Let $\mathrm{A}$ and $\mathrm{B}$ be two independent events of a random experiment. If the probability that…
Let $\mathrm{A}$ and $\mathrm{B}$ be two independent events of a random experiment. If the probability that both A and B occur is $\frac{1}{6}$ and the probability that neither of them occur is $\frac{1}{3}$, then the probability of occurrence of $\mathrm{A}$ is
0 or 1
$\frac{1}{2}$ or $\frac{1}{4}$
$\frac{1}{2}$ or $\frac{1}{3}$
$\frac{1}{2}$ or $\frac{1}{7}$
Solution
Since A \& B are independent events so
$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$
Now, $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{6}$
$\Rightarrow \mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=\frac{1}{6}$
Let $P(A)=x, P(B)=y$
Since $P(A \cup B)^{\prime}=\frac{1}{3}$
$\begin{aligned}
& \Rightarrow 1-P(A \cup B)=\frac{1}{3} \\
& \Rightarrow P(A \cup B)=\frac{2}{3} \\
& \Rightarrow P(A)+P(B)-P(A \cap B)=\frac{2}{3} \\
& \Rightarrow x+y-\frac{1}{6}=\frac{2}{3} \Rightarrow x+y=\frac{5}{6}
\end{aligned}$
Now, $x\left(\frac{5}{6}-x\right)=\frac{1}{6} \Rightarrow 5 x-6 x^2=1$
$\begin{aligned}
& \Rightarrow 6 x^2-5 x+1=0 \\
& \Rightarrow(2 x-1)(3 x-1)=0 \\
& \Rightarrow x=\frac{1}{2} \text { or } \frac{1}{3} \Rightarrow P(A)=\frac{1}{2} \text { or } \frac{1}{3}
\end{aligned}$