JEE Main 2002 — Probability Question with Solution

Given $$P\left(A\right)=\frac{1}{2}$$, $$P\left(B\right)=\frac{1}{3}$$, $$P\left(C\right)=\frac{1}{4}$$

So, $$P\left(\overline{A}\right)=\frac{1}{2}$$ (Probablity that the problem can't be solve by A)
$$P\left(\overline{B}\right)=\frac{2}{3}$$ (Probablity that the problem can't be solve by B)
and $$P\left(\overline{C}\right)=\frac{3}{4}$$ (Probablity that the problem can't be solve by C)

Now the probablity that the problem is solved by any one student of A, B and C = 1 - the probablity that the problem is solved by none of the students of A, B and C

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

$$=1-\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}$$
$$=$$ $$\frac{3}{4}$$