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

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(B) Total number of marbles $= 4 + 5 + 3 = 12$.Total ways to pick $3$ marbles from $12$ is given by $^{12}C_{3} = \frac{12 	imes 11 	imes 10}{3 	im

Vedclass · Accepted Answer

$\frac{37}{44}$

Vedclass · Answer

(B) Total number of marbles $= 4 + 5 + 3 = 12$.Total ways to pick $3$ marbles from $12$ is given by $^{12}C_{3} = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.Number of non-blue marbles $= 4 + 3 = 7$.Ways to pick $3$ marbles such that none is blue $= ^{7}C_{3} = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$.Probability that none is blue $= \frac{35}{220} = \frac{7}{44}$.Probability that at least one is blue $= 1 - P(	ext{none is blue}) = 1 - \frac{7}{44} = \frac{37}{44}$.

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

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