JEE Main 2022 — Probability Question with Solution

X	0	1	2	3
$$P(x)$$	$$\frac{1-d}{4}$$	$$\frac{1+2d}{4}$$	$$\frac{1-4d}{4}$$	$$\frac{1+3d}{4}$$

We know, $$0\le P(x)\le 1$$

$$\therefore$$ $$0\le \frac{1-d}{4}\le 1$$

$$\Rightarrow 0\le 1-d\le 4$$

$$\Rightarrow -1\le -d\le 3$$

$$\Rightarrow 1\ge d\ge -3$$

Also,

$$0\le \frac{1+2d}{4}\le 1$$

$$\Rightarrow 0\le 1+2d\le 4$$

$$\Rightarrow -1\le 2d\le 3$$

$$\Rightarrow -\frac{1}{2}\le d\le \frac{3}{2}$$

Also,

$$0\le \frac{1-4d}{4}\le 1$$

$$\Rightarrow 0\le 1-4d\le 4$$

$$\Rightarrow -1\le -4d\le 3$$

$$\Rightarrow 1\ge 4d\ge -3$$

$$\Rightarrow \frac{1}{4}\ge d\ge -\frac{3}{4}$$

And,

$$0\le \frac{1+3d}{4}\le 1$$

$$\Rightarrow 0\le 1+3d\le 4$$

$$\Rightarrow -1\le 3d\le 3$$

$$\Rightarrow -\frac{1}{3}\le d\le 1$$

Common range is $$=-\frac{1}{3}$$ to $$\frac{1}{4}$$

$$\therefore$$ $$d\in \left[-\frac{1}{3},\frac{1}{4}\right]$$

$$\therefore$$ Minimum value of $$d=-\frac{1}{3}$$

We know, mean

$$E(x)=\sum x.P(x)$$

$$=0\times \frac{1-d}{4}+1\times \frac{1+2d}{4}+2\times \frac{1-4d}{4}+3\times \frac{1+3d}{4}$$

$$=\frac{1+2d+2-8d+3+9d}{4}$$

$$=\frac{6+3d}{4}$$

For $$d=-\frac{1}{3}$$, $$E(x)=\frac{6+3\times -\frac{1}{3}}{4}=\frac{5}{4}$$

$$\therefore$$ $$60E(x)=60\times \frac{5}{4}=75$$