If A and B are two independent events such that $P(\overline{A} ∩ B)= \frac{2}{15}$ and $ P(A ∩ \overline{B})=\frac{1}{6}$, then P(B), is

$\frac{4}{5}\,  or \, \frac{1}{6}$

Let P(A) = x and P(B) =y. Since A and B are independent events. Therefore,

$P(\overline{A} ∩ B)=\frac{2}{15}$

$⇒ P(\overline{A})P(B)=\frac{2}{15}$

$⇒ \begin{Bmatrix} 1-P(A)\end{Bmatrix} P(B)= \frac{2}{15}$

$⇒ (1-x)y=\frac{2}{15}$

$⇒ y - xy = \frac{2}{15}$ ...........(i)

and, $P(A ∩ \overline{B})=\frac{1}{6}$

$⇒ P(A) P(\overline{B})=\frac{1}{6}$

$⇒ x(1-y)=\frac{1}{6}⇒x-xy=\frac{1}{6}$ ..............(ii)

Subtracting (i) from (ii), we get

$x-y = \frac{1}{30}⇒ x =\frac{1}{30}+y$

Putting this value of x in (i), we get

$y-y\left(\frac{1}{30}+y\right)=\frac{2}{15}$

$⇒ 30y - y - 30 y^2 = 4$

$⇒ 30y^2 -29y + 4= 0$

$ ⇒ (6y -1) (5y-4) = 0 ⇒ y=\frac{1}{6} $ or, $ y =\frac{4}{5}$

$∴ P(B)=\frac{1}{6} $ or, $ P(B)=\frac{4}{5}$