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,
A$\frac{3}{28}$
B$\frac{1}{14}$
C$\frac{9}{64}$
D$\frac{33}{56}$
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A$\frac{3}{28}$
We have,
The total number of batteries = 8
The number of dead batteries = 3
Let A be the event of selecting the first dead battery and B be the event of selecting the second dead battery.
Now,
P(both dead batteries are selected) $=\text{P}(\text{A}\cap\text{B})$
$=\text{P(A)}\times\text{P}(\text{B}|\text{A})$
$=\frac{3}{8}\times\frac{2}{7}$
$=\frac{3}{28}$
Hence, the correct alternative is option (a).
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