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  1. #1
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    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.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Vedicmaths View Post
    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)}
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  3. #3
    Moo
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    (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}
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    Thank you so much for your help!

    That was a great celebration!

    Regards,
    Vedic
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