An urn contains 5 red marbles,4 black marbles | vedclass

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(D) Total number of marbles $= 5 + 4 + 3 = 12$. We need to draw $4$ marbles from $12$ marbles. The total number of ways to draw $4$ marbles is $^{12}C

Vedclass · Accepted Answer

$490$

Vedclass · Answer

(D) Total number of marbles $= 5 + 4 + 3 = 12$. We need to draw $4$ marbles from $12$ marbles. The total number of ways to draw $4$ marbles is $^{12}C_{4} = \frac{12 	imes 11 	imes 10 	imes 9}{4 	imes 3 	imes 2 	imes 1} = 495$. We want to find the number of ways such that at most $3$ marbles are red. This is equivalent to: (Total ways) - (Ways where all $4$ marbles are red). The number of ways to choose $4$ red marbles from $5$ red marbles is $^{5}C_{4} = 5$. Therefore,the number of ways to draw $4$ marbles such that at most $3$ are red is $495 - 5 = 490$.

An urn contains $5$ red marbles,$4$ black marbles and $3$ white marbles. The number of ways in which $4$ marbles can be drawn so that at most three of them are red is

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