Ten bulbs are drawn successively, with replacement, from a lot containing $10 \%$ defective bulbs, then the…
Ten bulbs are drawn successively, with replacement, from a lot containing $10 \%$ defective bulbs, then the probability that there is at least one defective bulb, is
$1-\left(\frac{1}{10}\right)^{10}$
$\quad 1-\left(\frac{3}{10}\right)^{10}$
$1-\left(\frac{9}{10}\right)^{10}$
$1-\left(\frac{7}{10}\right)^{10}$
Solution
Let X : be the number of defective bulbs
$\therefore \quad$ Possible values of X is 1 .
Here,
$\mathrm{n}=$ number of bulbs picked $=10$.
Let P (probability of getting defective bulb)
$\begin{aligned}
& =10 \%=\frac{1}{10} \\
\therefore \quad & q=1-p=1-\frac{1}{10}=\frac{9}{10}
\end{aligned}$
$\therefore \quad$ Probability that at least one bulb is defective $=1-\mathrm{P}$ (getting 0 defective bulb)
$=1-\mathrm{P}(\mathrm{X}=0)$
$=1-{ }^{10} \mathrm{C}_0(\mathrm{p})^0(\mathrm{q})^{10-0}$
$=1-1 \times 1 \times\left(\frac{9}{10}\right)^{10}$
$=1-\left(\frac{9}{10}\right)^{10}$