Question

From a lot of 30 bulbs that includes 6 defective bulbs, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Answer 1

Let X denote the number of defective bulbs in a sample of 4 bulbs drawn successively with replacement .
Then, X follows a binomial distribution with the following parameters: n=4, $\mathit{\text{p}}=\frac{6}{30}=\frac{1}{5}\mathrm{and}q=\frac{4}{5}$
$$\begin{gathered} \mathrm{Then},\mathrm{the}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(X=r)={}^{4}C_r{\left(\frac{1}{5}\right)}^r{\left(\frac{4}{5}\right)}^{4-r},r=0,1,2,3,4 \end{gathered}$$

$$\begin{gathered} P\left(X=0\right)={\left(\frac{4}{5}\right)}^4 \\ =\frac{256}{625} \\ P\left(X=1\right)=4{\left(\frac{1}{5}\right)}^1{\left(\frac{4}{5}\right)}^3 \\ =\frac{256}{625} \\ P\left(X=2\right)=6{\left(\frac{1}{5}\right)}^2{\left(\frac{4}{5}\right)}^2 \\ =\frac{96}{625} \\ P\left(X=3\right)=4{\left(\frac{1}{5}\right)}^3{\left(\frac{4}{5}\right)}^1 \\ =\frac{16}{625} \\ \mathit{}P\left(X=4\right)={\left(\frac{1}{5}\right)}^4 \\ =\frac{1}{625} \end{gathered}$$



X 0 1 2 3 4

P(X) $\frac{256}{625}\frac{256}{625}\frac{96}{625}\frac{16}{625}\frac{1}{625}$