A basket contains 4 red,5 blue,and 3 green ma | vedclass

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(C) Total number of marbles $= 4 + 5 + 3 = 12$.We need to select $3$ marbles out of $12$. The total number of ways to select $3$ marbles is given by $

Vedclass · Accepted Answer

$\frac{9}{110}$

Vedclass · Answer

(C) Total number of marbles $= 4 + 5 + 3 = 12$.We need to select $3$ marbles out of $12$. The total number of ways to select $3$ marbles is given by ${}^{12}C_{3} = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.The number of ways to select $2$ red marbles out of $4$ is ${}^{4}C_{2} = \frac{4 	imes 3}{2 	imes 1} = 6$.The number of ways to select $1$ green marble out of $3$ is ${}^{3}C_{1} = 3$.The number of favorable outcomes is ${}^{4}C_{2} 	imes {}^{3}C_{1} = 6 	imes 3 = 18$.Required Probability $= \frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{18}{220} = \frac{9}{110}$.

$A$ basket contains $4$ red,$5$ blue,and $3$ green marbles. If three marbles are picked at random,what is the probability that $2$ are red and $1$ is green?

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