Question

If $A$ and $B$ are two independent events such that $ P \left(\bar{A} \cap B \right)= \frac{2}{15}$ and $P \left(A \cap \bar{B} \right) =\frac{1}{6}$, then find $P (A)$ and $P (B)$ respectively.

• A. $\frac{5}{6}$, $\frac{4}{5}$
• B. $\frac{1}{5}$, $\frac{1}{7}$
• C. $\frac{1}{6}$, $\frac{1}{7}$
• D. $\frac{1}{7}$, $\frac{1}{7}$

Answer 1

Answer: (A)

Since $A$ and $B$ are independent events, therefore, $A^c$ and $B$ are independent and also $A$ and $B^c$ are independent.
$\therefore P\left(\bar{A} \cap B\right) = P\left(A^{c} \cap B \right) = P \left(A^{c}\right) \,P\left(B\right) $
$ =\left(1 - P\left(A\right)\right) P\left(B\right)$
and $P\left(A \cap \bar{B}\right) = P\left(A \cap B^{c}\right) = P\left(A\right)\, P\,\left(B^{c}\right)$
$= P \left(A\right) \left(1 - P\left(B\right)\right)$
$\left(\because \bar{A} = A^{c} \,\text{and}\, \bar{B} = B^{c}\right)$
$\Rightarrow \left(1 - P \left(A\right)\right) P\left(B\right) = P \left(\bar{A} \cap B \right)= \frac{2}{15}\quad\ldots\left(1\right)$
and $P\left(A\right) \left(1-P\left(B\right)\right) = P \left(A \cap \bar{B} \right) =\frac{1}{6}\quad \ldots \left(2\right)$
Subtracting $\left(2\right)$ from $\left(1\right)$, we obtain
$P\left(B\right)-P\left(A\right) = \frac{2}{15} - \frac{1}{6}$
$\Rightarrow P \left(B\right) - P\left(A\right) =\frac{4-5}{30}$
$\Rightarrow P \left( B \right) - P \left(A \right) =-\frac{1}{30}\quad \ldots \left(3\right)$
Substituting this value of $P \left(B\right)$ in $\left(1\right)$, we get
$\left(1 - P \left( A \right)\right)\left\{P\left(A\right)-\frac{1}{30}\right\} = \frac{2}{15}$
$\Rightarrow P\left(A\right)+\frac{1}{30}P\left(A\right)-\left(P\left(A\right)\right)^{2}-\frac{1}{30} = \frac{2}{15}$
$\Rightarrow \left(P\left(A\right)\right)^{2} - \frac{31}{30}P\left(A\right)+\frac{1}{30}+\frac{2}{15} = 0$
$\Rightarrow 30\left(P\left(A\right)\right)^{2} - 31 P\left(A\right)+5 = 0$
$\Rightarrow P\left(A\right) = \frac{31\pm\sqrt{961-600}}{60}$
$= \frac{31\pm19}{60} = \frac{5}{6}, \frac{1}{5}$
Case I. When $P \left(A\right) = \frac{5}{6}$, then from $\left(3\right)$,
$P\left(B\right) = P \left(A \right)-\frac{1}{30} = \frac{5}{6}-\frac{1}{30} = \frac{25-1}{30} = \frac{4}{5}$
$\Rightarrow P\left(A\right) = \frac{5}{6}$ and $P\left(B\right) = \frac{4}{5}$
Case II. When $P \left(A\right) = \frac{1}{5}$, then from $\left(3\right)$,
$P\left(B\right) = P \left( A\right) -\frac{1}{30}=\frac{1}{5}-\frac{1}{30}=\frac{6-1}{30}=\frac{1}{6}$
$\Rightarrow P\left(A\right) = \frac{1}{5}$ and $P \left( B\right) = \frac{1}{6}$.