Of the three independent events E1, E2 and E3, the probability that only E1 occurs is \alpha, only E2 occurs is \beta and only E3 occurs is \gamma. Let the probability p that none of events E1, E2 or E3 occurs satisfy the equations (\alpha –2\beta) p = αβ and (\beta – 3\gamma) p = 2βγ. All the given probabilities are assumed to lie in the interval (0, 1). Then \frac{Probability \textrm{ }\textrm{ } of \textrm{ }\textrm{ } occurence \textrm{ }\textrm{ } of \textrm{ }\textrm{ } E_{1}}{Probability \textrm{ }\textrm{ } of \textrm{ }\textrm{ } occurence \textrm{ }\textrm{ } of \textrm{ }\textrm{ } E

Engineering

Mathematics

Conditional Probability

Question

Of the three independent events E₁, E₂ and E₃, the probability that only E₁ occurs is $α$ , only E₂ occurs is $β$ and only E₃ occurs is $γ$ . Let the probability p that none of events E₁, E₂ or E₃ occurs satisfy the equations ( $α$ –2 $β$ ) p = $αβ$ and ( $β$ – 3 $γ$ ) p = 2 $βγ$ .

All the given probabilities are assumed to lie in the interval (0, 1).

Then $\frac{Probability of occurence of E_{1}}{Probability of occurence of E_{3}} =$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Let the probability of occurrence of event E₁, E₂ and E₃ are a, b and c.

$∴$ P(E₁) = a, P(E₂) = b and P(E₃) = c

Given,

P (only E₁ occur) = $P (E_{1} {\bar{E}}_{2} {\bar{E}}_{3})$ = a (1 – b) (1 – c) ……. (1)

P (only E₂ occur) = $P (E_{1} {\bar{E}}_{2} {\bar{E}}_{3})$ = (1 – a) b (1 – c) ……. (2)

P (only E₃ occur) = $P (E_{1} {\bar{E}}_{2} {\bar{E}}_{3})$ = (1 – a) (1 – b) c ……. (3)

P (none of event occur) = p = $P (E_{1} {\bar{E}}_{2} {\bar{E}}_{3})$

$\Rightarrow$ P = (1 – a) (1 – b) (1 – c) …….(4)

Given, ( $α$ – 2 $β$ ) p = $αβ$

$∴ (\frac{1}{β} - \frac{2}{α}) p = 1$

Put value of p, $α, β$ from equation (1), (2), (3) and (4)

$we get \frac{1}{b} = \frac{2}{a}$ …….(5)

Given, $(β - 3 γ) p = 28 γ \Rightarrow (\frac{1}{γ} - \frac{3}{2 β}) p = 2$

Put value of $γ, β$ and p, we get

$\frac{1}{c} = \frac{3}{b}$ …….(6)

From (5) and (6), we get $\frac{a}{c}$ = 6.

Please subscribe our Youtube channel to unlock this solution.