Probability (Bayes' Theorem)

Question

The probability that a particular day is a rainy day is $3/4$. Two persons whose credibility are $4/5$ and $2/3$ claim that the day was a rainy day. What is the probability that it was actually a rainy day?

I think that I need to apply Bayes' theorem here, but don't know how? This question had come in my exam.

Answer 1

Most likely, the question wants you to assume independence between the two people. Additionally, I'm to assume that the idea of "credibility" simply means the probability that they say one thing given it is true. Let us denote raining by $R$, not raining by $N$ and the $i$th person saying it rains by $C_i$ ($i=1,2$). We seek the probability that it is raining given both people say it is raining, i.e., $$P(R|C_1 \cap C_2).$$ In painstaking detail, here is how we work the problem out,

1. By 'credibility,' $P(C_1|R) = \frac{4}{5}$, $P(C_2|R) = \frac{2}{3}$, $P(C_1|N) = \frac{1}{5}$, $P(C_2|N) = \frac{1}{3}$
2. By independence, $P(C_1C_2|R) =P(C_1|R)P(C_2|R) = \frac{4}{5}\cdot\frac{2}{3} = \frac{8}{15}.$
3. By independence, $P(C_1C_2|N) =P(C_1|N)P(C_2|N) = \frac{1}{5}\cdot\frac{1} {3} = \frac{1}{15}.$
4. By probability of raining, $P(R) = \frac{3}{4}$ and $P(N) = \frac{1}{4}.$
5. By Law of Total Probability, $P(C_1C_2) = P(C_1C_2|R)P(R)+P(C_1C_2|N)P(N) = \frac{25}{60}$
6. By Bayes', $P(C_1C_2R) = P(C_1C_2|R)P(R) = \frac{8}{15}\cdot \frac{3}{4} = \frac{2}{5}.$
7. By Bayes', $P(R|C_1C_2) = \frac{P(C_1C_2R)}{P(C_1C_2)} = \frac{2/5}{25/60} = \frac{24}{25}.$

This means answer "D" would seem to be the one you should go with.