Question

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

Answer 1

It is given that A and B are independent events.

${\displaystyle P(\overline{A}\cap B)=\frac{2}{15}}$

${\displaystyle \therefore P\left(\overline{A}\right) P\left(B\right)=\frac{2}{15} .....\left(1\right)}$

Also, ${\displaystyle (P\cap \overline{B})=\frac{1}{6}}$

${\displaystyle \therefore P\left(A\right) P\left(\overline{B}\right)=\frac{1}{6}}$

${\displaystyle \Rightarrow P\left(A\right)=\frac{1}{6[1-P\left(B\right)]} .....\left(2\right)}$

From (1), we have

${\displaystyle [1-P\left(A\right)]P\left(B\right)=\frac{2}{15}}$

${\displaystyle [1-\frac{1}{6[1-P\left(B\right)]}]P\left(B\right)=\frac{2}{15}}$

${\displaystyle \left\{\frac{6-6P\left(B\right)-1}{6[1-P\left(B\right)]}\right\}P\left(B\right)=\frac{2}{15}}$

${\displaystyle 5 P\left(B\right)-6{\left[P\left(B\right)\right]}^2=\frac{12[1-P\left(B\right)]}{15}}$

${\displaystyle 25P\left(B\right)-30{\left[P\left(B\right)\right]}^2=4-4P\left(B\right)}$

${\displaystyle 30{\left[P\left(B\right)\right]}^2-29P\left(B\right)+4=0}$

${\displaystyle 30{\left[P\left(B\right)\right]}^2-24P\left(B\right)-5P\left(B\right)+4=0}$

${\displaystyle 6P\left(B\right)[5P\left(B\right)-4]-1[5P\left(B\right)-4]=0}$

${\displaystyle [5P\left(B\right)-4] [6P\left(B\right)-1]=0}$

${\displaystyle P\left(B\right)=\frac{4}{5}, \frac{1}{6}}$

For P(B) = 4/5, using (2), we have

${\displaystyle P\left(A\right)=\frac{1}{6[1-P\left(B\right)] } }$

${\displaystyle =\frac{1}{6[1-\frac{4}{5}]} }$

${\displaystyle =\frac{5}{6}}$

For P(B) = 1/6, using (2), we have

${\displaystyle P\left(A\right)=\frac{1}{6[1-16] } }$

${\displaystyle =\frac{1}{5}}$

${\displaystyle \therefore P\left(A\right)=\frac{5}{6}, P\left(B\right)=\frac{4}{5} \text{or} P\left(A\right)=\frac{1}{5}, P\left(B\right)=\frac{1}{6}}$