If given $P(B\mid A) =4/5$, $P(B\mid A^\complement)= 2/5$ and $P(B)= 1/2$, what is the probability of $A$?

Question

If given $P(B\mid A) =4/5$, $P(B\mid A^\complement)= 2/5$ and $P(B)= 1/2$, what is the probability of $A$?

I know I need to apply Bayes theorem here to figure this out, but I'm struggling a bit to understand how.

So far I've considered this formula: $$P(B\mid A) = \dfrac{P (B \cap A) }{ P (B \cap A) + P(B^\complement \cap A)}$$

From this formula, I understand that $P(B \cap A) = P(A) \cdot P(B\mid A)$ so I plug in the given values but then only find that $P(B^\complement |A)$ is $2/25$. But this does not get me any closer to my goal, $P(A)$.

I imagine my understanding of this is quite backward. Any pointers would be helpful.

Thank you

Answer 1

You know that $$P(B)=P(B|A)P(A)+P(B|A^C)P(A^C)$$ and $P(A^C)=1-P(A)$. From there, it's just plugging in and solving for $P(A)$.

Answer 2

All you need is the definition of conditional probability: $$P(X|Y)=P(X\cap Y)/P(Y).$$ We have that $$P(A^c\cap B)=P(B|A^c)P(A^c)=\frac{2}{5}(1-P(A))$$ and $$P(A \cap B)=P(B|A)P(A)=\frac{4}{5}P(A).$$ Hence $$\frac{1}{2}=P(B)=P(A^c\cap B)+P(A \cap B)=\frac{2}{5}(1-P(A))+\frac{4}{5}P(A),$$ and, after solving it, we easily find that $P(A)=1/4$.

Answer 3

Use the Venn diagram. Let the blue side be the even that $A$ occurs. Also inside the ellipse be the event that $B$ occurs. Then the $\mathbb{P}(B|A)=\frac45$ means that the light blue section forms $4/5$ of the whole blue side. Similarly, the yellow section forms $2/5$ of the non-blue section.

Finally, let $x=P(A)$ and you have

$$\frac 45 x + \frac 25 (1-x)=\frac 12,$$

which implies that $x=\frac 14$.

Answer 4

Using venn diagram one of easy way to find solution(https://i.stack.imgur.com/ZiYVm.jpg)