Thread: probability help!

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.

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)

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

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

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

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

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

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

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

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

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

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

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

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

Thank you so much for your help!

That was a great celebration!

Regards,
Vedic