Let there be three independent events $E_{1},E_{2}$ and $E_{3}$ . The probabolity that only $E_{1}$ occurs is $\alpha$ , only $E_{2}$ occurs is $\beta$ and only $E_{3}$ occurs is $\gamma$ . Let 'p' denote the probability of none of events occurs that satisfies the equation $\left ( \alpha -2\beta \right )p=\alpha \beta$ and $\left ( \beta -3\gamma \right )p=2\beta \gamma$ . All the given probabilities are assumed to lie in the interval $\left ( 0,1 \right ).$ Then, $\frac{\text{Probability of occurrence of E}_{1}}{\text {Probability of occurrence of E}_{3}}$ is equal to _______________.
Option: 1 5
Option: 2 6
Option: 3 7
Option: 4 8

Answers (1)

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

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

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

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

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

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

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

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

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

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

Exams

Colleges

Predictors

Resources

Quick links

Quick Links

Exams

Colleges

Resources

Colleges

Resources

Exams

Exams

Colleges

Predictor

Resources

Other Exams

Exams

Resources

Upcoming Events

Exams

Ranking

Products & Resources

NCERT Solutions

Exams

Resources

Exams

Colleges

Predictors

Resources

Exams

Colleges

Predictors

Resources

Online Courses

Products

Engineering Preparation

Medical Preparation

Exams

Colleges

Predictors

Resources

Resources

Exams

Colleges

Predictors

Exams

Predictors

Colleges

Resources

Exams

Colleges

Resources

Exams

Resources

Similar Questions

Latest Asked Questions

Most Viewed Questions

Preparation Products

Knockout JEE Main April 2021 (One Month)

Knockout JEE Main May 2021

Test Series JEE Main May 2021

Knockout JEE Main May 2022

JEE Main Rank Booster 2021

Ask Question

Welcome Back :)

Register to post Answer

Welcome Back :)