Question

There are 4 boys and 2 girls in room no. 1 and 5 boys and 3 girls in room no. 2. Find the probability that a girl from one of the two rooms laughed loudly was from room no.2.

Answer 1

Hint: We will find the probability of a girl from room no. 1 and then the probability of a girl from room no. 2. And then by using the Bayes theorem, we will find the probability that the girl who laughed loudly was from room no. 2. Bayes theorem is given as, \[P\left( A/B \right)=\dfrac{P\left( B/A \right)P\left( A \right)}{P\left( A \right)P\left( B/A \right)+P\left( A \right)P\left( B/A \right)}\], here A and B are the events, $P\left( A/B \right)$ is the probability of A given B, $P\left( B/A \right)$ is the probability of B given A and P(A), P(B) are the independent probabilities of A and B respectively.

Complete step-by-step answer:
It is given in the question that there are 4 boys and 2 girls in room no. 1 and 5 boys and 3 girls in room no. 2. And we have been asked to find the probability that a girl from one of the two rooms laughed loudly was from room no.2.
Let us assume that the probability that the girl who laughed loudly from room 1 is $P\left( {{E}_{1}} \right)$ and the probability that the girl who laughed loudly from room 2 is $P\left( {{E}_{2}} \right)$. To find the probability that, from which of the two rooms, the girl laughed loudly, we will have the total number of outcomes as 2, that is room 1 and room 2. And the favourable outcome as 1, that is either room 1 or room2, so by applying these values in the formula of probability, that is, $\text{probability=}\dfrac{\text{favourable outcomes}}{\text{total outcomes}}$ we will get $P\left( {{E}_{1}} \right)=\dfrac{1}{2}$ and $P\left( {{E}_{2}} \right)=\dfrac{1}{2}$.
Now, we will find the probability that it is a girl who laughed loudly from room 1. In room 1 we have 4 boys and 2 girls, so the total number of outcomes is (4 + 2) = 6 and the number of favourable outcomes is 2. So, we will get, \[P\left( ~G/{{E}_{1}} \right)=\dfrac{2}{6}\].
Similarly, we will find the probability that it is a girl who laughed loudly from room 2. In room 2 we have 5 boys and 3 girls, so the total number of outcomes is (5 + 3) = 8 and the number of favourable outcomes is 3. So, we will get, \[P\left( ~G/{{E}_{2}} \right)=\dfrac{3}{8}\].
We know that the Bayes theorem is given by, \[P\left( A/B \right)=\dfrac{P\left( B/A \right)P\left( A \right)}{P\left( A \right)P\left( B/A \right)+P\left( A \right)P\left( B/A \right)}\], where A and B are the events, $P\left( A/B \right)$ is the probability of A given B, $P\left( B/A \right)$ is the probability of B given A and P(A), P(B) are the independent probabilities of A and B respectively.
Now, we will apply Bayes theorem to find the probability that the girl who laughed loudly was from room 2. So, we will get,
\[P\left( {{E}_{2}}/G \right)=\dfrac{P\left( G/{{E}_{2}} \right)P\left( {{E}_{2}} \right)}{P\left( {{E}_{1}} \right)P\left( G/{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( G/{{E}_{2}} \right)}\]
$\begin{align}
  & P\left( {{E}_{2}}/G \right)=\dfrac{\left( \dfrac{3}{8} \right)\left( \dfrac{1}{2} \right)}{\left( \dfrac{1}{2} \right)\left( \dfrac{2}{6} \right)+\left( \dfrac{1}{2} \right)\left( \dfrac{3}{8} \right)} \\
 & P\left( {{E}_{2}}/G \right)=\dfrac{\dfrac{3}{16}}{\dfrac{1}{6}+\dfrac{3}{16}} \\
 & P\left( {{E}_{2}}/G \right)=\dfrac{\dfrac{3}{16}}{\dfrac{17}{48}} \\
 & P\left( {{E}_{2}}/G \right)=\dfrac{9}{17} \\
\end{align}$
Thus, the probability that the girl who laughed loudly was from room 2 is $\dfrac{9}{17}$.

Note: The most common mistake that the students can make in this question is by writing the wrong values into the Bayes theorem formula. They may substitute the value of \[P\left( ~G/{{E}_{1}} \right)\] as $\dfrac{3}{8}$ and \[P\left( ~G/{{E}_{2}} \right)\] as $\dfrac{2}{6}$, which is wrong. Also, some students write the wrong formula for Bayes theorem, like \[P\left( {{E}_{2}}/G \right)=\dfrac{P\left( G/{{E}_{2}} \right)P\left( {{E}_{2}} \right)}{P\left( {{E}_{1}} \right)+P\left( {{E}_{2}} \right)}\], but this is wrong and the correct formula is \[P\left( {{E}_{2}}/G \right)=\dfrac{P\left( G/{{E}_{2}} \right)P\left( {{E}_{2}} \right)}{P\left( {{E}_{1}} \right)P\left( G/{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( G/{{E}_{2}} \right)}\]. So, the students must be careful while solving this question.