Of the three independent events \(E_{1}, E_{2}\) and \(E_{3}\), the probability that only \(E_{1}\) occurs is \(\alpha\), only \(E_{2}\) occurs is \(\beta\) and only \(E_{3}\) occurs is \(\gamma\). Let the probability \(p\) that none of events \(E_{1}, E_{2}\) or \(E_{3}\) occurs satisfy the equations \((\alpha-2 \beta) p=\alpha \beta\) and \((\beta-3 \gamma) p=2 \beta \gamma\). All the given probabilities are assumed to lie in the interval \((0,1)\). \[ \text { Then } \frac{\text { Probability of occurrence of } E_{1}}{\text { Probability of occurrence of } E_{3}}= \]

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Probability XII

Of the three independent events

E_{1}, E_{2}

and

E_{3}

, the probability that only

E_{1}

occurs is

α

, only

E_{2}

occurs is

β

and only

E_{3}

occurs is

γ

. Let the probability

p

that none of events

E_{1}, E_{2}

E_{3}

occurs satisfy the equations

(α - 2 β) p = α β

and

(β - 3 γ) p = 2 β γ

. All the given probabilities are assumed to lie in the interval

(0, 1)

Then \frac{Probability of occurrence of E _{1}}{Probability of occurrence of E _{3}} =

Answer: 6

Let

P (E_{1}) = p_{1}

P (E_{2}) = p_{2}

P (E_{3}) = p_{3}

. Then

α = p_{1} (1 - p_{2}) (1 - p_{3})

β = p_{2} (1 - p_{1}) (1 - p_{3})

γ = p_{3} (1 - p_{1}) (1 - p_{2})

, and

p = (1 - p_{1}) (1 - p_{2}) (1 - p_{3})

. Thus

\frac{α}{p} = \frac{p _{1}}{1 - p _{1}}

\frac{β}{p} = \frac{p _{2}}{1 - p _{2}}

\frac{γ}{p} = \frac{p _{3}}{1 - p _{3}}

. From

(α - 2 β) p = α β

\frac{p}{β} - \frac{2 p}{α} = 1

. From

(β - 3 γ) p = 2 β γ

\frac{p}{γ} - \frac{3 p}{β} = 2

. Solving these gives

\frac{p}{γ} - \frac{6 p}{α} = 5

, which leads to

\frac{p _{1}}{p _{3}} = 6

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Integer answer