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Let $B_{f} (i = 1, 2, 3)$ be three independent events in a sample space. The probability that only B1 occur is $α,$ only $B_{2}$ occurs is $β$ and only $B_{3}$ occurs is $γ .$ Let p be the probability that none of the events $B_{f}$ occurs and these 4 probabilities satisfy the equations $(α - 2 β) p = α β$ and $(β - 3 γ) p = - 2 β γ$ (All the probabilities are assumed to lie in the interval (0, 1). Then $\frac{P (B_{1})}{P (B_{3})}$ is equal to _________.

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Detailed Solution

Let x, y, z be probability of B1, B2, B3 respectively $\Rightarrow x (1 - y) (1 - z) = α \Rightarrow y (1 - x) (1 - z) = β$

$\Rightarrow z (1 - x) (1 - y) = γ \Rightarrow (1 - x) (1 - z) = P$

Putting in the given relation we get x = 2y and y = 3z $\Rightarrow x = 6 z \Rightarrow \frac{x}{z} = 6$

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