An urn contains 9 red,7 white and 4 black bal | vedclass

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(C) Total number of balls $= 9 + 7 + 4 = 20$.The number of ways to choose $2$ balls out of $20$ is given by $^{20}C_{2} = \frac{20 	imes 19}{2 	imes

Vedclass · Accepted Answer

$\frac{63}{190}$

Vedclass · Answer

(C) Total number of balls $= 9 + 7 + 4 = 20$.The number of ways to choose $2$ balls out of $20$ is given by $^{20}C_{2} = \frac{20 	imes 19}{2 	imes 1} = 190$.The number of ways to draw one white ball out of $7$ is $^{7}C_{1} = 7$.The number of ways to draw one red ball out of $9$ is $^{9}C_{1} = 9$.The number of favorable outcomes is $^{7}C_{1} 	imes ^{9}C_{1} = 7 	imes 9 = 63$.Therefore,the required probability $= \frac{63}{190}$.

An urn contains $9$ red,$7$ white and $4$ black balls. If two balls are drawn at random,then the probability that one is white and the other is red is

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