0
$\begingroup$

Events $B$ and $C$ are dependent on event $A$. $B$ and $C$ are mutually exclusive and collectively exhaustive. $Prob (A \cap B) = 0.3$ and $Prob( A \cap C) = 0.2$ . What is $Prob(C | A)$.

I know: $$Prob(C|A) = Prob( A \cap C) \div Prob(A)$$

How to calculate $Prob(A)$?

  • 2
    Hint: Argue and use that we can write $A=(A\cap B)\cup (A\cap C)$. – 2017-01-23

1 Answers 1

0

Hint:

$$Pr(E\cup F)=Pr(E)+Pr(F)-Pr(E\cap F)$$

$B$ and $C$ are mutually exclusive implies $Pr(B\cap C)=0$

$B$ and $C$ are collectively exhaustive implies $B\cup C=\Omega$ where $\Omega$ is the entire sample space.

$A\cap \Omega = A$ since $A$ is by definition a subset of the sample space.

$Pr(A)=Pr(A\cap \Omega)=\dots$