recently I bumped into following puzzle and I would like to validate(or correct) my results as I asked several people and got several different answers.
You are planning a picnic with your friends and you are wondering whether it will rain. You look at three different weather forecasts: two of them say it will be rainy, third says it won't. You know that each forecast only has a 3/4 chance of being correct. What is the probability that it will rain during your picnic? You can assume that the prior probability of rain is 1/5.
My answer to this question is 35% but I found myself that I am failing in basic intuition whether this result can be correct.
Did I missed something important in my consideration? Trying to get an intuition behind equations (Bayesian theorem, Joint Probability, conditional probability etc.)
Thanks
$$P(R)=1/5$$ For each forecast $i=1,2,3$ $$P(H_i = R|R)= 3/4$$ and $$P(H_i=R|\bar{R})=1/4$$
We need to find $$P(R|H_1=R, H_2= R, H_3 = \bar{R}) = \frac{P(R) P(H_1=R, H_2=R, H_3=\bar{R}|R)}{P(H_1=R, H_2=R, H_3=\bar{R})} = \frac{P(R) P(H_1=R, H_2=R, H_3=\bar{R}|R)}{P(R) P(H_1=R, H_2=R, H_3=\bar{R}|R)+P(\bar{R}) P(H_1=R, H_2=R, H_3=\bar{R}|\bar{R})} =\frac{1/5 \times 3/4\times 3/4\times 1/4}{1/5 \times 3/4\times 3/4\times 1/4 + 4/5 \times 1/4\times 1/4\times 3/4} = \frac{9}{9 + 12} = 0.428$$