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Question: There are \(4\) boys and \(2\) girls in room ‘A’ and \(5\) boys and \(3\) girls in room ‘B’. A girl ...

There are 44 boys and 22 girls in room ‘A’ and 55 boys and 33 girls in room ‘B’. A girl from one of the two rooms laughed loudly. What is the probability that the girl who laughed was from room B?

Explanation

Solution

Hint : We will find the probability in girls in the room A and Room B. Thereafter, we will find the girls laughed in room A and room B. Then we will apply Bayes theorem: P(AB)=P(BA)PAP(B)P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {\dfrac{B}{A}} \right)PA}}{{P(B)}}

Complete step-by-step answer :
Number of boys in room A =4 = 4
Number of girls in room A =2 = 2
P(girls laugh in a room ) =12 = \dfrac{1}{2}
Total number of candidate in room A =2+4 = 2 + 4
Total number of candidate in room A =6 = 6
By using the formula of probability P(E)=favourable  outcomestotal  number  of  outcomesP(E) = \dfrac{{favourable\,\,outcomes}}{{total\,\,number\,\,of\,\,outcomes}}
P(girls laughed in room A) == favourable  outcometotal  number  of  outcome\dfrac{{favourable\,\,outcome}}{{total\,\,number\,\,of\,\,outcome}}
P(girls laughed in room A) =26 = \dfrac{2}{6}
P(girls laughed in room A) =13 = \dfrac{1}{3}
Now,
Number of girls in room B =3 = 3
Number of boys in room B =5 = 5
Total candidates in room B=3+5B = 3 + 5
Total candidates in room B =8 = 8
Now, by using the formula of probability P(E)=favourable  outcometotal  number  of  outcomeP(E) = \dfrac{{favourable\,\,outcome}}{{total\,\,number\,\,of\,\,outcome}}
P(girls laughed in room B) =38 = \dfrac{3}{8}
Now, we will apply Bay’s theorem
P(the girl who laughed was from room B) =12×3812×38×12×13 = \dfrac{{\dfrac{1}{2} \times \dfrac{3}{8}}}{{\dfrac{1}{2} \times \dfrac{3}{8} \times \dfrac{1}{2} \times \dfrac{1}{3}}}
P(the girls who laughed was from room B) =316316+16 = \dfrac{{\dfrac{3}{{16}}}}{{\dfrac{3}{{16}} + \dfrac{1}{6}}}
We will take LCM of 6,16−486,16 - 48, we have
P(the girls who laughed was from room B) =3163×3+1×848 = \dfrac{{\dfrac{3}{{16}}}}{{\dfrac{{3 \times 3 + 1 \times 8}}{{48}}}}
P(the girls who laughed was from room B) =3169+848 = \dfrac{{\dfrac{3}{{16}}}}{{\dfrac{{9 + 8}}{{48}}}}
P(the girls who laughed was from room B) =3161748 = \dfrac{{\dfrac{3}{{16}}}}{{\dfrac{{17}}{{48}}}}
P(the girls who laughed was from room B) =316×4817 = \dfrac{3}{{16}} \times \dfrac{{48}}{{17}}
P(the girls who laughed was from room B) =917 = \dfrac{9}{{17}}
Hence, the required probability is 917\dfrac{9}{{17}}

Note : Students must keep in mind that first you find the probability of girls laughed in a room .After that find the probability of girls laughed in room A and room B separately.