JEE Main 2021 — Probability Question with Solution

Let P(E₁) = x, P(E₂) = y and P(E₃) = z

$$\alpha$$ = P$$\left(E_1\cap {\overline{E}}_2\cap {\overline{E}}_3\right)$$ = $$P\left(E_1\right).P\left({\overline{E}}_2\right).P\left({\overline{E}}_3\right)$$

$$\Rightarrow$$ $$\alpha$$ = x(1 $$-$$ y) (1 $$-$$ z) ......(i)

Similarly

β = (1 – x).y(1 – z) ...(ii)

$$\gamma$$ = (1 – x)(1 – y).z ...(iii)

p = (1 – x)(1 – y)(1 – z) ...(iv)

From (i) and (iv)

$$\frac{x}{1-x}=\frac{\alpha }{p}$$

$$\Rightarrow$$ x = $$\frac{\alpha }{\alpha +p}$$

From (iii) and (iv)

$$\frac{z}{1-z}=\frac{\gamma }{p}$$

$$\Rightarrow$$ z = $$\frac{\gamma }{\gamma +p}$$

$$\therefore$$ $$\frac{P\left(E_1\right)}{P\left(E_3\right)}=\frac{x}{z}=\frac{\frac{\alpha }{\alpha +p}}{\frac{\gamma }{\gamma +p}}$$ $$=\frac{\frac{\gamma +p}{\gamma }}{\frac{\alpha +p}{\alpha }}=\frac{1+\frac{p}{\gamma }}{1+\frac{p}{\alpha }}$$ ..(v)

Also given,

($$\alpha$$ $$-$$ 2$$\beta$$)p = $$\alpha$$$$\beta$$ $$\Rightarrow$$ $$\alpha$$p = ($$\alpha$$ + 2p)$$\beta$$ ....(vi)

$$\beta$$ $$-$$ 3$$\gamma$$)p = 2$$\beta$$$$\gamma$$ $$\Rightarrow$$ 3$$\gamma$$p = (p - 2$$\gamma$$)$$\beta$$ .....(vii)

From (vi) and (vii),

$$\frac{\alpha }{3\gamma }=\frac{\alpha +2p}{p-2\gamma }$$

$$\Rightarrow$$ p$$\alpha$$ - 6p$$\gamma$$ = 5$$\gamma$$$$\alpha$$

$$\Rightarrow$$ $$\frac{p}{\gamma }-\frac{6p}{\alpha }=5$$

$$\Rightarrow$$ $$\frac{p}{\gamma }+1=6\left(\frac{p}{\alpha }+1\right)$$ ....(viii)

Now from (v) and (viii),

$$\frac{P\left(E_1\right)}{P\left(E_3\right)}$$ = 6