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 such that at most $3$ are red.This is equivalent to: (Total ways to draw $4

Vedclass · Accepted Answer

$490$

Vedclass · Answer

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