From a lot containing 25 items, 5 of which are defective, 4 are chosen at random. Let X be the number of defectives found. Obtain the probability distribution of X if the items are chosen without replacement.

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Solution

Let X denote the number of defective items in a sample of 4 items drawn from a bag containing 5 defective items and 20 good items. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
$P (X = 0) = P (no defective item) = \frac{^{20} C_{4}}{^{25} C_{4}} = \frac{4845}{12650} = \frac{969}{2530} P (X = 1) = P (1 defective item) = \frac{^{5} C_{1} \times^{20} C_{3}}{^{25} C_{4}} = \frac{5700}{12650} = \frac{114}{253} P (X = 2) = P (2 defective items) = \frac{^{5} C_{2} \times^{20} C_{2}}{^{25} C_{4}} = \frac{1900}{12650} = \frac{38}{253} P (X = 3) = P (3 defective items) = \frac{^{5} C_{3} \times^{20} C_{1}}{^{25} C_{4}} = \frac{200}{12650} = \frac{4}{253} P (X = 4) = P (4 defective items) = \frac{^{5} C_{4}}{^{25} C_{4}} = \frac{5}{12650} = \frac{1}{2530}$

Thus, the probability distribution of X is given by

X P(X)

0 $\frac{969}{2530}$

1 $\frac{114}{253}$

2 $\frac{38}{253}$

3 $\frac{4}{253}$

4 $\frac{1}{2530}$

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X	P(X)
0	$\frac{969}{2530}$
1	$\frac{114}{253}$
2	$\frac{38}{253}$
3	$\frac{4}{253}$
4	$\frac{1}{2530}$