Question

A box contains 13 bulbs, out of which 5 are defective. 3 bulbs are randomly drawn, one by one without replacement, from the box. Find the probability distribution of the number of defective bulbs.

Answer 1

Let X denote the number of defective bulbs in a sample of 3 bulbs drawn from a bag containing 13 bulbs. 5 bulbs in the bag turn out to be defective. Then, X can take the values 0, 1, 2 and 3.

\[P\left( X = 0 \right) = P\left( \text{ no defective bulb } \right) = \frac{{}^8 C_3}{{}^{13} C_3} = \frac{56}{286} = \frac{28}{143}\]
\[P\left( X = 1 \right) = P\left( 1 \text{ defective bulb}  \right) = \frac{{}^5 C_1 \times^8 C_2}{{}^{13} C_3} = \frac{140}{286} = \frac{70}{143}\]
\[P\left( X = 2 \right) = P\left( 2 \text{ defective bulbs } \right) = \frac{{}^5 C_2 \times^8 C_1}{{}^{13} C_3} = \frac{80}{286} = \frac{40}{143}\]
\[P\left( X = 3 \right) = P\left( 3 \text{ defective bulbs } \right) = \frac{{}^5 C_3}{{}^{13} C_3} = \frac{10}{286} = \frac{5}{143}\]

Thus, the probability distribution of X is given by

X	P(X)
0	\[\frac{28}{143}\]
1	\[\frac{70}{143}\]
2	\[\frac{40}{143}\]
3	\[\frac{5}{143}\]