If 2% of the pens are defective, out of a lot of pens in a stationary shop. What is the probability that a sample of 10 pens, exactly 2 pens, will be defective ?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 45\cdot\displaystyle\frac{49^{8}}{50^{10}} \)
Each pen is defective with probability \( p=\frac{1}{50} \). For a sample of 10 pens the number of defective pens is binomial:
\( \displaystyle P(X=2)={}^{10}C_{2}\left(\frac{1}{50}\right)^{2}\left(\frac{49}{50}\right)^{8} \)
Here \( {}^{10}C_{2}=45 \) and \( \left(\frac{1}{50}\right)^{2}\left(\frac{49}{50}\right)^{8}=\frac{49^{8}}{50^{10}} \), so
\( \displaystyle P(X=2)=45\cdot\frac{49^{8}}{50^{10}} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{45}{50}\left(\displaystyle\frac{49}{50}\right)^{8} \) — Incorrect. It has only one factor \( \frac{1}{50} \); two defective pens need \( \left(\frac{1}{50}\right)^{2} \).
Option (3) \( \left(\displaystyle\frac{49}{50}\right)^{10}\cdot 45 \) — Incorrect. \( \left(\frac{49}{50}\right)^{10} \) is the probability that all ten pens are good.
Option (4) \( \left(\displaystyle\frac{49}{50}\right)^{2}\cdot\displaystyle\frac{45}{50} \) — Incorrect. The good pens need the power 8 and the defective pens the power 2, not 2 and 1.