Question

Mark the correct alternative in the following question:

A flash light has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, then the probability that both are dead is

$\left(\mathrm{a}\right)\frac{3}{28}\left(\mathrm{b}\right)\frac{1}{14}\left(\mathrm{c}\right)\frac{9}{64}\left(\mathrm{d}\right)\frac{33}{56}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{The}\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{batteries}=8 \\ \mathrm{The}\mathrm{number}\mathrm{of}\mathrm{dead}\mathrm{batteries}=3 \\ \mathrm{Let}A\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{selecting}\mathrm{the}\mathrm{first}\mathrm{dead}\mathrm{battery}\mathrm{and} \\ B\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{selecting}\mathrm{the}\mathrm{second}\mathrm{dead}\mathrm{battery} \\ \mathrm{Now}, \\ \mathrm{P}\left(\mathrm{both}\mathrm{dead}\mathrm{batteries}\mathrm{are}\mathrm{selected}\right)=\mathrm{P}\left(A\cap B\right) \\ =\mathrm{P}\left(A\right)\times \mathrm{P}\left(B|A\right) \\ =\frac{3}{8}\times \frac{2}{7} \\ =\frac{3}{28} \end{gathered}$$

Hence, the correct alternative is option (a).