Three defective bulbs are mixed with 8 good ones. If three bulbs are drawn one by one with replacement, the probabilities of getting exactly 1 defective, more than 2 defective, no defective and more than 1 defective respectively are:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{576}{1331}, \frac{27}{1331}, \frac{512}{1331}\) and \(\displaystyle\frac{243}{1331}\)
There are 11 bulbs, 3 of them defective. The draws are with replacement, so the 3 draws are independent with the same chance of a defective each time. The number X of defective bulbs follows a binomial distribution with
\( \displaystyle n = 3,\quad p = \frac{3}{11},\quad q = \frac{8}{11} \)
\( \displaystyle P(X = r) = {}^{3}C_{r}\,\frac{3^{r}\,8^{3-r}}{11^{3}},\quad 11^3 = 1331 \)
Exactly 1 defective:
\( \displaystyle P(X = 1) = \frac{3 \times 3 \times 64}{1331} = \frac{576}{1331} \)
More than 2 defective means X = 3:
\( \displaystyle P(X = 3) = \frac{27}{1331} \)
No defective:
\( \displaystyle P(X = 0) = \frac{512}{1331} \)
More than 1 defective means X = 2 or X = 3:
\( \displaystyle P(X = 2) = \frac{3 \times 9 \times 8}{1331} = \frac{216}{1331} \)
\( \displaystyle P(X \ge 2) = \frac{216 + 27}{1331} = \frac{243}{1331} \)
In the order asked, the probabilities are 576/1331, 27/1331, 512/1331 and 243/1331.