There are 4 boys and 2 girls in room number 1 and 5 boys and 3 girls in room number 2. A girl from one of the two rooms laughed loudly. What is the probability that the girl who laughed was from room number 2.

I am finding this question very confusing. One thing that I have figured out is we have to use Bayes' theorem.

If I take $E_1, E_2$ and $A$ as following:

$E_1=$Event in which the girl is from room number 1,

$E_2=$Event in which the girl is from room number 2,

$A=$Event in which a girl from one of the two rooms laughed loudly.

Then, we have to find $P(E_2/A)$






If I consider $E_1, E_2$ and $A$ as the following,

$E_1=$Event in the person is from room number 1,

$E_2=$Event in the person is from room number 2,

$A=$Event in which a girl from one of the two rooms laughed loudly.

then I am getting 3/5.

Which one is correct?

  • 1
    $\begingroup$ Are you sure the question isn't trying to trick you into using Bayes' theorem by giving you irrelevant information (namely the number of boys, unless you've got some information about how laughter - provoking they are)? $\endgroup$
    – timtfj
    Dec 25, 2018 at 19:10
  • $\begingroup$ This question was given in the exercise for Bayes' theorem. Many such questions can be solved without using the theorem directly but Bayes' theorem provides an organized approach as per my knowledge. If you use Bayes' theorem then you have to take the number of boys into account to get 3/5 which @Daniel Mathias also got. Also, you do not need information about "how laughter-provoking they are". The language of the question is a bit tricky. They are actually asking you to find the probability that a person is from room number 2 given that she is a girl. $\endgroup$
    – MrAP
    Dec 25, 2018 at 20:52

1 Answer 1


A girl laughed.

There are $5$ girls.

$3$ of the $5$ are in room $2$.


  • $\begingroup$ While I agree with your answer, you do not provide a very rigorous reasoning (in particular, no mention of events and conditional probability). $\endgroup$
    – Sambo
    Dec 25, 2018 at 18:05
  • 2
    $\begingroup$ @sambo I disagree. I think it is perfectly rigorous. $\endgroup$
    – user608030
    Dec 25, 2018 at 18:07
  • $\begingroup$ @Zachary Perhaps rigorous was not the right word. I meant the answer isn't formulated with usual probability theory vocabulary, like the question was. It seems like OP wants to produce an answer this way, and this explanation, while clear, doesn't help in that regard. $\endgroup$
    – Sambo
    Dec 25, 2018 at 18:18
  • 1
    $\begingroup$ @Sambo I think the problem is meant as an exercise in ignoring irrelevant information (the number of boys)—or is maybe followed by other versions in which the boys become relevant. $\endgroup$
    – timtfj
    Dec 25, 2018 at 19:16

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