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

[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.

Please help. Thanks.

 

[mathman]

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.

Please help. Thanks.

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

 

[_joey]

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.

Please help. Thanks.

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

 

[statdad]

You could cut the final proof down a little:

<br /> \begin{align*}<br /> P(A) + P(B) - P(A \cap B) &amp; \le 1 \\<br /> P(A) + P(B) &amp; \le 1 \\<br /> P(A) &amp; \le 1 - P(B) \\<br /> P(A) &amp; \le P(B&#039;)<br /> \end{align*}<br />

 

[_joey]

You could cut the final proof down a little:

<br /> \begin{align*}<br /> P(A) + P(B) - P(A \cap B) &amp; \le 1 \\<br /> P(A) + P(B) &amp; \le 1 \\<br /> P(A) &amp; \le 1 - P(B) \\<br /> P(A) &amp; \le P(B&#039;)<br /> \end{align*}<br />

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.

 

[statdad]

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.

 

[mathman]

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

 

[statdad]

mathman, there is no doubt about that, and if i implied anything else, my apologies. my point was meant as an add-on to the longer approach give by others.