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

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(C) Total number of marbles $= 6 + 4 + 3 + 2 = 15$.Number of ways to choose $4$ marbles out of $15$ is $^{15}C_4 = \frac{15 	imes 14 	imes 13 	imes

Vedclass · Accepted Answer

$\frac{69}{91}$

Vedclass · Answer

(C) Total number of marbles $= 6 + 4 + 3 + 2 = 15$.Number of ways to choose $4$ marbles out of $15$ is $^{15}C_4 = \frac{15 	imes 14 	imes 13 	imes 12}{4 	imes 3 	imes 2 	imes 1} = 1365$.Number of non-red marbles $= 15 - 4 = 11$.Number of ways to choose $4$ marbles such that none is red is $^{11}C_4 = \frac{11 	imes 10 	imes 9 	imes 8}{4 	imes 3 	imes 2 	imes 1} = 330$.Probability of getting no red marble $= \frac{330}{1365} = \frac{22}{91}$.Probability of getting at least one red marble $= 1 - P(	ext{no red}) = 1 - \frac{22}{91} = \frac{69}{91}$.

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 at least one is red?

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