An urn contains 10 green,12 yellow,and 11 whi | vedclass

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(D) Total number of balls = $10 + 12 + 11 = 33$.Number of ways to select $2$ balls from $33$ is $^33C_2 = \frac{33 	imes 32}{2 	imes 1} = 528$.Numbe

Vedclass · Accepted Answer

$\frac{9}{16}$

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(D) Total number of balls = $10 + 12 + 11 = 33$.Number of ways to select $2$ balls from $33$ is $^33C_2 = \frac{33 	imes 32}{2 	imes 1} = 528$.Number of balls that are not white = $10 + 12 = 22$.Number of ways to select $2$ balls such that none are white = $^22C_2 = \frac{22 	imes 21}{2 	imes 1} = 231$.Probability of selecting no white balls = $\frac{231}{528} = \frac{7}{16}$.Probability of selecting at least one white ball = $1 - P(	ext{no white ball}) = 1 - \frac{7}{16} = \frac{9}{16}$.

An urn contains $10$ green,$12$ yellow,and $11$ white balls. What is the probability that when $2$ balls are selected,at least one is white?

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