Given that events A and B are such that P(A)=frac{1}{2}, P(A cup B)=frac{3}{5} and mathrm{P}(mathrm{

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14.Probability

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Given that the events $A$ and $B$ are such that $P(A)=\frac{1}{2}, P(A \cup B)=\frac{3}{5}$ and $\mathrm{P}(\mathrm{B})=p .$ Find $p$ if they are mutually exclusive.

A

$\frac {1}{10}$

B

$\frac {1}{10}$

C

$\frac {1}{10}$

D

$\frac {1}{10}$

Solution

It is given that $P(A)=\frac{1}{2},\, P(A \cup B)=\frac{3}{5}$ and $P(B)=p$

When $A$ and $B$ are mutually exclusive, $A \cap B=\phi$

$\therefore P(A \cap B)=0$

It is known that, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$\Rightarrow \frac{3}{5}=\frac{1}{2}+p-0$

$\Rightarrow p=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$

Standard 11

Mathematics

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