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

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(B) Total number of balls $= 10 + 12 + 11 = 33$.The number of ways to select $2$ balls out of $33$ is given by the combination formula $nC_r = \frac{n

Vedclass · Accepted Answer

$\frac{15}{176}$

Vedclass · Answer

(B) Total number of balls $= 10 + 12 + 11 = 33$.The number of ways to select $2$ balls out of $33$ is given by the combination formula $nC_r = \frac{n!}{r!(n-r)!}$.Total ways $= 33C_2 = \frac{33 	imes 32}{2 	imes 1} = 33 	imes 16 = 528$.The number of ways to select $2$ green balls out of $10$ green balls is $10C_2$.Favorable ways $= 10C_2 = \frac{10 	imes 9}{2 	imes 1} = 45$.The required probability is the ratio of favorable outcomes to total outcomes.Probability $= \frac{45}{528}$.Dividing both numerator and denominator by $3$,we get $\frac{15}{176}$.

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

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