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(C) Total number of components = $10$.Number of defective components = $4$.Number of non-defective components = $10 - 4 = 6$.We need to select $3$ com

Vedclass · Accepted Answer

$\frac{3}{10}$

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(C) Total number of components = $10$.Number of defective components = $4$.Number of non-defective components = $10 - 4 = 6$.We need to select $3$ components such that exactly $2$ are defective and $1$ is non-defective.The number of ways to choose $2$ defective components from $4$ is ${}^{4}C_{2} = \frac{4 	imes 3}{2 	imes 1} = 6$.The number of ways to choose $1$ non-defective component from $6$ is ${}^{6}C_{1} = 6$.The total number of ways to choose $3$ components from $10$ is ${}^{10}C_{3} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$.The required probability is $\frac{{}^{4}C_{2} 	imes {}^{6}C_{1}}{{}^{10}C_{3}} = \frac{6 	imes 6}{120} = \frac{36}{120} = \frac{3}{10}$.

$A$ box contains $10$ identical electronic components,of which $4$ are defective. If $3$ components are selected at random from the box in succession,without replacement,what is the probability that exactly two components of the selected components are defective?

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