An urn contains 9 red balls and p green balls. If the probability of picking a red ball is thrice that of picking a green ball, then p is equal to

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Solution

The correct option is D 3
Probability of picking a red ball = $\frac{9}{9 + p}$

Probability of picking a green ball = $\frac{p}{9 + p}$

Given,
$\frac{9}{9 + p} = \frac{3 p}{9 + p}$
$3 p = 9$
$p = 3$

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An urn contains 9 red balls and p green balls. If the probability of picking a red ball is thrice that of picking a green ball, then p is equal to

The correct option is D 3Probability of picking a red ball = 99+pProbability of picking a green ball = p9+pGiven,99+p=3p9+p3p=9p=3

The correct option is D 3
Probability of picking a red ball = $\frac{9}{9 + p}$

Probability of picking a green ball = $\frac{p}{9 + p}$

Given,
$\frac{9}{9 + p} = \frac{3 p}{9 + p}$
$3 p = 9$
$p = 3$