Question

There are 4 boys and 2 girls in room A and 5 boys and 3 girls in room B. A girl from one of the two rooms laughed loudly. What is the probability that the girl who laughed loudly was from room B.

Answer 1

Hint: Now first we will find the probability that the laughter comes from room A and room B respectively. Now let us consider that the laughter comes from room A. in this case we will find the probability that the girl laughed. Similarly we will consider that the laughter comes from room B and then we will calculate the probability that the person who laughed was a girl. Now we will use Bayes theorem to find the probability that the girl who laughed loudly was from room B.

Complete step-by-step answer:
Now we have 14 people in total. Among the 14, 6 people are in room A and 8 people are in room B.
Let ${{E}_{a}}$ and ${{E}_{b}}$ be the events where the laughter comes from room A and room B respectively.
Now we know that probability is given by \[\dfrac{\text{number of required cases}}{\text{number of total cases}}\]
Then we can say that $P\left( {{E}_{a}} \right)=\dfrac{6}{14}$ and $P\left( {{E}_{b}} \right)=\dfrac{8}{14}$
Now let us say that B is the probability that the laughter was from room B and A be the probability that a girl laughed.
Now by Bayes theorem which says the probability of A happening given that B happens is $P\left( B|A \right)=\dfrac{P\left( B \right).P\left( B|A \right)}{P\left( A \right)}..........................\left( 1 \right)$
Now we know that $P\left( B \right)=\dfrac{8}{14}..................\left( 2 \right)$
Probability that the laughter is from room A and the person who laughs in room A is a girl + probability that the laughter is from room B and the person who laughs in room B is a girl.
Hence \[P\left( A \right)=\dfrac{8}{14}\times \dfrac{3}{8}+\dfrac{6}{14}\times \dfrac{2}{6}................\left( 3 \right)\]
Now we know that there are 8 people in room B and 3 of them are girls.
Hence the probability that the laughter comes from room B and a girl laughs is $\dfrac{2}{6}$
Hence we have $P\left( B|A \right)=\dfrac{2}{6}................\left( 4 \right)$ .
Now from equation (1), equation (2), equation (3) and equation (4) we get,
\[\begin{align}
  & P\left( B|A \right)=\dfrac{\dfrac{8}{14}\times \dfrac{3}{8}}{\dfrac{8}{14}\times \dfrac{3}{8}+\dfrac{6}{14}\times \dfrac{2}{6}} \\
 & \Rightarrow P\left( B|A \right)=\dfrac{\dfrac{3}{14}}{\dfrac{3}{14}+\dfrac{2}{14}} \\
 & \Rightarrow P\left( B|A \right)=\dfrac{\dfrac{3}{14}}{\dfrac{5}{14}}=\dfrac{3}{5} \\
\end{align}\]
Hence the probability that the girl who laughed was in room B is $\dfrac{3}{5}$ .

Note: We are given that there are 4 boys and 2 girls in room A and 5 boys and 3 girls in room B.
Hence there are a total of 5 girls. Among those 5 are 2 are in room A and 3 are in room B. Hence we directly can say that the probability of the girl who laughed was in room B is $\dfrac{3}{5}$