Given two independent events A and B such P(A)=0.3,, P(B)=0.6 . Find P(A and not B)

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14.Probability

easy

Given two independent events $A$ and $B$ such $P(A)=0.3,\, P(B)=0.6 .$ Find $P(A $ and not $B)$

A

$0.12$

B

$0.12$

C

$0.12$

D

$0.12$

Solution

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 $\mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})^{\prime}$

$=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$

$=0.3-0.18$

$=0.12$

Standard 11

Mathematics

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