# Let there be three independent events and . The probabolity that only occurs is , only occurs is and only  occurs  is . Let 'p' denote the probability of none of events occurs that satisfies the equation  and . All the given probabilities are assumed to lie in the interval  Then, is equal to _______________. Option: 1 5 Option: 2 6 Option: 3 7 Option: 4 8

$\text { Let } P \left( E _{1}\right)= P _{1} ; P \left( E _{2}\right)= P _{2} ; P \left( E _{3}\right)= P _{3}$

$P \left( E _{1} \cap \overline{ E }_{2} \cap \overline{ E }_{3}\right)=\alpha= P _{1}\left(1- P _{2}\right)\left(1- P _{3}\right) \ldots \ldots .(1)$

$P \left(\overline{ E }_{1} \cap E _{2} \cap \overline{ E }_{3}\right)=\beta=\left(1- P _{1}\right) P _{2}\left(1- P _{3}\right) \ldots \ldots.(2)$

$P \left(\overline{ E }_{1} \cap \overline{ E }_{2} \cap E _{3}\right)=\gamma=\left(1- P _{1}\right)\left(1- P _{2}\right) P _{3} \ldots .\ldots(3)$

$P \left(\overline{ E }_{1} \cap \overline{ E }_{2} \cap \overline{ E }_{3}\right)= P =\left(1- P _{1}\right)\left(1- P _{2}\right)\left(1- P _{3}\right) \ldots \ldots(4)$

${\color{Blue} \text { Given that, }(\alpha-2 \beta) P =\alpha \beta }$

\begin{aligned} &\Rightarrow\left( P _{1}\left(1- P _{2}\right)\left(1- P _{3}\right)-2\left(1- P _{1}\right) P _{2}\left(1- P _{3}\right)\right) P = P _{1} P _{2} \left(1- P _{1}\right)\left(1- P _{2}\right)\left(1- P _{3}\right)^{2} \\ &\Rightarrow \left( P _{1}\left(1- P _{2}\right)-2\left(1- P _{1}\right) P _{2}\right)= P _{1} P _{2} \\ &\Rightarrow \left( P _{1}- P _{1} P _{2}-2 P _{2}+2 P _{1} P _{2}\right)= P _{1} P _{2} \\ &\Rightarrow P _{1}=2 P _{2} \ldots \ldots .(5) \end{array}

${\color{Blue} \text { and similarly, }(\beta-3 \gamma) P =2 \beta \gamma }$

\begin{aligned} &\Rightarrow\left( \left(1- P _{1}\right)P _{2}\left(1- P _{3}\right)-3\left(1- P _{1}\right) \left(1- P _{2}\right)P _{3}\right) P = 2P _{3} P _{2} \left(1- P _{3}\right)\left(1- P _{2}\right)\left(1- P _{1}\right)^{2} \\ &\Rightarrow \left( P _{2}\left(1- P _{3}\right)-3\left(1- P _{2}\right) P _{3}\right)= 2P _{3} P _{2} \\ &\Rightarrow \left( P _{2}- P _{3} P _{2}-3 P _{3}+3 P _{3} P _{2}\right)= 2P _{3} P _{2} \\ &\Rightarrow P _{2}=3 P _{3} \ldots \ldots .(6) \end{array}

$\text { So, } P _{1}=6 P _{3} \Rightarrow \frac{ P _{1}}{ P _{3}}=6$

## Most Viewed Questions

### Preparation Products

##### Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
##### Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
##### Test Series JEE Main May 2021

Unlimited Chapter Wise Tests, Unlimited Subject Wise Tests, Unlimited Full Mock Tests, Get Personalized Performance Analysis Report,.

₹ 6999/- ₹ 2999/-
##### Knockout JEE Main May 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Live Classes, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 24999/-