Question

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 $P\left(A\right)$ and $P\left(B\right).$

Answer 1

Given $P(\overline{A}\cap B)=\frac{2}{15}\Rightarrow P\left(\overline{A}\right)P\left(B\right)=\frac{2}{15}$--- (1)

Since events $A$ and B are independent, $\overline{A}$ and B are also independent.
$P(A\cap \overline{B})=\frac{1}{6}\Rightarrow P\left(A\right)P\left(\overline{B}\right)=\frac{1}{6}$--- (2)

As events A and B are independent, A, $\overline{B}$ are also independent.

Let P(A) = x, P(B) = y
$\Rightarrow P\left(\overline{A}\right)=1-x$ and $P\left(\overline{B}\right)=1-y$

From (1) and (2), we get
$(1-x)y=\frac{2}{15}$ i.e., $y-xy=\frac{2}{15}$--- (3)
and $x(1-y)=\frac{1}{6}$ i.e., $x-xy=\frac{1}{6}$ -- (4)

Subtracting (3) and (4), we get
$x-y=\frac{1}{6}-\frac{2}{15}\Rightarrow x-y=\frac{1}{30}\Rightarrow x=y+\frac{1}{30}$ ----(5)

Putting this value of x in (3), we get
$y-(y+\frac{1}{30})y=\frac{2}{15}\Rightarrow \frac{29}{30}y-y^2=\frac{2}{15}$

$\Rightarrow 30y^2-29y+4=0\Rightarrow (5y-4)(6y-1)=0$

$\Rightarrow y=\frac{4}{5},\frac{1}{6}$

From (5), when $y=\frac{4}{5},x=\frac{4}{5}+\frac{1}{30}=\frac{5}{6}$
when $y=\frac{1}{6},x=\frac{1}{6}+\frac{1}{30}=\frac{1}{5}$

Hence, if $P\left(A\right)=\frac{5}{6}$, then $P\left(B\right)=\frac{4}{5}$
and if $P\left(A\right)=\frac{1}{5}$, then $P\left(B\right)=\frac{1}{6}$