# If mutually exclusive, prove Pr(A) <= Pr(B')

1. Sep 7, 2010

### bjersey

I'm having issues proving the following which should be simple:

If A and B are mutually exclusive, prove Pr(A) <= Pr(B')

From the statement about being mutually exclusive, I know A $$\cap$$ B = $$\phi$$

Therefore we have P(A $$\cap$$ B) = Pr(A) + Pr(B)

Also, A = A $$\cap$$ B'
and B = A' $$\cap$$ B

But I'm having a hard time putting all of this together.

2. Sep 7, 2010

### mathman

A = A $$\cap$$ B' implies A is a subset of B'.

3. Sep 17, 2010

### _joey

Here's my solution.

Pr(A union B) <= Pr(Entire Sample Space)=1 from probability axiom; and
Pr(A union B) = Pr(A)+Pr(B)-Pr(A intersection B)

Hence, Pr(A intersection B) >= Pr(A) + Pr(B) -1 ........(3)

But for two mutually exclusive events Pr(A intersection B) = P(empty) = 0. Also, Pr(B) = 1- Pr(B')

It follows from (3) that

0>=Pr(A)+1-Pr(B')-1

Therefore, Pr(B')>=Pr(A)

As required

4. Sep 17, 2010

You could cut the final proof down a little:

\begin{align*} P(A) + P(B) - P(A \cap B) & \le 1 \\ P(A) + P(B) & \le 1 \\ P(A) & \le 1 - P(B) \\ P(A) & \le P(B') \end{align*}

5. Sep 17, 2010

### _joey

The inequality in (3) is well known named after Bonferonni. It has many applications. Its derivation is good to know too. Anyway, your answer is neat.

6. Sep 18, 2010

Yes, Bonferroni's inequality is important (classical example is in the first study of multiple comparisons), but the original question wasn't about that; usually you want the derivations to be as straightforward as possible.

there is nothing wrong with the earlier solution, but the mix of mathematics and english makes its reading awkward. learning when the written word can be safely removed from the mathematical work is an important step as well.

7. Sep 18, 2010

### mathman

I think my comment is the simplest. A is a subset of B', therefore P(A) ≤ P(B').

8. Sep 18, 2010