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(B) The total number of items is $10$,with $3$ defective and $7$ non-defective items. $A$ sample of $4$ items is drawn without replacement. The random

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$\frac{4}{5}$

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(B) The total number of items is $10$,with $3$ defective and $7$ non-defective items. $A$ sample of $4$ items is drawn without replacement. The random variable $X$ follows a hypergeometric distribution.The probability mass function is given by $P(X = x) = \frac{^3C_x 	imes ^7C_{4-x}}{^{10}C_4}$.The total number of ways to choose $4$ items from $10$ is $^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$.We need to find $P(0 For $x = 1$: $P(X = 1) = \frac{^3C_1 	imes ^7C_3}{210} = \frac{3 	imes 35}{210} = \frac{105}{210}$.For $x = 2$: $P(X = 2) = \frac{^3C_2 	imes ^7C_2}{210} = \frac{3 	imes 21}{210} = \frac{63}{210}$.Therefore,$P(0 < X < 3) = \frac{105}{210} + \frac{63}{210} = \frac{168}{210} = \frac{4}{5}$.

$A$ sample of $4$ items is drawn at random without replacement from a lot of $10$ items containing $3$ defective items. If $X$ denotes the number of defective items in the sample,then $P(0 < X < 3)$ is equal to:

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