Bayesian Inference Toy Problem

Question

Problem statement: Consider a probabilistic model where there are two states of the world, framed as complimentary events: $A$: All chocolates are black and $A^C$: 50% of chocolates are black. Let $p$ be the prior probability $P(A)$ that all chocolates are black. Assume we make an observation of a chocolate with probability $q$, independent of $A$. Also assume $0 \lt p,q \lt 1$. Given the event $B$: a black chocolate is observed, what is $P(A|B)$?

I used Bayes' Rule to expand $P(A|B)$: $$ P(A|B) = \frac{P(B|A)P(A)}{P(B)} \\ = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|A^C)P(A^C)} \\ = \frac{p^2}{p^2 + (1-p)^2} $$

I'm slightly unclear about the conditional probabilities $P(B|A)$ and $P(B|A^C)$.

• $P(B|A)$ reads as "the probability that a black chocolate is observed given that all chocolates are black." I would interpret this to have a probability of $p$ since $A$ occurs with probability $p$.
• $P(B|A^C)$ reads as "the probability that a black chocolate is observed given that 50% of chocolates are black." I would interpret this to have probability $1-p$ since $A^C$ occurs with probability $1-p$.

Are these interpretations to solve $P(A|B)$ correct?

Answer 1

When you calculate conditional probabilities $P(B|A),P(B|A^c)$ etc remember that $A$ and $A^c$ respectively have already occurred. You'll need to revisit your calculations.

Just to give you an example of what I mean: Let's say you have 20 people in a room and 1 of them is a man with a beard and 19 are women (without a beard). If you choose one person at random then $P(man) = 1/20$, $P(beard) = 1/20$ but $P(beard|Man) = 1$. The problem in the way you calculate conditional probabilities in your example is that you account that only 1 man in the room but you shouldn't!! Given that the chosen person is a man, then the conditional probability of him having a beard is 100% . I hope that's clear now.