Given two independent events $A$ and $B$ such $P(A)=0.3,\, P(B)=0.6 .$ Find $P(A $ and not $B)$
Given two independent events $A$ and $B$ such $P(A)=0.3,\, P(B)=0.6 .$ Find $P(A $ and not $B)$
It is given that $\mathrm{P}(\mathrm{A})=0.3, \,\mathrm{P}(\mathrm{B})=0.6$
Also, $A$ and $B$ are independent events.
$\mathrm{P}(\mathrm{A}$ and not $\mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})^{\prime}$
$=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=0.3-0.18$
$=0.12$
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