If $\mathbb{P}(A)=\mathbb{P}(B)=1$, then $\mathbb{P}(A\cap B)=1$?

Question

If $\mathbb{P}(A)=\mathbb{P}(B)=1$, then $\mathbb{P}(A\cap B)=1$?

We can use this:

$\mathbb{P}(A\cap B) = \mathbb{P}(A|B)\mathbb{P}( B)=\mathbb{P}(A|B)$

Then, we have to find counter examples for dependent events $A, B$ such that $\mathbb{P}(A|B)<1$. Are there any examples?

There are answers that provide proofs of "yes". Intuition leads to that answer too: if each of $A$ and $B$ is sure to happen (no matter what) then both are sure to happen. — Ethan Bolker, May 11 '17 at 19:05

score 6 · Accepted Answer · answered May 11 '17 at 18:51

6

From $$\mathbb{P} (A\cup B) = \mathbb{P} (A)+\mathbb{P} (B)-\mathbb{P} (A\cap B)$$ and $1=\mathbb{P} (A)\leq \mathbb{P} (A\cup B)\leq 1$, we get $$ 1 = 2 - \mathbb{P} (A\cap B)$$ so $\mathbb{P} (A\cap B)=1$.

answered May 11 '17 at 18:51

Clement C.

67,323

score 1 · Answer 2 · answered May 11 '17 at 18:50

1

$P(A\cap B)\ge P(A)+P(B)-1=1$ since $P(A)=P(B)=1$. Hence $P(A\cap B)=1$ since $P(A\cap B )\le1$

answered May 11 '17 at 18:50

bah bhai bah!!! – Arpan1729 May 12 '17 at 02:48

If $\mathbb{P}(A)=\mathbb{P}(B)=1$, then $\mathbb{P}(A\cap B)=1$?

2 Answers2

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