0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ If all chocolates are dark, how is it possible to observe a chocolate that isn't dark? What's the probability of drawing a red ball from an urn given that 50% of the (otherwise identical) balls are red? (In your formulation, is "black chocolate" the same as "dark chocolate?") $\endgroup$ – Matthew Gunn Feb 12 at 3:11
  • $\begingroup$ @MatthewGunn Sorry for the confusion––I meant for black and dark to be synonymous. I updated the question. $\endgroup$ – Shrey Feb 12 at 3:18
  • $\begingroup$ Please add the self-study tag. $\endgroup$ – Xi'an Feb 12 at 5:44
  • 1
    $\begingroup$ To repeat @MatthewGunn's comment, "the probability that a black chocolate is observed given that all chocolates are black" is not the probability that all chocolates are black. $\endgroup$ – Xi'an Feb 12 at 5:47
  • $\begingroup$ What is the relevance of $q$? It is mentioned but not used. I'm not sure of the meaning of "we make an observation of a chocolate with probability $q$, independent of $A$". $\endgroup$ – CarbonFlambe Feb 14 at 16:49
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.