# 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 $A$?

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

• @Robert Z is it the actual question or the formulas I used to solve the problem? Aug 11, 2019 at 16:18
• Recall that "The person who asked can mark one answer as accepted". See math.stackexchange.com/tour Aug 12, 2019 at 17:22

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 $$x=\frac 14$$.

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 $$P(A)$$.

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 $$P(A)=1/4$$.

• @KatieMelosto Any further doubt? Aug 12, 2019 at 9:08

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