# Easy probability theory proof

$A$ 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?

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Your solution is perfectly good. There are often many ways to solve a problem, and the ones chosen by the authors of textbooks (or perhaps by some poor student who was employed to produce solutions) are often not the best. –  Robert Israel May 7 '12 at 15:54
Thank you very much! :) –  Straightfw May 7 '12 at 15:54
$\TeX$ assumes the use of decimal points. Commas are interpreted as punctuation; hence the wrong spacing in your numbers. –  joriki May 7 '12 at 16:08