Of the three independent events $E_{1}, E_{2}$ and $E_{3}$ the probability that only $E_{1}$ occurs $α$ , only $E_{2}$ occurs is $β$ and $E_{3}$ occurs is $γ .$ Let the probability $p$ that none of the events $E_{1}, E_{2}$ OR $E_{3}$ occurs satisfy the equation $(α - 2 β) p = α β$ and $(β - 3 γ) p = 2 β γ$ . All the given probabilities are assumed to lie in the interval (0,1) then $\frac{p r o b a b i l i t y o f o c c u r e n c e o f E_{1}}{p r o b a b i l i t y o f o c c u r e n c e o f E_{3}} i s$

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Detailed Solution

$l e t P (E_{1}) = y, P (E_{2}) = y, p (E_{3}) = z$

$α = x (1 - y) (1 - z), α = \frac{p x}{1 - x} {∴ p = (1 - x) (1 - y) (1 - z)}$

Similar $β = \frac{p y}{1 - y}, γ = \frac{p z}{1 - z}$

$n o w (α - 2 β) p = α β$

$\begin{array}{l} \Rightarrow \frac{p}{β} - \frac{2 p}{α} = 1 \\ \Rightarrow \frac{1 - y}{y} - \frac{2 (1 - x)}{x} = 1 \\ \Rightarrow x = 2 y \end{array}$

$(β - 3 γ) p = 2 β γ$

$\begin{array}{l} \Rightarrow \frac{p}{γ} - \frac{3 p}{β} = 2 \\ \Rightarrow \frac{1 - z}{z} - \frac{3 (1 - y)}{y} = 2 \\ \Rightarrow y = 3 z \end{array}$

$\begin{array}{l} ∴ x = 6 z \\ \Rightarrow \frac{x}{z} = 6 \end{array}$

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