If A and B are independent events of a random | vedclass

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

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) Given that $A$ and $B$ are independent events, we have $P(A \cap B) = P(A)P(B) = \frac{1}{6}$.Also, $P(\bar{A} \cap \bar{B}) = P(\bar{A})P(\bar{B}) = \frac{1}{3}$.Since $P(\bar{A}) = 1 - P(A)$ and $P(\bar{B}) = 1 - P(B)$, we have $(1 - P(A))(1 - P(B)) = \frac{1}{3}$.Expanding this, $1 - (P(A) + P(B)) + P(A)P(B) = \frac{1}{3}$.Substituting $P(A)P(B) = \frac{1}{6}$, we get $1 - (P(A) + P(B)) + \frac{1}{6} = \frac{1}{3}$.$P(A) + P(B) = 1 + \frac{1}{6} - \frac{1}{3} = \frac{6+1-2}{6} = \frac{5}{6}$.Let $x = P(A)$ and $y = P(B)$. Then $x + y = \frac{5}{6}$ and $xy = \frac{1}{6}$.The quadratic equation $t^2 - (x+y)t + xy = 0$ becomes $t^2 - \frac{5}{6}t + \frac{1}{6} = 0$.$6t^2 - 5t + 1 = 0 \Rightarrow (2t - 1)(3t - 1) = 0$.Thus, $t = \frac{1}{2}$ or $t = \frac{1}{3}$.Therefore, $P(A)$ can be $\frac{1}{2}$ or $\frac{1}{3}$. Since $\frac{1}{3}$ is given in option $B$ and $\frac{1}{2}$ is given in option $D$, both are valid.

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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