An URN contains 6 blue,4 red,3 green and 2 ye | vedclass

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(D) Total number of marbles $= 6 + 4 + 3 + 2 = 15$.We need to select $4$ marbles out of $15$. The total number of ways to select $4$ marbles is given

Vedclass · Accepted Answer

$\frac{24}{455}$

Vedclass · Answer

(D) Total number of marbles $= 6 + 4 + 3 + 2 = 15$.We need to select $4$ marbles out of $15$. The total number of ways to select $4$ marbles is given by ${}^{15}C_{4} = \frac{15 	imes 14 	imes 13 	imes 12}{4 	imes 3 	imes 2 	imes 1} = 1365$.The number of ways to select $1$ yellow marble from $2$ is ${}^{2}C_{1} = 2$.The number of ways to select $2$ red marbles from $4$ is ${}^{4}C_{2} = \frac{4 	imes 3}{2 	imes 1} = 6$.The number of ways to select $1$ blue marble from $6$ is ${}^{6}C_{1} = 6$.The number of favorable outcomes $= {}^{2}C_{1} 	imes {}^{4}C_{2} 	imes {}^{6}C_{1} = 2 	imes 6 	imes 6 = 72$.Required Probability $= \frac{72}{1365} = \frac{24}{455}$.

An $URN$ contains $6$ blue,$4$ red,$3$ green and $2$ yellow marbles. If four marbles are picked at random,what is the probability that one is yellow,two are red and one is blue?

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