Question

If $A,B,C$ are three independent events of an experiment such that, $P(A\cap B^C\cap C^C)=\frac{1}{4},P(A^C\cap B\cap C^C)=\frac{1}{8},P(A^C\cap B^C\cap C^C)=\frac{1}{4}$, then find $P\left(A\right),P\left(B\right),P\left(C\right)$

Answer 1

Let $P\left(A\right)=x,P\left(B\right)=y,P\left(C\right)=z$

Since these are independent events, we can write

$P(A\cap B^c\cap C^c)=P\left(A\right)\times P\left(B^c\right)\times P\left(C^c\right)=\frac{1}{4}$

$\therefore x(1-y)(1-z)=\frac{1}{4}...\left(1\right)$

Similarly, $(1-x)y(1-z)=\frac{1}{8}...\left(2\right)$

And $(1-x)(1-y)(1-z)=\frac{1}{4}...\left(3\right)$

Dividing equation (3) by (1), we have

$\frac{1-x}{x}=1$ or $1-x=x$

$\therefore x=\frac{1}{2}$

Dividing equation (1) by (2), we have

$\frac{1-y}{y}=2$

$\therefore 1-y=2y$ or $y=\frac{1}{3}$

Substituting these values in equation (1), we have

$\frac{1}{2}\times \frac{2}{3}\times (1-z)=\frac{1}{4}$

$\therefore z=1-\frac{3}{4}=\frac{1}{4}$

Thus, $P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{1}{3},P\left(C\right)=\frac{1}{4}$