Question

The probability that atleast one of the two events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.3, evaluate $P\left(\overline{A}\right)+P\left(\overline{B}\right)$.

Answer 1

We know that, $A\cup B$ denotes the occurence of atleast one of A and B and $A\cap B$ denotes the occurence of both A and B, simultaneously.

Thus, $P(A\cup B)=0.6$ and $P(A\cap B)=0.3$

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

$\Rightarrow 0.6=P\left(A\right)+P\left(B\right)-0.3$

$\Rightarrow P\left(A\right)+P\left(B\right)=0.9$

$\Rightarrow [1-P(\overline{A}\left)\right]+[1-P(\overline{B}\left)\right]=0.9$ $[\because P(A)=1-P(\overline{A}\left)andP\right(B)=1-P(\overline{B}\left)\right]$

$\Rightarrow P\left(\overline{A}\right)+P\left(\overline{B}\right)=2-0.9=1.1$