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(D) Total number of marbles in the urn $= 4 + 5 + 2 + 3 = 14$.Total number of ways to select $2$ marbles out of $14$ is given by ${}^{14}C_2 = \frac{1

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(D) Total number of marbles in the urn $= 4 + 5 + 2 + 3 = 14$.Total number of ways to select $2$ marbles out of $14$ is given by ${}^{14}C_2 = \frac{14 	imes 13}{2 	imes 1} = 91$.The event 'both are red or at least $1$ is red' is equivalent to the event 'at least $1$ is red'.Number of ways to select $0$ red marbles $= {}^{12}C_2 = \frac{12 	imes 11}{2 	imes 1} = 66$.Number of ways to select at least $1$ red marble $= 	ext{Total ways} - 	ext{Ways to select } 0 	ext{ red marbles} = 91 - 66 = 25$.Therefore,the required probability $= 25/91$.

Study the given information carefully to answer the question that follows. An urn contains $4$ green,$5$ blue,$2$ red,and $3$ yellow marbles. If $2$ marbles are drawn at random,what is the probability that both are red or at least $1$ is red?

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