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Question: Answered & Verified by Expert
$A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur is $\frac{1}{6}$ and the probability that neither of them occurs is $\frac{1}{3}$. Then the probability of the two events are respectively
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Options:
  • A $\frac{1}{2}$ and $\frac{1}{3}$
  • B $\frac{1}{5}$ and $\frac{1}{6}$
  • C $\frac{1}{2}$ and $\frac{1}{6}$
  • D $\frac{2}{3}$ and $\frac{1}{4}$
Solution:
1029 Upvotes Verified Answer
The correct answer is: $\frac{1}{2}$ and $\frac{1}{3}$
$P(A \cap B)=P(A) \cdot P(B)=\frac{1}{6}$
$P(\bar{A} \cap \bar{B})=\frac{1}{3}=1-P(A \cup B)$
$\Rightarrow \frac{1}{3}=1-[P(A)+P(B)]+\frac{1}{6} \Rightarrow P(A)+P(B)=\frac{5}{6}$.
Hence $P(A)$ and $P(B)$ are $\frac{1}{2}$ and $\frac{1}{3}$.

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