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

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

Vedclass · Accepted Answer

$\frac{18}{455}$

Vedclass · Answer

(C) Total number of marbles = $6 + 4 + 3 + 2 = 15$.We need to choose $3$ marbles out of $15$. The total number of ways to select $3$ marbles is given by $^{15}C_{3} = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1} = 455$.The number of ways to choose $2$ red marbles from $4$ red marbles is $^{4}C_{2} = \frac{4 	imes 3}{2 	imes 1} = 6$.The number of ways to choose $1$ green marble from $3$ green marbles is $^{3}C_{1} = 3$.The number of favorable outcomes is $^{4}C_{2} 	imes ^{3}C_{1} = 6 	imes 3 = 18$.Therefore,the probability is $\frac{18}{455}$.

An urn contains $6$ blue,$4$ red,$3$ green,and $2$ yellow marbles. If three marbles are picked at random,what is the probability that $2$ are red and $1$ is green?

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