Two bolts,two nuts,and three needles are in a | vedclass

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(C) Total number of items = $2 	ext{ (bolts)} + 2 	ext{ (nuts)} + 3 	ext{ (needles)} = 7 	ext{ items}$.Total number of ways to choose $2$ parts fr

Vedclass · Accepted Answer

$\frac{6}{21}$

Vedclass · Answer

(C) Total number of items = $2 	ext{ (bolts)} + 2 	ext{ (nuts)} + 3 	ext{ (needles)} = 7 	ext{ items}$.Total number of ways to choose $2$ parts from $7$ is given by $n(S) = {}^{7}C_{2} = \frac{7 	imes 6}{2 	imes 1} = 21$.Let $E$ be the event that one part is a bolt and one part is a needle.The number of ways to choose $1$ bolt from $2$ and $1$ needle from $3$ is $n(E) = {}^{2}C_{1} 	imes {}^{3}C_{1} = 2 	imes 3 = 6$.Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{6}{21}$.

Two bolts,two nuts,and three needles are in a box. Two parts are chosen at random from the box. What is the probability that one is a bolt and one is a needle?

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