Homework - Probability of raining

Question

When I was doing my homework in probability, I encounter this question.

"The weather is raining 70% of the time. There are two people that provide weather forecast, A and B, which their prediction are wrong 10% and 40% respectively. Their predictions performance is independent. Given A predict it will rain and B predict it will not, find the probability that the weather is raining."

I can divide the situation into 4 case, which are ‘both A and B are right’, ‘A is right and B is wrong’, ‘A is wrong and B is right’ and ‘both A and B are wrong’, and then use P(A is right and B is wrong) as the answer. This computation is rather simple, but I think I should not ignore ‘the weather is raining 70% of the time’ in the calculation.

Can anyone please tell me if I am missing something in the calculation. This question is related to the topic of conditional probability, total probability theorem, Bayes’ Rule and independence. Thank you.

Answer 1

Let $R$ be the event it rains, $A$ and $B$ be the events that $A$ and $B$ predict it rains. We want $P(R|\,A\cap B^c)$, which you can use Baye's rule to find: $$ P(R|\,A\cap B^c) = \frac{P(A\cap B^c|R)\cdot P(R)}{P(A\cap B^c)}\\ \qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\,\,\,\,\,=\frac{P(A\cap B^c|R)\cdot P(R)}{P(A\cap B^c|R)\cdot P(R)+P(A\cap B^c|R^c)\cdot P(R^c)} $$ You can now find each of the terms in the last expression.

Answer 2

Call $W$, $A$ and $B$ the weather and its predictions by $A$ and $B$, each being $r$ (rain) or $s$ (no rain). Then, $$P(W=r,A=r,B=s)=P(W=r)P(A\ \text{true})P(B\ \text{false}),$$ and $$P(W=s,A=r,B=s)=P(W=s)P(A\ \text{false})P(B\ \text{true}),$$ hence $$P(W=r\mid A=r,B=s)=\frac{P(W=r,A=r,B=s)}{P(W=r,A=r,B=s)+P(W=s,A=r,B=s)}$$ is $$\frac{P(W=r)P(A\ \text{true})P(B\ \text{false})}{P(W=r)P(A\ \text{true})P(B\ \text{false})+P(W=s)P(A\ \text{false})P(B\ \text{true})}.$$ Exercise: List the hypotheses missing from the statement of the problem, that are necessary to justify the computations above.