$A$ and $B$ are random occurrences in $\Omega$. Prove that if $P(A)=0{,}9$ and $P(B)=0{,}7$, then $P(A\cap B')\leq0{,}3$, where $B'$ is a complementary event of $B$.
I thought of something like this: $P(B')=1-P(B)=0{,}3$ and as the intersection of events can't be greater than any of the events it's taken from, $P(A\cap B')\leq0{,}3$ q.e.d.
Is it OK? When I look at the answer the author gave to this task, it includes using the formula for the probability of the sum of occurrences. Is it needed or my solution works as well?