JEE Main 2014 — Probability Question with Solution

$$P(\overline{A\cup B})=\frac{1}{6}$$

or, $$1-P(A\cup B)=\frac{1}{6}$$

$$\therefore$$ $$P(A\cup B)=1-\frac{1}{6}=\frac{5}{6};$$ $$P(A\cap B)=\frac{1}{4}$$; and $$P(\overline{A})=\frac{1}{4};$$

$$\therefore$$ $$P(A)=1-P(\overline{A})=1-\frac{1}{4}=\frac{3}{4}$$

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

or, $$\frac{5}{6}=\frac{3}{4}+P(B)-\frac{1}{4}$$

or, $$P(B)=\frac{5}{6}-\frac{1}{2}=\frac{1}{3}$$

Now, $$P(A).P(B)=\frac{3}{4}.\frac{1}{3}=\frac{1}{4}=P(A\cap B)$$

i.e., events A and B are mutually independent.

Since the probability of A and B are different, so they are not equally likely events.

Therefore, (A) is the correct option.