1

If $B⊂A, P(A)=0.6, P(B)=0.4,$ what is $P(A∣B)?$

$A. 2/5$

$B. 3/5$

$C. 1/3$

$D. 2/3$

now since B is a subset of A. I know $P(A \cap B) = 0.4 $

So I thought the answer is $1$ because $P(A | B)$ = $\cfrac {P(A \cap B)}{P(B)} = 1$

But the final answer is D: $\frac 23$

The reasoning behind the answer is as follows

$P(A\cap B)=P(B)\cdot P(A|B)= \cfrac{P(A\cap B)}{P(A)}=\frac 23.$

But I don't understand why you divide by $P(A)$ in the above. Isn't conditional probability formula $P(A | B)$ = $\cfrac {P(A \cap B)}{P(B)}$ ?

3 Answers3

3

It's a simply typo in the question. The solution they show is the probability

$$\mathbb{P}[B|A]$$

That's all. Your answer is correct.

tommik
  • 32,733
  • 4
  • 15
  • 34
  • 1
    It's worse than that. Where do they get the equality $$P(B)\cdot P(A|B)=\frac{P(A\cap B)}{P(A)}\ ??$$ – bof Sep 28 '20 at 04:57
  • Thank you all, I was afraid I was crazy. I love you all. –  Sep 28 '20 at 05:02
1

B is the subevent/subset of A, that means if B occurs then A occurs, so P(A|B)=1.

If we compute P(B|A) then P(B|A)= P(AB)/P(A) = P(B)/P(A) =0.4/0.6=2/3.

kelffon
  • 11
0

By drawing a Venn-diagram, it is clear that if B occurs then with a probability of 1 A occurs. If the question is about $P(B|A)$ then $P(B|A) = \frac{P(A|B)P(B)}{P(A)} = \frac{1*2/3}{3/5} = 2/3$

MOHAMMED
  • 169
  • 5