Question

Let $A$ and $B$ be two independent event then show that $P(A\cap B)=1-P\left(A^{\prime }\right)P\left(B^{\prime }\right)$.

Answer 1

Given $A$ and $B$ are independent then $P(A\cap B)=P\left(A\right).P\left(B\right)$......(1).

We know

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

or, $P(A\cup B)=P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)$ [ Using (1)]

or, $P(A\cup B)=P\left(A\right)\{1-P(B\left)\right\}+P\left(B\right)$

or, $P(A\cup B)=P\left(A\right)\{1-P(B\left)\right\}+1-\{1-P(B\left)\right\}$

or, $P(A\cup B)=P\left(A\right)P\left(B^{\prime }\right)+1-P\left(B^{\prime }\right)$

or, $P(A\cup B)=1-P\left(A^{\prime }\right)P\left(B^{\prime }\right)$