A box contains 2 white balls,3 black balls,an | vedclass

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(A) The total number of balls in the box is $2 + 3 + 4 = 9$.We need to select $3$ balls such that at least one is black.This can be calculated by subt

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(A) The total number of balls in the box is $2 + 3 + 4 = 9$.We need to select $3$ balls such that at least one is black.This can be calculated by subtracting the number of ways to select $3$ balls with no black balls from the total number of ways to select $3$ balls.Total ways to select $3$ balls from $9$ is $^{9}C_{3} = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.Number of ways to select $3$ balls such that no black ball is chosen (i.e.,selecting from the $2$ white and $4$ red balls,total $6$ non-black balls) is $^{6}C_{3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$.Therefore,the number of ways to select at least one black ball is $84 - 20 = 64$.

$A$ box contains $2$ white balls,$3$ black balls,and $4$ red balls. In how many ways can $3$ balls be drawn from the box if at least one black ball is to be included in the draw?

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