A box contains 8 batteries, of which 3 are de | vedclass

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(A) Total batteries $= 8$, Defective $= 3$, Non-defective $= 5$. Two batteries are selected. $X$ can take values $0, 1, 2$.Total ways to select $2$ ba

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$X = x$$0$$1$$2$$P(X = x)$$\frac{10}{28}$$\frac{15}{28}$$\frac{3}{28}$

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(A) Total batteries $= 8$, Defective $= 3$, Non-defective $= 5$. Two batteries are selected. $X$ can take values $0, 1, 2$.Total ways to select $2$ batteries $= ^8C_2 = \frac{8 	imes 7}{2} = 28$.For $X = 0$ (no defective): $P(X=0) = \frac{^3C_0 	imes ^5C_2}{28} = \frac{1 	imes 10}{28} = \frac{10}{28}$.For $X = 1$ (one defective): $P(X=1) = \frac{^3C_1 	imes ^5C_1}{28} = \frac{3 	imes 5}{28} = \frac{15}{28}$.For $X = 2$ (two defective): $P(X=2) = \frac{^3C_2 	imes ^5C_0}{28} = \frac{3 	imes 1}{28} = \frac{3}{28}$.Thus, the distribution is given by option $A$.

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$\lambda$	$2\lambda$	$3\lambda$	$4\lambda$

$X = x_i$	$0$	$1$	$2$	$3$
$P(X = x_i)$	$\frac{1}{8}$	$\frac{3}{8}$	$3K$	$K$

$X$	$-3$	$-1$	$0$	$1$	$3$	$5$	$7$	$9$
$F(X)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

$A$ box contains $8$ batteries, of which $3$ are defective. If a person randomly selects $2$ batteries from this box, find the probability distribution of the number of defective batteries $X$.

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$X = x$	$0$	$1$	$2$
$P(X = x)$	$\frac{10}{28}$	$\frac{15}{28}$	$\frac{3}{28}$