Given two independent events $A$ and $B$ such $P(A)=0.3,\,P(B)=0.6 .$ Find $P($ neither $A$or $B)$
Given two independent events $A$ and $B$ such $P(A)=0.3,\,P(B)=0.6 .$ Find $P($ neither $A$or $B)$
$P($ neither $A$ nor $B)$ $=P\left(A^{\prime} \cap B^{\prime}\right)$
$=\mathrm{P}\left((\mathrm{A} \cup \mathrm{B})^{\prime}\right)$
$=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
$=1-0.72$
$=0.28$
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- [IIT 1993]
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