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 has six bulbs, out 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. $\dfrac{{115}}{{330}}$
B. $\dfrac{{115}}{{360}}$
C. $\dfrac{{113}}{{330}}$
D. None of these

Answer 1

Hint:In this question we have to find the probability of a dead bulb when selected randomly from the boxes. So we will first find the probability of bulbs being selected from all the three boxes and then we find the probability of dead bulb from each of the boxes. After this we will use the law of total probability.

Formula used:
The formula of total probability states:
$P(B) = P({A_1})P\left( {\dfrac{B}{{{A_1}}}} \right) + ... + P({A_n})P\left( {\dfrac{B}{{{A_n}}}} \right)$
where $n$ is the number of number of items from which selection can be done and $B$ is the term whose probability is to be calculated.

Complete step by step answer:
Let us assume that ${A_1},{A_2},{A_3}$ denote the events of selecting bulbs from the three boxes. And let us assume $B$ denotes the event that the bulbs that are selected are dead.
We know the general formula of probability is
$P = \dfrac{{n(E)}}{{n(S)}}$
where $n(E)$ is the number of favourable outcomes and $n(S)$ is the total number of possible outcomes.
Now we have a total of $3$ boxes and we have to select one bulb from each box.
So for each box we can say that the total possible outcome is $3$ i.e.
$n(S) = 3$
And since we have to select one bulb from each box, so this is our favourable outcome:
$n(E) = 1$

Now the probability of ${A_1}$ is :
$\dfrac{{n(E)}}{{n(S)}} = \dfrac{1}{3}$
Similarly the probability of ${A_2}$ is
$\dfrac{{n(E)}}{{n(S)}} = \dfrac{1}{3}$
And the probability of ${A_3}$ by putting in the formula is
$\dfrac{{n(E)}}{{n(S)}} = \dfrac{1}{3}$
We can write that:
$P({A_1}) = P({A_2}) = P({A_3}) = \dfrac{1}{3}$
We will now find the probability of a dead bulb from each of the boxes.
In the first box, total number of bulbs are $10$ and dead bulbs are $4$, it can be written as
${A_1} = 10,B = 4$
Here we have
$n(E) = B$ , it is the number of favourable outcome
And
$n(S) = {A_1}$ is the total number of possible outcomes.
So by applying the formula of probability we can write:
$P\left( {\dfrac{B}{{{A_1}}}} \right) = \dfrac{4}{{10}}$.

Now in the second box we have by applying the same method as above we have,
$n(E) = B = 1$ , it is the number of favourable outcome
And
$n(S) = {A_1} = 6$ is the total number of possible outcomes.
So it gives us
$P\left( {\dfrac{B}{{{A_2}}}} \right) = \dfrac{1}{6}$
Now in the third box we can write that:
$n(E) = B = 3$ , it is the number of favourable outcome
And
$n(S) = {A_1} = 8$ is the total number of possible outcomes.
Again it gives the value
$P\left( {\dfrac{B}{{{A_2}}}} \right) = \dfrac{3}{8}$
Now we have to find the probability of dead bulb i.e. $B$ from each of the boxes, i.e.
${A_1},{A_2},{A_3}$

By comparing it with the formula of the law of total probability we have,
$n = 3$, we have to find $P(B)$
So we can write that:
$P(B) = P({A_1})P\left( {\dfrac{B}{{{A_1}}}} \right) + P({A_2})P\left( {\dfrac{B}{{{A_2}}}} \right) + P({A_3})P(\left( {\dfrac{B}{{{A_3}}}} \right)$
By putting all the values from the above we can write
$P(B) = \dfrac{1}{3} \times \dfrac{4}{{10}} + \dfrac{1}{3} \times \dfrac{1}{6} + \dfrac{1}{3} \times \dfrac{3}{8}$
On multiplication,
$\therefore P(B) = \dfrac{{113}}{{360}}$

Hence the correct option is B.

Note:We should note that total probability is used when we do not know the probability of an event but we know that its occurrence under several disjoint scenarios and the probability of each scenario. The total probability rule also called the law of total probability breaks up the probability calculations into distinct parts. It is used to find the probability of an event, say $B$, when we do not know much about B’s probabilities to calculate it directly. So we take an event A to calculate the probability of B. We should note that if $P\left( {\dfrac{B}{A}} \right) = B$ then A and B are independent events and $P\left( {\dfrac{B}{A}} \right)$ is the conditional probability which gives the probability of occurrence of event B when event A has already occurred.