JEE Main 2017 — Probability Question with Solution

Let P(E) = x and P(F) = y

Now, $$P(E\cap F)=\frac{1}{12}$$

$$\Rightarrow P(E)P(F)=\frac{1}{12}$$

$$\Rightarrow xy=\frac{1}{12}$$

Also, $$P(E^{\prime }\cap F^{\prime })=\frac{1}{2}$$

$$\Rightarrow (1-P(E))(1-P(F))=\frac{1}{2}$$

$$\Rightarrow (1-x)(1-y)=\frac{1}{2}$$

$$\Rightarrow 1-x-y+xy=\frac{1}{2}$$

$$\Rightarrow x+y=1+xy-\frac{1}{2}$$

$$\Rightarrow x+y=1+\frac{1}{12}-\frac{1}{2}$$

$$\Rightarrow x+y=\frac{1}{2}+\frac{1}{12}=\frac{7}{12}$$ ........ (1)

Now,

$$(x-y{)}^2=(x+y{)}^2-4xy$$

$$\Rightarrow (x-y{)}^2=\frac{49}{144}-\frac{1}{3}\Rightarrow \frac{49-98}{144}\Rightarrow \frac{1}{144}$$

$$\Rightarrow (x-y)=\frac{1}{12}\Rightarrow x-y=\frac{1}{2}$$ ........ (2)

From Eqs. (1) and (2), we get

$$(x+y)(x-y)=\frac{7}{12}+\frac{1}{12}$$

$$\Rightarrow x=\frac{4}{12};y=\frac{3}{12}$$

$$\Rightarrow \frac{x}{y}=\frac{4}{3}=\frac{P(E)}{P(F)}$$