Question

Five defective bolts are accidently mixed with twenty good ones. If four bolts are drawn at random from this lot, find the probability distribution of the number of defective bolts.

Answer 1

Let X denote the number of defective bolts in a sample of 4 bolts drawn from a bag containing 5 defective bolts and 20 good bolts. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
$$\begin{gathered} P\left(X=0\right) \\ =P\left(\mathrm{no}\mathrm{defective}\mathrm{bolts}\right) \\ =\frac{{}^{20}{C}_4}{{}^{25}{C}_4} \\ =\frac{4845}{12650} \\ =\frac{969}{2530} \\ P\left(X=1\right) \\ =P\left(1\mathrm{defective}\mathrm{bolt}\right) \\ =\frac{{}^5{C}_1{\times }^{20}{C}_3}{{}^{25}{C}_4} \\ =\frac{5700}{12650} \\ =\frac{114}{253} \\ P\left(X=2\right) \\ =P\left(2\mathrm{defective}\mathrm{bolts}\right) \\ =\frac{{}^5{C}_2{\times }^{20}{C}_2}{{}^{25}{C}_4} \\ =\frac{1900}{12650} \\ =\frac{38}{253} \\ P\left(X=3\right) \\ =P\left(3\mathrm{defective}\mathrm{bolts}\right) \\ =\frac{{}^5{C}_3{\times }^{20}{C}_1}{{}^{25}{C}_4} \\ =\frac{200}{12650} \\ =\frac{4}{253} \\ P\left(X=4\right) \\ =P\left(4\mathrm{defective}\mathrm{bolts}\right) \\ =\frac{{}^5{C}_4}{{}^{25}{C}_4} \\ =\frac{5}{12650} \\ =\frac{1}{2530} \end{gathered}$$

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	$\frac{1}{2530}$