If $A$ and $B$ are two independent events such that $P(\overline{A} \cap B) = \frac{2}{15}$ and $P(A \cap \overline{B}) = \frac{1}{6}$, then find $P(A)$ and $P(B)$.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $P(A) = \frac{1}{5}, P(B) = \frac{1}{6}$ or $P(A) = \frac{5}{6}, P(B) = \frac{4}{5}$ ##
$P(A \cap B) = \frac{2}{15}$
$\text{or} \quad P(A)P(B) = \frac{2}{15}$
$\text{and} \quad P(A \cap \overline{B}) = \frac{1}{6} $
$\text{or} \quad P(A)P(\overline{B}) = \frac{1}{6}$
$∴\quad [1 - P(A)]P(B) = \frac{2}{15}$
$\text{or} \quad P(B) - P(A)P(B) = \frac{2}{15}\quad ...(i) $
$\text{Similarly,}$
$P(A)[1 - P(B)] = \frac{1}{6} $
$\text{or} \quad P(A) - P(A)P(B) = \frac{1}{6}\quad ...(ii) $
From (i) and (ii),
$P(A) - P(B) = \frac{1}{6} - \frac{2}{15} = \frac{5 - 4}{30} = \frac{1}{30}$
Let \( P(A) = x \), \( P(B) = y \)
$∴\quad x = \frac{1}{30} + y$
By (i), $y - \left( \frac{1}{30} + y \right)y = \frac{2}{15}$
$∴\quad 30y^2 - 29y + 4 = 0$
$(5y - 4)(6y - 1) = 0$
On solving, we get
$y = \frac{1}{6} \quad \text{or} \quad y = \frac{4}{5}$
$∴\quad x = \frac{1}{5} \quad \text{or} \quad x = \frac{5}{6}$
Hence, $P(A) = \frac{1}{5}, \quad P(B) = \frac{1}{6}$
Or $P(A) = \frac{5}{6}, \quad P(B) = \frac{4}{5}$