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Q.

The chance of happening of an event is the square of the chance of a second event but the odds against the first are cube of the odds against the second. If the chances of events are P₁, P₂ respectively, then

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a

$P_{1} = \frac{1}{9}$

b

$P_{1} = \frac{1}{6}$

c

$P_{2} = \frac{1}{3}$

d

$P_{2} = \frac{1}{4}$

answer is A, C.

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Detailed Solution

From given data, $P_{1} = P_{2}^{2} \to$
$\frac{\bar{P_{1}}}{P_{1}} = {(\frac{\bar{P_{2}}}{P_{2}})}^{3} \to (2) . Also \bar{P_{1}} = 1 - P_{1}, \bar{P_{2}} = 1 - P_{2}$
From (1), (2), eliminating P₁ we get
$P_{2}^{3} (1 - P_{2}^{2}) = P_{2}^{2} {(1 - P_{2})}^{3}$
Solving we get $P_{2} = \frac{1}{3}, P_{1} = \frac{1}{9}$

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