probability help!

**Vedicmaths** · Oct 19th 2008, 04:39 PM

I need help solving this particular problem!

Please help!

Let (S, P) be a probability space, and let A,B be any events. Prove the following statements.
(A) If A is a subset of B, then
P(B − A) = P(B) − P(A) and P(A) <= P(B).
(B) If A and B are independent, then A and S − B are independent.

**Moo** · Oct 20th 2008, 11:44 AM

Hello,

Originally Posted by Vedicmaths

I need help solving this particular problem!

Please help!

Let (S, P) be a probability space, and let A,B be any events. Prove the following statements.
(A) If A is a subset of B, then
P(B − A) = P(B) − P(A) and P(A) <= P(B).

I'll try this one... (not sure if I can use some of the formula, so you'll have to tell me)

$B-A=B \cap A^c$

$A \subseteq B \implies B^c \subseteq A^c$ . Hence $\underbrace{B \cup B^c}_{=S} \subseteq B \cup A^c$

So $P(B \cup A^c) \geqslant P(B \cup B^c)=1 \implies P(B \cup A^c)=1$

From the formula $P(F \cup G)=P(F)+P(G)-P(F \cap G)$ , we have :
${\color{red}1}=P(B \cup A^c)=P(B)+P(A^c)-P(B \cap A^c)=P(B)+{\color{red}1}-P(A)-P(B \cap A^c)$

$0=P(B)-P(A)-P(B \cap A^c)$

Therefore $\boxed{P(B-A)=P(B \cap A^c)=P(B)-P(A)}$

**Moo** · Oct 20th 2008, 12:04 PM

(B) If A and B are independent, then A and S − B are independent.

$\text{A and B are independent } \Leftrightarrow P(A \cap B)=P(A) P(B)$

$A=A \cap S=A \cap (B \cup B^c)=(A \cap B) \cup (A \cap B^c)$

$\implies P(A)=P(A \cap B)+P(A \cap B^c)-P\underbrace{(A \cap B \cap A \cap B^c)}_{\emptyset \text{ because } B \cap B^c=\emptyset}$

So $\boxed{P(A)=P(A \cap B)+P(A \cap B^c)}$

Hence
$\begin{aligned} P(A \cap B^c) &=P(A)-P(A \cap B) \hfill \\<br /> &=P(A)-P(A)P(B) \hfill \\<br /> &=P(A)(1-P(B)) \hfill \\<br /> \boxed{P(A \cap B^c)} &=\boxed{P(A)P(B^c)} \quad \square \hfill \end{aligned}$

**Vedicmaths** · Oct 20th 2008, 07:01 PM

Thank you so much for your help!

That was a great celebration!

Regards,
Vedic

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