30% buy the premium variety, out of those who buy Brand B, 40% buy the premium variety, while out of those who buy Brand C, 60% buy the premium variety. Given that a randomly selected customer has bought the premium variety of tea, the probability that he/she has bought Brand B equals

A.\(\frac{14}{41}\)
B.\(\frac{13}{31}\)
C.\(\frac{15}{51}\)
D.\(\frac{16}{61}\)

Exam Year(s)

IIT JAM - 2026,

Answer: A

The probability is \(\frac{14}{41}\).

Let \(Pr\) be the event that the customer buys premium. By the theorem of total probability:

\[ \begin{aligned} P(Pr) &= 0.4(0.3) + 0.35(0.4) + 0.25(0.6) \\ &= 0.12 + 0.14 + 0.15 = 0.41 \end{aligned} \]

Then by Bayes' theorem:

\[ \begin{aligned} P(B \mid Pr) &= \frac{0.35 \times 0.4}{0.41} \\ &= \frac{0.14}{0.41} = \frac{14}{41} \end{aligned} \]

About 34%, a little under the 35% share of Brand B overall, because Brand B's premium rate (40%) is close to the average premium rate (41%). The other fractions do not arise from this data.

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In 1989 there were three candidates for the position of Principal, Mr. Chatterji, Mr. Ayangar and Dr. Singh, whose chances of getting the appointment are in the proportion 4 : 2 : 3 respectively. The probability that Mr. Chatterji if selected would introduce co-education in the college is 0.3. The probabilities of Mr. Ayangar and Dr. Singh doing the same are respectively 0.5 and 0.8. What is the probability that there was co-education in the college in 1990 ?

A.\(\frac{32}{55}\)
B.\(\frac{32}{45}\)
C.\(\frac{23}{55}\)
D.\(\frac{23}{45}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The probability is \(\frac{23}{45}\).

The chances 4 : 2 : 3 give selection probabilities of 4/9, 2/9 and 3/9. By the theorem of total probability:

\[ \begin{aligned} P &= \tfrac{4}{9}(0.3) + \tfrac{2}{9}(0.5) + \tfrac{3}{9}(0.8) \\ &= \frac{1.2 + 1.0 + 2.4}{9} = \frac{4.6}{9} \\ &= \frac{46}{90} = \frac{23}{45} \end{aligned} \]

The other fractions come from wrong denominators: 55 does not arise from the 4 + 2 + 3 = 9 split, and 32/45 is not the weighted sum. A quick check: the answer, about 0.51, must lie between the smallest and largest individual probabilities, 0.3 and 0.8.

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