JEE Main 2021 — Mathematics Probability & Statistics

Let $P({B}_{1})={p}_{1},P({B}_{2})={p}_{2},P({B}_{3})={p}_{3}$P(B1​)=p1​,P(B2​)=p2​,P(B3​)=p3​given that ${p}_{1}(1-{p}_{2})(1-{p}_{3})=\alpha \ldots (i)$p1​(1−p2​)(1−p3​)=α…(i)

$${p}_{2}(1-{p}_{1})(1-{p}_{3})=\beta \ldots (\mathrm{ii})$$p2​(1−p1​)(1−p3​)=β…(ii)

$${p}_{3}(1-{p}_{1})(1-{p}_{2})=\gamma \ldots (\mathrm{iii})$$p3​(1−p1​)(1−p2​)=γ…(iii)

and $(1-{p}_{1})(1-{p}_{2})(1-{p}_{3})=p\ldots (\mathrm{iv})$(1−p1​)(1−p2​)(1−p3​)=p…(iv)

$\Rightarrow \frac{{p}_{1}}{1-{p}_{1}}=\frac{\alpha }{p},\frac{{p}_{2}}{1-{p}_{2}}=\frac{\beta }{p}$⇒1−p1​p1​​=pα​,1−p2​p2​​=pβ​ and $\frac{{p}_{3}}{1-{p}_{3}}=\frac{\gamma }{p}$1−p3​p3​​=pγ​

Also $\beta =\frac{\alpha p}{\alpha +2p}=\frac{3\gamma p}{p-2\gamma }$β=α+2pαp​=p−2γ3γp​

$$\Rightarrow \alpha p-2\alpha \gamma =3\alpha \gamma +6p\gamma$$⇒αp−2αγ=3αγ+6pγ

$$\Rightarrow \alpha p-6p\gamma =5\alpha \gamma$$⇒αp−6pγ=5αγ

$$\Rightarrow \frac{{p}_{1}}{1-{p}_{1}}-\frac{6{p}_{3}}{1-{p}_{3}}=\frac{5{p}_{1}{p}_{3}}{(1-{p}_{1})(1-{p}_{3})}$$⇒1−p1​p1​​−1−p3​6p3​​=(1−p1​)(1−p3​)5p1​p3​​

$$\Rightarrow {p}_{1}-6{p}_{3}=0$$⇒p1​−6p3​=0

$$\Rightarrow \frac{{p}_{1}}{{p}_{3}}=6$$⇒p3​p1​​=6