Incorporating evidence
You roll a weighted die and intend to observe the number of spots X. You know that, due to the weighting, higher numbers are more likely: in particular, P(X = x) = (x + 2) / 33 for x from 1 to 6.
The die rolls under the table, so that you can't see it. However, a friend tells you that the number of spots is even. You know that your friend is not that reliable: he only tells the truth about this sort of thing with probability 3/4 , and lies with probability 1 /4 . What is the probability that X = 5 given the evidence above? Hint: use Bayes rule to incorporate the noisy observation of the die roll. You can use the version where you don't need to know the denominator ahead of time; instead you can calculate the denominator using the fact that probabilities must sum to 1.
I'm going along the first comment and calculate the joint as P(S|X=5) * P(X=5) which equals 1/4 * 7/33. Given everything computed so far I've reached the final answer P(X=5|S) = 7/69. Is this correct?