Event A and B are independent, A and C are mutually exclusive, B and C are independent. Given:
- $\Pr(A) = 1/2$
- $\Pr(B) = 1/4$
- $\Pr(C) = 1/8$
Find $P(A \cup B \cup C)$ ?
The teacher gave the answer to be $23/32$. I got $47/64$, not quite sure if I'm doing the right thing.
Ps. $Pr(A \cap B \cap C)$ is ?
$\endgroup$ 62 Answers
$\begingroup$$$P(A\cup B\cup C)=P(A\cup C)+P(B)-P(B\cap (A\cup C))=P(A)+P(B)+P(C)-P((A\cap B) \cup (B\cap C))=P(A)+P(B)+P(C)-P(A)P(B)-P(B)P(C)+P((A\cap B) \cap (B\cap C)$$
The last term is $0$, so we get
$$P(A\cup B\cup C)=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}-\frac{1}{8}-\frac{1}{32}=\frac{23}{32}$$
$\endgroup$ 0 $\begingroup$A venn diagram should certainly help you solve this problem!
One of the property of Independent events is that the probability of their intersection is a product of their individual probabilities. So, $P(A \cap B)$ is $P(A) \times P(B)$. Whereas for mutually exclusive events, the probability of intersection is $0$ as they can't both occur simultaneously!
$$ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$$
Can you take from here? It shouldn't be so hard to get the right answer.
$\endgroup$ 1
