It is known that a box of 8 batteries contains 3 defective pieces and a person randomly selects two…
It is known that a box of 8 batteries contains 3 defective pieces and a person randomly selects two batteries from the box. If $X$ is the number of defective batteries selected,
then $P(X \leq 1)=$
$\frac{55}{64}$
$\frac{14}{28}$
$\frac{10}{28}$
$\frac{13}{28}$
Solution
We have probability of selecting defective batteries $=\frac{3}{8}$
$\begin{aligned}
\therefore & p=\frac{3}{8} \text { and } q=\frac{5}{8} \\
\therefore & P(X=0)+P(X=1) \\
&=\left[{ }^{2} C_{0}\left(\frac{3}{8}\right)^{0} \times\left(\frac{5}{8}\right)^{2}\right]+\left[{ }^{2} C_{1}\left(\frac{3}{8}\right)^{1} \times\left(\frac{5}{8}\right)^{1}\right] \\
&=\frac{25}{64}+\frac{2 \times 3 \times 5}{64}=\frac{55}{64}
\end{aligned}$