So I've been stuck on Bayes' Theorem for a while. I got some help from another extremely patient user but I'm just not seeming to get it.
The exact prompt: Consider an election with two candidates, Candidate B and Candidate X. Every voter is invited to participate in an exit poll, where they are asked whom they voted for; some accept and some refuse. For a randomly selected voter, let B be the event that they voted for B, and A be the event that they are willing to participate in the exit poll. Suppose that P(A|B) = 0.7 but P(A|Bc) = 0.3. In the exit poll,60%of the respondents say they voted for B (assume that they are all honest), suggesting a comfortable victory for B. Find P(B),the true proportion of people who voted for B.
My known:
P(A | B) = .7
P(A | ~B) = .3
P(B | A) = .6
I need to figure out P(B) but I can't seem to do that without knowing P(A n ~B).
My work is as follows: https://i.stack.imgur.com/dPstW.jpg
I feel like I'm missing a fundamental concept of sets. If anyone could point me in the right direction, I'd really appreciate it. Please be patient with me
By Bayes' Theorem, we have $$P(B|A) = \frac{P(A|B)P(B)}{P(A)}= \frac{P(A|B)P(B)}{P(A|B)P(B) + P(A|B^c)P(B^c)}$$ Unlike in a more "traditional" Bayes' Theorem exercise, here we're asked to find not $P(B|A)$ but $P(B)$. To make the equations a bit easier to read, let's introduce some shorthand. Let $p \equiv P(B)$ and define $\alpha \equiv P(A|B)$ and $\beta \equiv P(A|B^c)$ along with $\gamma \equiv P(B|A)$. Then the preceding equality becomes: $$\gamma = \frac{\alpha p}{\alpha p + \beta (1 - p)} = \frac{1}{1 + \displaystyle\frac{\alpha}{\beta} \frac{(1 - p)}{p}}$$ Now, we are given $\alpha, \beta$, and $\gamma$. Can you see how we can use this to solve for $p$?