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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)

best_answer

 

\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

Posted by

Suraj Bhandari

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