Question

There are three boxes, each containing a different number of light bulbs. The first box has 10 bulbs, of which four are dead, the second box has six bulbs, of which one is dead, and the third box has eight bulbs of which three are dead. What is the probability of a dead bulb being selected when a bulb is chosen at random from one of the three boxes?

• A. $\frac{115}{330}$
• B. $\frac{113}{360}$
• C. $\frac{113}{330}$
• D. None of these

Answer 1

The correct option is B $\frac{113}{360}$

Let $A_1,A_2,A_3$ denotes the events of selecting bulbs from bags $1,2and3$ respectively.

Let $B$ denotes the event the bulbs selected are dead.

$P\left(A_1\right)=P\left(A_2\right)=P\left(A_3\right)=\frac{1}{3}$

Also $P\left(B|A_1\right)=\frac{4}{10},P\left(B|A_2\right)=\frac{1}{6},P\left(B|A_3\right)=\frac{3}{8}$

By law of total probability,

$P\left(B\right)=P\left(A_1\right)P\left(B|A_1\right)+P\left(A_2\right)P\left(B|A_2\right)+P\left(A_3\right)P\left(B|A_3\right)$

Substituting the values we get,

$P\left(B\right)=\frac{1}{3}\times \frac{4}{10}+\frac{1}{3}\times \frac{1}{6}+\frac{1}{3}\times \frac{3}{8}$

$\Rightarrow P\left(B\right)=\frac{113}{360}$

Thus the probability of a dead bulb being selected when a bulb is chosen at random from one of the three boxes is $\frac{113}{360}$.