Probability 3 Question 41
41. $A$ 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 $\frac{1}{3}$. Find the probability of the occurrence of $A$.
(1984, 2M)
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Answer:
Correct Answer: 41. $\frac{1}{3}$ or $\frac{1}{2}$
Solution:
- Given, $P(A) \cdot P(B)=\frac{1}{6}, P(\bar{A}) \cdot P(\bar{B})=\frac{1}{3}$
$$ \therefore \quad[1-P(A)][1-P(B)]=\frac{1}{3} $$
$$ \begin{aligned} & \text { Let } \quad P(A)=x \text { and } P(B)=y \\ & \Rightarrow \quad(1-x)(1-y)=\frac{1}{3} \quad \text { and } \quad x y=\frac{1}{6} \\ & \Rightarrow \quad 1-x-y+x y=\frac{1}{3} \text { and } x y=\frac{1}{6} \\ & \Rightarrow \quad x+y=\frac{5}{6} \quad \text { and } x y=\frac{1}{6} \\ & \Rightarrow \quad x \frac{5}{6}-x=\frac{1}{6} \\ & \Rightarrow \quad 6 x^{2}-5 x+1=0 \\ & \Rightarrow \quad(3 x-1)(2 x-1)=0 \\ & \Rightarrow \quad x=\frac{1}{3} \text { and } \frac{1}{2} \\ & \therefore \quad P(A)=\frac{1}{3} \text { or } \frac{1}{2} \end{aligned} $$
