If $\mathbb P(A)=0.61, \mathbb P(B)=0.56$ and $\mathbb P(A\cup B)=0.81$. Find $\mathbb P(A \mid A\cap B)$

Question

I’ve figured out what $(A\cap B)$ is, which is $0.36.$

I tried putting 0.61 over 0.36 but the answer is over 1 which is obviously incorrect. How do I answer this question?

Answer 1

Obverse that $A \cap B \subseteq A\implies P(A|A\cap B) = 1$. Or maybe you want to find $P(A|A \cup B)$ ?

Answer 2

Given that you are in $A\cap B$ you are in $A$ with probability 1

Answer 3

If $A$ and $B$ both occur, then certainly $A$ occurs, so the probability is $1$. If you want to use the formula,

$$\Pr(A|A\cap B)=\frac{\Pr(A\cap(A\cap B))}{\Pr(A\cap B)}=\frac{\Pr(A\cap B)}{\Pr(A\cap B)}=1$$