QJEEAdv13Ma0b44: JEE Advanced 2013 Mathematics - Class 12 JEEAdv13Ma0b44 Solution: Of the three independent events and the probability that only…

Given, three independent events $\mathrm{E}_1, \mathrm{E}_2$ and $\mathrm{E}_3$ .

Probability that only

$\mathrm{E}_1 \text { occurs }=\mathrm{P}\left(\mathrm{E}_1 \cap \overline{\mathrm{E}}_2 \cap \overline{\mathrm{E}}_3\right)=\alpha$

Probability that only

$\mathrm{E}_2 \text { occurs }=\mathrm{P}\left(\mathrm{E}_1 \cap \overline{\mathrm{E}}_2 \cap \overline{\mathrm{E}}_3\right)=\beta$

Probability that only

$\mathrm{E}_3 \text { occurs }=P\left(\bar{E}_1 \cap \bar{E}_2 \cap E_3\right)=\gamma$

Probability that none of $E_1, E_2$ or $E_3$ occurs

$=P\left(\overline{\mathrm{E}}_1 \cap \overline{\mathrm{E}}_2 \cap \overline{\mathrm{E}}_3\right)=\rho$

$\begin{aligned}& \text { Let } \mathrm{P}\left(\mathrm{E}_1\right)=x, \mathrm{P}\left(\mathrm{E}_2\right)=y \text { and } \mathrm{P}\left(\mathrm{E}_3\right)=z \\& \alpha=\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_3\right) \\& \Rightarrow \quad \alpha=x(1-y)(1-z) \quad \text{... (i)}\\& \beta=\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_3\right) \\& \Rightarrow \quad \beta=(1-x) \cdot(y)(1-z) \quad \text{... (ii)}\\& \gamma=\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right) \cdot \mathrm{P}\left(\mathrm{E}_3\right) \\& \Rightarrow \quad \gamma=(1-x)(1-y) z \quad \text{... (iii)}\\& \rho=\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_3\right) \\& \Rightarrow \quad \rho=(1-x)(1-y)(1-z) \quad \text{... (iv)}\\\end{aligned}$

Given $(\alpha-2 \beta) \rho=a \beta$ and $(\beta-3 \gamma) \rho=2 b \gamma$

$\begin{aligned}\Rightarrow \quad & {[x(1-y)(1-z)-2(1-x) y(1-z)](1-x) } \\& (1-y)(1-z) \\& =x(1-y)(1-z) \cdot(1-x) y(1-z) \text { and } \\& {[(1-x) y(1-z)-3(1-x)(1-y) z](1-x) } \\& (1-y)(1-z) \\& =2(1-x) y(1-z) \cdot(1-x)(1-y) z \\\Rightarrow \quad & {[x-x y-2 y+2 x y](1-x)(1-y)(1-z)^2 } \\& =x y(1-x)(1-y)(1-z)^2 \text { and } \\& {[y-y z-3 z+3 y z](1-x)^2(1-y)(1-z) } \\& =2 y z(1-x)^2(1-y)(1-z) \\\Rightarrow \quad & (x-2 y+x y)=x y \text { and }(y-3 z+2 y z)=2 y z \\\Rightarrow \quad & x=2 y \text { and } y=3 z \\\Rightarrow \quad & \frac{x}{2}=y=3 z \\\Rightarrow \quad & \frac{x}{Z}=6\end{aligned}$

$\Rightarrow \frac{\text { Probability of occurrence of } E_1}{\text { Probability of occurrence of } E_3}=\frac{x}{Z}=6$

Hints :

If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are independent events, then probability of occurrence of only event A

$=\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\overline{\mathrm{B}}) \cdot \mathrm{P}(\overline{\mathrm{C}})$ .