Question

Mark the correct alternative in the following question:
A box contains 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement at most one is defective?

Options

• \[\left( \frac{9}{10} \right)^5 \]

• \[\frac{1}{2} \left( \frac{9}{10} \right)^4 \]

• \[\frac{1}{2} \left( \frac{9}{10} \right)^5\]

• \[\left( \frac{9}{10} \right)^5 + \frac{1}{2} \left( \frac{9}{10} \right)^4\]

Answer 1

\[\text{ We have,}  \]

\[p = \text{ probability of getting a defective pen in a trial } = \frac{10}{100} = \frac{1}{10} \text{ and } \]

\[q = \text{ probability of getting a good pen in a trial } = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}\]

\[\text{ Let X denote a success of getting a defective pen . Then, }  \]

\[\text{ X follows a binomial distribution with parameters n = 5 and } p = \frac{1}{10}\]

\[ \therefore P\left( X = r \right) =^{5}{}{C}_r p^r q^\left( 5 - r \right) = ^{5}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^\left( 5 - r \right) = ^{5}{}{C}_r \left( \frac{1}{10} \right)^5 9^\left( 5 - r \right) = \frac{^{5}{}{C}_r 9^\left( 5 - r \right)}{{10}^5}, \text{ where r }  = 0, 1, 2, 3, 4, 5\]

\[\text{ Now } , \]

\[\text{ Required probability } = P\left( X \leq 1 \right)\]

\[ = P\left( X = 0 \right) + P\left( X = 1 \right)\]

\[ = \frac{^{5}{}{C}_0 9^5}{{10}^5} + \frac{^{5}{}{C}_1 9^4}{{10}^5}\]

\[ = \frac{^{5}{}{C}_0 9^5}{{10}^5} + \frac{^{5}{}{C}_1 9^4}{{10}^5}\]

\[ = \left( \frac{9}{10} \right)^5 + \frac{5 \times 9^4}{{10}^5}\]

\[ = \left( \frac{9}{10} \right)^5 + \frac{9^4}{2 \times {10}^4}\]

\[ = \left( \frac{9}{10} \right)^5 + \frac{1}{2} \left( \frac{9}{10} \right)^4\]