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(B) Total number of marbles $= 4 + 5 + 3 = 12$.Total ways to pick $3$ marbles from $12$ is $n(S) = {}^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes

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$\frac{37}{44}$

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(B) Total number of marbles $= 4 + 5 + 3 = 12$.Total ways to pick $3$ marbles from $12$ is $n(S) = {}^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.To find the probability that at least $1$ marble is blue,we use the complement rule: $P(	ext{at least } 1 	ext{ blue}) = 1 - P(	ext{no blue marble})$.Number of non-blue marbles $= 4 	ext{ (red)} + 3 	ext{ (green)} = 7$.Ways to pick $3$ marbles such that none are blue is $n(E') = {}^{7}C_3 = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$.Probability of picking no blue marbles $P(E') = \frac{35}{220} = \frac{7}{44}$.Therefore,the probability of picking at least $1$ blue marble is $1 - \frac{7}{44} = \frac{37}{44}$.

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 at least $1$ is blue?

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