Mathematics Asked on January 29, 2021
Let $P (A) = 0.7$, $P(B^c) = 0.4$ and $P(B ~text{and} ~C) = 0.48$
a. Find $P (A ~text{or}~ B)$ when $A$ and $B$ are independent
$P(B) = 1 – P(B^c) = 1 – 0.4 = 0.6$
Seeing as $A$ and $B$ are independent $P(A ~text{and}~ B) = 0.7 times 0.6 = 0.42$
$P (A~text{ or}~ B) = 0.7 + 0.6 – 0.42 = 0.88$
b. Is it possible that $A$ and $C$ are mutually exclusive if they are independent?
The answer in the book is no, they cannot be mutually exclusive, but I don’t understand why? I thought events could be both mutually exclusive AND independent. Why is the answer no?
If anyone can explain why, (and double-check if my work for a. is right). I would greatly appreciate it.
Your answer for the first part is correct.
If $A$ and $C$ are mutually exclusive and independent then $0=P(emptyset)=P(A cap C) =P(A)P(C)$ so either $P(A)=0$ or $P(C)=0$.
Answered by Kavi Rama Murthy on January 29, 2021
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