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(A) The total number of ways to select $2$ batteries from $8$ is given by $^{8}C_{2} = \frac{8 	imes 7}{2 	imes 1} = 28$. Let $X$ be the number of d

Vedclass · Accepted Answer

$\frac{25}{28}$

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(A) The total number of ways to select $2$ batteries from $8$ is given by $^{8}C_{2} = \frac{8 	imes 7}{2 	imes 1} = 28$. Let $X$ be the number of defective batteries. The number of defective batteries is $3$ and non-defective is $5$. We need to find $P(X \leq 1) = P(X=0) + P(X=1)$. $P(X=0)$ is the probability of selecting $0$ defective and $2$ non-defective batteries: $P(X=0) = \frac{^{3}C_{0} 	imes ^{5}C_{2}}{^{8}C_{2}} = \frac{1 	imes 10}{28} = \frac{10}{28}$. $P(X=1)$ is the probability of selecting $1$ defective and $1$ non-defective battery: $P(X=1) = \frac{^{3}C_{1} 	imes ^{5}C_{1}}{^{8}C_{2}} = \frac{3 	imes 5}{28} = \frac{15}{28}$. Therefore,$P(X \leq 1) = \frac{10}{28} + \frac{15}{28} = \frac{25}{28}$.

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$2k$	$4k$	$3k$	$k$

$X=x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$2K$	$0$	$0.3$	$K$	$0.1$

It is known that a box of $8$ batteries contains $3$ defective pieces and a person randomly selects two batteries from the box. If $X$ is the number of defective batteries selected,then $P(X \leq 1) = $

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