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$A$ and $B$ are events such that $P(A\cup B)=3/4, P(A\cap B)=1/4, P(\bar{A})=2/3\; \; then\; \; P(\bar{A}\cap B)$ is

Option 1)
5/12

Option 2)
3/8

Option 3)
5/8

Option 4)
1/4

Answers (1)

best_answer

As we learnt in

Algebra of events -

[ If A and B are mutually exclusive events then $A\cap B= \phi$

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

$\therefore P\left ( A\cup B \right )= P\left ( A \right )+P\left ( B \right )$

$\Rightarrow P\left ( A\cup B\cup C \right )= P\left ( A \right )+\left P( B \right )+P\left ( C \right )$

$\Rightarrow P\left ( \overline{A}\cap \overline{B} \right )= 1-P\left ( A\cup B \right )$

$\Rightarrow P\left ( \overline{A}\cup \overline{B} \right )= 1-P\left ( A\cap B \right )$

$\Rightarrow P \left ( A \right )=P\left ( A\cap B \right )+P\left ( A\cap \overline{B} \right )$

$\Rightarrow P \left ( B\right )=P\left ( B\cap A \right )+P\left ( B\cap \overline{A} \right ) ]$

-

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$\frac{3}{4}=\frac{1}{3}+(P(B)-\frac{1}{4}$

$\Rightarrow P(B)=\frac{2}{3}$

Now $\bar{A}\cap B$ is represented as shaded areas in figure

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

$=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$

Option 1)

5/12

This option is correct

Option 2)

3/8

This option is incorrect

Option 3)

5/8

This option is incorrect

Option 4)

1/4

This option is incorrect

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