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?

score 3 · Answer 1 · answered Mar 20 '21 at 01:09

3

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

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

answered Mar 20 '21 at 01:09

saulspatz

Your explanation makes a lot of sense, but I don’t understand the working out. Are you able to explain it? – Somxr Mar 20 '21 at 01:38
1

Do you know the formula for conditional probability? It's $$\Pr(X|Y)=\frac{\Pr(X\cap Y)}{\Pr(Y)}$$ I'm just applying it with $A=X, Y=A\cap B$. – saulspatz Mar 20 '21 at 02:31
But do you see the first A in Pr(∩(∩))? Why does it disappear? Apologies for asking too much. You’ve been very helpful – Somxr Mar 20 '21 at 02:56
1

When you have a question like this, I suggest you draw a Venn diagram. That will be more convincing than the formal proof, which is $$A\cap(A\cap B)=(A\cap A) \cap B = A\cap B$$ – saulspatz Mar 20 '21 at 06:16

score 1 · Answer 2 · answered Mar 20 '21 at 00:41

1

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

answered Mar 20 '21 at 00:41

score 1 · Answer 3 · answered Mar 20 '21 at 00:47

1

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

answered Mar 20 '21 at 00:47

tommik

What can be more straightforward than this? :-) – Star Bright Mar 20 '21 at 01:40

3 Answers3