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

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(D) Total number of marbles $= 6 + 4 + 3 + 2 = 15$.The number of ways to choose $2$ marbles out of $15$ is given by $^{15}C_2 = \frac{15 	imes 14}{2

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$\frac{1}{7}$

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(D) Total number of marbles $= 6 + 4 + 3 + 2 = 15$.The number of ways to choose $2$ marbles out of $15$ is given by $^{15}C_2 = \frac{15 	imes 14}{2 	imes 1} = 105$.The number of ways to choose $2$ blue marbles out of $6$ blue marbles is given by $^{6}C_2 = \frac{6 	imes 5}{2 	imes 1} = 15$.The probability of picking both blue marbles is $\frac{^{6}C_2}{^{15}C_2} = \frac{15}{105} = \frac{1}{7}$.

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 both are blue?

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