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(D) Total number of marbles $= 4 + 5 + 3 = 12$.$n(S) =$ Total ways to pick $3$ marbles out of $12 = {}^{12}C_{3} = \frac{12 	imes 11 	imes 10}{3 	i

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$\frac{1}{44}$

Vedclass · Answer

(D) Total number of marbles $= 4 + 5 + 3 = 12$.$n(S) =$ Total ways to pick $3$ marbles out of $12 = {}^{12}C_{3} = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.$n(E) =$ Favourable outcomes (either all green or all red).Ways to pick $3$ green marbles from $3 = {}^{3}C_{3} = 1$.Ways to pick $3$ red marbles from $4 = {}^{4}C_{3} = 4$.Total favourable outcomes $n(E) = 1 + 4 = 5$.Required probability $= \frac{n(E)}{n(S)} = \frac{5}{220} = \frac{1}{44}$.

Study the given information carefully and answer the questions that follow: $A$ basket contains $4$ red,$5$ blue and $3$ green marbles. If $3$ marbles are picked at random,what is the probability that either all are green or all are red?

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