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.

Now, P(X=0)=P(no defective bulb)$=\frac{{}^8{C}_3}{{}^{13}{C}_3}=\frac{56}{286}=\frac{28}{143}$
P(X=1)=P(1 defective bulb)$=\frac{{}^5{C}_1{\times }^8{C}_2}{{}^{13}{C}_3}=\frac{140}{286}=\frac{70}{143}$
P(X=2)=P(2 defective bulbs)$=\frac{5_2^C\times 8_1^C}{{}^{13}{C}_3}=\frac{80}{286}=\frac{40}{143}$
P(X=3)=P(3 defective bulbs)$=\frac{{}^5{C}_3}{{}^{13}{C}_3}=\frac{10}{286}=\frac{5}{143}$

Thus, the probability distribution of X is given by
$\begin{array}{cc}X& P\left(X\right)\\ 0& \frac{28}{143}\\ 1& \frac{70}{143}\\ 2& \frac{40}{143}\\ 3& \frac{5}{143}\end{array}$