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

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(C) Total number of marbles $= 6 + 4 + 3 + 2 = 15$.The total number of ways to pick $2$ marbles from $15$ is given by ${}^{15}C_{2} = \frac{15 	imes

Vedclass · Accepted Answer

$\frac{4}{105}$

Vedclass · Answer

(C) Total number of marbles $= 6 + 4 + 3 + 2 = 15$.The total number of ways to pick $2$ marbles from $15$ is given by ${}^{15}C_{2} = \frac{15 	imes 14}{2 	imes 1} = 105$.Probability of picking $2$ yellow marbles: There are $2$ yellow marbles,so the number of ways is ${}^{2}C_{2} = 1$. Thus,$P(	ext{both yellow}) = \frac{1}{105}$.Probability of picking $2$ green marbles: There are $3$ green marbles,so the number of ways is ${}^{3}C_{2} = 3$. Thus,$P(	ext{both green}) = \frac{3}{105}$.Since these events are mutually exclusive,the required probability is $P(	ext{both yellow or both green}) = \frac{1}{105} + \frac{3}{105} = \frac{4}{105}$.

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

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