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

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(D) 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

$\frac{1}{44}$

Vedclass · Answer

(D) Total number of marbles $= 4 + 5 + 3 = 12$.Total ways to pick $3$ marbles from $12$ is given by $^{12}C_{3} = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.Probability that all $3$ marbles are green $= \frac{^{3}C_{3}}{^{12}C_{3}} = \frac{1}{220}$.Probability that all $3$ marbles are red $= \frac{^{4}C_{3}}{^{12}C_{3}} = \frac{4}{220} = \frac{1}{55}$.Since these are mutually exclusive events,the required probability is the sum of the individual probabilities:Required probability $= \frac{1}{220} + \frac{4}{220} = \frac{5}{220} = \frac{1}{44}$.

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

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