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

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(B) Total number of balls $= 10 + 12 + 11 = 33$.Number of non-white balls (green + yellow) $= 10 + 12 = 22$.When selecting $2$ balls without replaceme

Vedclass · Accepted Answer

$\frac{7}{16}$

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(B) Total number of balls $= 10 + 12 + 11 = 33$.Number of non-white balls (green + yellow) $= 10 + 12 = 22$.When selecting $2$ balls without replacement,the probability that none is white is the probability that both selected balls are non-white.Probability $= \frac{22}{33} 	imes \frac{21}{32}$.Simplifying the fractions: $\frac{22}{33} = \frac{2}{3}$ and $\frac{21}{32}$ remains as is.Probability $= \frac{2}{3} 	imes \frac{21}{32} = \frac{1}{1} 	imes \frac{7}{16} = \frac{7}{16}$.

An urn contains $10$ green,$12$ yellow,and $11$ white balls. What is the probability that when $2$ balls are selected,none is white?

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