Probability of 2 independent events and 1 mutually exclusive

Question

Let $\Omega$ be the sample space for an experiment $E$ and let $A,B,C\subset \Omega$. If events $A,B$ are independent, events $A,C$ are disjoint, and events $B,C$ are independent, find $\Pr(B)$ if $\Pr(A) = 0.2$, $\Pr(C) = 0.4$, and $\Pr(A \cup B \cup C) = 0.8$.

I know that $\Pr(B)$ is between .2 and .4 but how can I determine where ($B$ intersects $A$) and ($B$ intersects $C$)?

Answer 1

Here goes....

$$0.8=\Pr(A\cup B \cup C) = \Pr(A\cup B) + \Pr(C) - \Pr( (A\cup B) \cap C)\\ = \Pr(A)+\Pr(B)-\Pr(A\cap B) + \Pr(C) -\Pr( (A\cap C) \cup (B\cap C) ) =\\ \Pr(A)+\Pr(B)-\Pr(A)\Pr(B)+\Pr(C) - \Pr(B\cap C)\\ = \Pr(A)+\Pr(B)-\Pr(A)\Pr(B)+\Pr(C) - \Pr(B) \Pr(C)= 0.6 +0.4\Pr(B). $$

Hence $\Pr(B)=1/2$.

Answer 2

The formula for the probability of three events is

$P(A\cup B \cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B \cap C)$

Now you can replace in the before formula, the conditions of problem say:

*$P(A\cap B)=P(A)P(B)$

*$P(B\cap C)=P(B)P(C)$

*$A\cap C=\emptyset$

God bless