Proof of: If $P(A) = P(B) = 1$ then $P(A \cap B) = 1$. Ok so I know this is obviously true but not sure if my method is right to proving it. Here's my go.
$P(A\cap B)= P(A) \cdot P(B)$ by definition.
$P(A)=P(B)=1$
Therefore $1\cdot 1=1$ so $P(A\cap B)=1$
I realise this is a really simply question but this answer seems too easy.
Thanks
 A: Well, you cannot say $P(A\cap B)= P(A) \cdot P(B)$ because you do not know if $A$ and $B$ are independent. 
Anyway, you could say $$P(A\cap B) = 1-P((A\cap B)^c)=1-P(A^c \cup B^c) \geq 1-(P(A^c)+ P(B^c))= 1-(0+0)=1$$
A: Note that $A\subset A\cup B$ and $B\subset A\cup B$ implies $P(A\cup B)\geq P(A)$ and $P(A\cup B)\geq P(B)$. Then $P(A\cup B)=1$. By Inclusion–exclusion principle 
$$
P(A\cup B)= P(A)+P(B)-P(A\cap B). 
$$
So
$$
1=1+1-P(A\cap B).
$$
A: We begin with $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
Since $P(A)=P(B)=1$, we can rearrange this as 
$$P(A\cap B)=2-P(A\cup B)$$
While we might not know $P(A\cup B)$, we know that $P(A\cup B)\le 1$ (true for all probabilities), and hence $$P(A\cap B)\ge 2-1=1$$
Since also $P(A\cap B)\le 1$ (true for all probabilities), we can be sure that $P(A\cap B)=1$.
A: $P(A|B)=\frac{P(A \bigcap B)}{P(B)}$, thus $P(A \bigcap B)=P(A|B)P(B)$. 
If $P(A)=1$, then that means that $A$ always happens, so it will still always happen when $B$ happens, so $P(A|B)=1$. Since $P(B)=1$, $P(A \bigcap B)=1$.
