Question

Let A and B be two independent events such that

$\mathbb{P}(\overline{A}/\overline{B}) = \frac{2}{3}$ then $\mathbb{P}(\overline{A} \cap B)$

$\mathbb{P}(1) + \mathbb{P}(2) = \frac{1}{2}$,

equals

• A. $\frac{1}{3}$
• B. $\frac{1}{6}$
• C. $\frac{1}{9}$
• D. $\frac{1}{18}$

Answer 1

Answer: (B)

$\frac{1}{6}$

Answer 2

Let A and B be two independent events. We are given $\mathbb{P}(\overline{A}/\overline{B}) = \frac{2}{3}$. Since A and B are independent, their complements $\overline{A}$ and $\overline{B}$ are also independent. For independent events, the conditional probability $\mathbb{P}(\overline{A}/\overline{B})$ is equal to the marginal probability $\mathbb{P}(\overline{A})$. So, $\mathbb{P}(\overline{A}) = \frac{2}{3}$. The probability of event A is $\mathbb{P}(A) = 1 - \mathbb{P}(\overline{A}) = 1 - \frac{2}{3} = \frac{1}{3}$.

The second condition given is "$\mathbb{P}(1) + \mathbb{P}(2) = \frac{1}{2}$". This notation is unusual in the context of events A and B. Assuming this is a typo and it refers to the probability of the union of events A and B, the condition is interpreted as $\mathbb{P}(A \cup B) = \frac{1}{2}$.

For any two events A and B, the probability of their union is given by $\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B)$. Since A and B are independent, the probability of their intersection is $\mathbb{P}(A \cap B) = \mathbb{P}(A)\mathbb{P}(B)$. Substituting this into the union formula, we get $\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A)\mathbb{P}(B)$.

We have $\mathbb{P}(A) = \frac{1}{3}$ and assumed $\mathbb{P}(A \cup B) = \frac{1}{2}$. Let $\mathbb{P}(B) = p$.

$\frac{1}{2} = \frac{1}{3} + p - \frac{1}{3}p$

$\frac{1}{2} = \frac{1}{3} + p(1 - \frac{1}{3})$

$\frac{1}{2} = \frac{1}{3} + p(\frac{2}{3})$

Subtract $\frac{1}{3}$ from both sides:

$\frac{1}{2} - \frac{1}{3} = \frac{2}{3}p$

$\frac{3 - 2}{6} = \frac{2}{3}p$

$\frac{1}{6} = \frac{2}{3}p$

Solve for $p$:

$p = \frac{1}{6} \times \frac{3}{2} = \frac{3}{12} = \frac{1}{4}$.

So, $\mathbb{P}(B) = \frac{1}{4}$.

We need to find $\mathbb{P}(\overline{A} \cap B)$. Since A and B are independent events, $\overline{A}$ and B are also independent events. The probability of the intersection of independent events $\overline{A}$ and B is the product of their individual probabilities:

$\mathbb{P}(\overline{A} \cap B) = \mathbb{P}(\overline{A})\mathbb{P}(B)$.

We found $\mathbb{P}(\overline{A}) = \frac{2}{3}$ and $\mathbb{P}(B) = \frac{1}{4}$.

$\mathbb{P}(\overline{A} \cap B) = \frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6}$.