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,

\[p = \frac{6}{30} = \frac{1}{5}\text{ and } q = \frac{4}{5}\]

\[\text{ Then, the distribution is given by } \]
\[P(X = r) = ^{4}{}{C}_r \left( \frac{1}{5} \right)^r \left( \frac{4}{5} \right)^{4 - r} , r = 0, 1, 2, 3, 4\]

\[P \left( X = 0 \right) = \left( \frac{4}{5} \right)^4 \]
\[ = \frac{256}{625} \]
\[P \left( X = 1 \right) = 4 \left( \frac{1}{5} \right)^1 \left( \frac{4}{5} \right)^3 \]
\[ = \frac{256}{625}\]
\[ P\left( X = 2 \right) = 6 \left( \frac{1}{5} \right)^2 \left( \frac{4}{5} \right)^2 \]
\[ = \frac{96}{625}\]
\[ P\left( X = 3 \right) = 4 \left( \frac{1}{5} \right)^3 \left( \frac{4}{5} \right)^1 \]
\[ = \frac{16}{625}\]
\[ P\left( X = 4 \right) = \left( \frac{1}{5} \right)^4 \]
\[ = \frac{1}{625}\]

X         0      1     2     3     4

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