Let A and B be two mutually exclusive events of a random experiment

Solution:

Given : A and B are mutually exclusive events

$\mathrm{P}(\operatorname{not} \mathrm{A})=\mathrm{P}\left({ }^{\bar{A}}\right)=0.65, \mathrm{P}(\mathrm{A}$ or $\mathrm{B})=0.65$

To find : P(B)

Formula used : $\mathrm{P}(\mathrm{A})=1-\mathrm{P}(\bar{A})$

$P(A$ or $B)=P(A)+P(B)-P(A$ and $B)$

For mutually exclusive events $A$ and $B, P(A$ and $B)=0$

$P(A)=1-P($ not $A)$

$P(A)=1-0.65$

$P(A)=0.35$

Substituting in the above formula we get,

$0.65=0.35+P(B)$

$P(B)=0.65-0.35$

$P(B)=0.30$

P(B) = 0.30