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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.

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    I'm certain that you need some assumptions about (in)dependence to answer this question. Is this the full description of the question? – 2017-02-28
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    I think it simply is the product $\frac{4}{5}\times \frac{2}{3}\times \frac{3}{4}$ – 2017-02-28
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    Yes, this is the complete question – 2017-02-28
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    Yeah, one of the option is 2/5. Can you explain it frederick – 2017-02-28
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    Unfortunately, @David is right, your question (wherever you got it from) is poorly stated and has no answer. You need some assumptions about independence of the people. You also need to define what exactly "credibility" means -- the probably that, given it is actually raining, the person says so correctly? Or just the probability that they are right overall? – 2017-02-28
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    This question is from JEE Mains paper. Joint Entrance Examination (JEE) is an all India common engineering entrance examination conducted for admission to various engineering colleges and courses all over the country. The test comprises of two stages - JEE Main and the JEE Advanced. The exams are of the objective pattern. – 2017-02-28
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    Is there an option for "not enough information"? – 2017-02-28
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    @6005 it doesn't say _given that it is actually raining_ so I think it should be the overall accuracy of the speaker. – 2017-02-28
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    (A) 2/5 (B) 21/25 (C) 3/4 (D) 24/25 – 2017-02-28
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    And I think my answer is wrong. – 2017-02-28
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    OK @frederick99,... but if that were the case, then that makes the question unanswerable because we don't know if the speaker is more likely to be right when it does rain, or when it doesn't rain. That will affect the probability. By the way, your calculation of the product $2/5$ is wrong by all accounts. You are saying that it is *less likely* to be raining after the two people claim it is raining, than it was before! – 2017-02-28
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    The answer they are probably looking for is (D), i.e. $\frac{(3/4) (2/3) (4/5)}{(3/4) (2/3) (4/5) + (1/4) (1/3) (1/5)}$. However, the question is terrible. There is no way you can assume the two people's predictions to be independent, since they did not state that you can (and this would not be true in real life). – 2017-02-28
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    Why can two people not predict independently, unless it is mentioned otherwise. – 2017-02-28
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    @frederick99 If CNN and NBC both make a predication as to whether it is raining, would you expect them to be independent from each other? No, you would not. It is more likely that they are both wrong or both right. Similarly here. – 2017-02-28
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    @6005 There is no saying if they were consulting each other. It could be that they did not even meet. In that case, the predictions _are_ independent. – 2017-02-28
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    @frederick99 Not so. It is not said how they are able to make an accurate prediction, but if they use the same data, or even if they use correlated data (which would be the case in real life), they will come up with dependent predictions. *Independent* doesn't mean that they didn't consult each other, it means their probabilities are distributed independently, which is a very strong assumption and usually a wrong one. – 2017-02-28
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    @6005 Will not the predictions depend on the data itself, rather than other's opinion? – 2017-02-28
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    @frederick99 Moreover, this is a mathematics question, you should never have to make assumptions that are necessary but omitted. The question is incomplete and very poorly worded at best. – 2017-02-28
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    @frederick99 Do you know what [independent](https://www.statlect.com/fundamentals-of-probability/independent-random-variables) means? Depending on the data means they will not be *independent* events. I.e., it will not be the case that $P(A \cap B) = P(A) P(B)$ (where $A$ is the event that person 1 predicts rain, and $B$ is the event that person 2 predicts rain). – 2017-02-28
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    Okay, I'm not sure if I get the dependence _as in english_ but David's solution is quite explanatory. _I think I should just stick to StackOverflow :)_ – 2017-02-28

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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.

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    @hanzala I see you've unmarked my answer as accepted. Is there something I can do to improve it? Are you simply hoping for more responces? – 2017-03-02
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    @hanzala no problem, I just wanted to make sure you didn't have any additional questions – 2017-03-02
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    $@David$ you explained it very well – 2017-03-02