From a lot of 15 bulbs which include 5 defective, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.
Solution
Let getting a defective bulb in a trial be a success.
We have,
\[p = \text{ probability of getting a defective bulb } = \frac{5}{15} = \frac{1}{3} \text{ and} \]
\[q = \text{ probability of getting non  defective bulb } = 1  p = 1  \frac{1}{3} = \frac{2}{3}\]
\[\text{ Let X denote the number of success in a sample of 4 trials . Then, } \]
\[ \text{ X follows binomial distribution with parameters n = 4 and } p = \frac{1}{3}\]
\[ \therefore P\left( X = r \right) =^{4}{}{C}_r p^r q^\left( 4  r \right) = ^{4}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^\left( 4  r \right) = \frac{^{4}{}{C}_r 2^\left( 4  r \right)}{3^4}, \text{ where } r = 0, 1, 2, 3, 4\]
\[\text{ i . e .} \]
\[P\left( X = 0 \right) = \frac{^{4}{}{C}_0 2^4}{3^4} = \frac{16}{81}, \]
\[P\left( X = 1 \right) = \frac{^{4}{}{C}_1 2^3}{3^4} = \frac{32}{81}, \]
\[P\left( X = 2 \right) = \frac{^{4}{}{C}_2 2^2}{3^4} = \frac{24}{81}, \]
\[P\left( X = 3 \right) = \frac{^{4}{}{C}_3 2^1}{3^4} = \frac{8}{81}, \]
\[P\left( X = 4 \right) = \frac{^{4}{}{C}_4 2^0}{3^4} = \frac{1}{81}\]
So, the probability distribution of X is given as follows:
X:  0  1  2  3  4 
P(X): 
\[\frac{16}{81}\]

\[\frac{32}{81}\]

\[\frac{24}{81}\]

\[\frac{8}{81}\]

\[\frac{1}{81}\]

Now,
\[\text{ Mean } , E\left( X \right) = 0 \times \frac{16}{81} + 1 \times \frac{32}{81} + 2 \times \frac{24}{81} + 3 \times \frac{8}{81} + 4 \times \frac{1}{81}\]
\[ = \frac{32 + 48 + 24 + 4}{81}\]
\[ = \frac{108}{81}\]
\[ = \frac{4}{3}\]
\[ \text{ Note: We can also calculate the mean of the binomial distribution by } \]
\[\text{ Mean } , E\left( X \right) = \text{ np } = 4 \times \frac{1}{3} = \frac{4}{3}\]