From a collection of 15 bulbs which includes 5 defective bulbs, a sample of 4 bulbs is drawn one by one with replacement. What is the probability that 3 or more bulbs drawn are defective?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{1}{9} \)
The draws are with replacement, so they are independent trials with the same probability of a defective bulb. Let X be the number of defective bulbs; X is binomial with
\( \displaystyle n=4 \)
\( \displaystyle p=\frac{5}{15}=\frac{1}{3} \)
\( \displaystyle q=\frac{2}{3} \)
\( \displaystyle P(X=r)=^{4}C_{r}\left(\frac{1}{3}\right)^{r}\left(\frac{2}{3}\right)^{4-r} \)
\( \displaystyle P(X=3)=4\times\frac{1}{27}\times\frac{2}{3}=\frac{8}{81} \)
\( \displaystyle P(X=4)=\left(\frac{1}{3}\right)^4=\frac{1}{81} \)
\( \displaystyle P(X\ge 3)=\frac{8}{81}+\frac{1}{81} \)
\( \displaystyle =\frac{9}{81}=\frac{1}{9} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{81} \) — Incorrect. \( \frac{1}{81} \) is P(X = 4) alone; “3 or more” also includes X = 3.
Option (2) \( \displaystyle\frac{8}{81} \) — Incorrect. \( \frac{8}{81} \) is P(X = 3) alone; “3 or more” also includes X = 4.
Option (4) \( \displaystyle\frac{10}{81} \) — Incorrect. The two probabilities add to \( \frac{9}{81} \), not \( \frac{10}{81} \).