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(D) Given that $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B) = \frac{1}{6}$.Also,$P(\bar{A} \cap \bar{B}) = P(\bar{A})P(\bar{B}) = (1 - P

Vedclass · Accepted Answer

None of these

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(D) Given that $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B) = \frac{1}{6}$.Also,$P(\bar{A} \cap \bar{B}) = P(\bar{A})P(\bar{B}) = (1 - P(A))(1 - P(B)) = \frac{1}{3}$.Expanding the second equation: $1 - P(A) - P(B) + P(A)P(B) = \frac{1}{3}$.Substituting $P(A)P(B) = \frac{1}{6}$: $1 - (P(A) + P(B)) + \frac{1}{6} = \frac{1}{3}$.$P(A) + P(B) = 1 + \frac{1}{6} - \frac{1}{3} = \frac{6 + 1 - 2}{6} = \frac{5}{6}$.Let $x = P(A)$ and $y = P(B)$. We have $x + y = \frac{5}{6}$ and $xy = \frac{1}{6}$.These are roots of the quadratic equation $t^2 - (x+y)t + xy = 0$,which is $t^2 - \frac{5}{6}t + \frac{1}{6} = 0$.Multiplying by $6$: $6t^2 - 5t + 1 = 0$.$(2t - 1)(3t - 1) = 0$.So,$t = \frac{1}{2}$ or $t = \frac{1}{3}$.Thus,$P(A)$ can be $\frac{1}{2}$ or $\frac{1}{3}$.Since neither $\frac{1}{2}$ nor $\frac{1}{3}$ are uniquely listed as the only correct option,the answer is $D$.

If $A$ and $B$ are independent events of a random experiment such that $P(A \cap B) = \frac{1}{6}$ and $P(\bar{A} \cap \bar{B}) = \frac{1}{3}$,then $P(A)$ is equal to (Here,$\bar{E}$ is the complement of the event $E$)

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