Question

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 .

 

Answer 1

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,

\[P\left( X = 0 \right)\]
\[ = P\left( \text{ no defective item } \right)\]
\[ = \frac{{}^{20} C_4}{{}^{25} C_4}\]
\[ = \frac{4845}{12650}\]
\[ = \frac{969}{2530}\]
\[P\left( X = 1 \right)\]
\[ = P\left( 1 \text{ defective item } \right)\]
\[ = \frac{{}^5 C_1 \times^{20} C_3}{{}^{25} C_4}\]
\[ = \frac{5700}{12650}\]
\[ = \frac{114}{253}\]
\[P\left( X = 2 \right)\]
\[ = P\left( 2 \text{ defective items } \right)\]
\[ = \frac{{}^5 C_2 \times^{20} C_2}{{}^{25} C_4}\]
\[ = \frac{1900}{12650}\]
\[ = \frac{38}{253}\]
\[P\left( X = 3 \right)\]
\[ = P\left( 3 \text{ defective items } \right)\]
\[ = \frac{{}^5 C_3 \times^{20} C_1}{{}^{25} C_4}\]
\[ = \frac{200}{12650}\]
\[ = \frac{4}{253}\]
\[P\left( X = 4 \right)\]
\[ = P\left( 4 \text{ defective items}  \right)\]
\[ = \frac{{}^5 C_4}{{}^{25} C_4}\]
\[ = \frac{5}{12650}\]
\[ = \frac{1}{2530}\]

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	\[\frac{1}{2530}\]