If A and B are two independent events such that P(A) > 0.5,,P(B)

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14.Probability

normal

If $A$ and $B$ are two independent events such that $P(A) > 0.5,\,P(B) > 0.5,\,P(A \cap \bar B) = \frac{3}{{25}},\,P(\bar A \cap B) = \frac{8}{{25}}$ , then $P(A \cap B)$ is

$\frac {12}{25}$

$\frac {14}{25}$

$\frac {18}{25}$

$\frac {24}{25}$

Solution

$\frac{3}{25}=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$

$\frac{8}{25}=P(B)-P(A \cap B)=P(B)-P(A) \cdot P(B)$

$P(B)-P(A)=\frac{1}{5}$

Let $P(A)=a, P(B)=b$

$a-a b=\frac{3}{25}, b-a b=\frac{8}{25}, b-a=\frac{1}{5}$

$\therefore a-a\left(a+\frac{1}{5}\right)=\frac{3}{25}$

$a-a^{2}-\frac{a}{5}=\frac{3}{25} \Rightarrow a-5 a^{2}=\frac{3}{5}$

$\Rightarrow 25 a^{2}-20 a+3=0$

$25 a^{2}-15 a-5 a+3=0$

$\Rightarrow(5 a-1)(5 a-3)=0$

$\therefore a=3 / 5$

$\therefore b=\frac{3}{5}+\frac{1}{5}=\frac{4}{5}$

$\therefore P(A \cap B) = \frac{4}{5} \cdot \frac{3}{5} = \frac{{12}}{{25}}$

Standard 11

Mathematics

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medium

(IIT-1989)

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