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Homework Help: Conditional Probabilities

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data
    A couple has two children. What is the probability that both are girls given that the oldest is a girl? What is the probability that both are girls given that one of them is a girl?

    DO NOT assume that female births and male births are equally likely. Assume identical twins are necessarily the same sex.

    Write your answer in terms of the probability of a female birth and the probability of identical twins (assume that identical twins are just as likely to be female as the overall female probability).

    2. Relevant equations
    Bayes Theorem?

    3. The attempt at a solution
    All right the problem as stated in the first paragraph without extra stipulations is easy, but I am unsure about how to complete solution now. Considering the question "What is the probability that both are girls given that the oldest is a girl" , if we let A = event that both are girls and B = oldest is a girl. Then by Bayes,

    P(A|B) = P(B|A)P(A) / P(B). But is this the correct way to go? Isn't P(B) just P(female birth)?
  2. jcsd
  3. Oct 12, 2009 #2


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    is there anything that says the sex of one child is dependent on another for a given couple?

    if they are independent events then things simplify
  4. Oct 12, 2009 #3


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    They are dependent if they are identical twins. There are two cases here. They are identical or they are not identical. Only the second is independent.
  5. Oct 12, 2009 #4
    I'm afraid to assume independence of probability here (in general I've been told it's a bad idea, but it seems a bit strong). By the way independence is not a property of events, but rather probabilities (we can have disjoint events, but these coincide with independent probabilities only in a specific case). I know what you meant and I probably have been confused about this distinction in the past myself.
  6. Oct 13, 2009 #5


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    nice point Dick, that does introduce the dependence

    as there is not a heap of events, i would start by drawing a probability tree, first branch is whether or not they have identical twins, 2nd barch is sex of 1st child (ends heres for twins branch) 3rd branch is sex of the 2nd.

    you know they don't have identical twins after having a single child as they only have 2 children.

    from there it should be easy to read off probabilities & relate to bayes, if needed
  7. Oct 13, 2009 #6
    Thanks lanedance, I drew out the tree as you suggested. I'm a little shaky on the rules of adding and multiplying probabilities (not thinking clearly at the moment), but I'll try to see what I come up with.
  8. Oct 13, 2009 #7


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    tips -
    - put the probabilty of the outcome of of each event on each branch
    - the probabilty of the final outcome will be all teh probabilties on each branch l;eading to it muliplied
    - check the sum of all the probabilties of final outcomes sums to 1
    - label each outcome as one of (MM, MF, FM, FF)

    now to get the total porbabilty of FF, add the probabilties of all the FF outcomes. This should give you all the probabilties you need to use Bayes theorem

    note also if A = event that both are girls and B = oldest is a girl
    then P(B|A) = 1
  9. Oct 13, 2009 #8
    Ah, so I don't really have to use Bayes? That was extremely useful, and I ended up with pq + (1-q)pp, where p = probability of female birth and q = probability of identical twins. Hopefully that's right. Thanks.
  10. Oct 13, 2009 #9


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    not sure how you have done it (might want to check that), but the use of Bayes is somewhat implicit
    for example is

    note also if A = event that both are girls and B = oldest is a girl
    then you are effectively using Bayes with

    P(A|B) = P(B|A)P(A) / P(B) = 1.P(A) / P(B) = P(A) / P(B) in this case as every case with 2 girls has the oldest a girl

    P(A) = sum porbs of all cases with FF
    P(B) = sum probs of all cases with Fx, where x is anything

    then for this case
    P(A|B) = P(A)/P(B)
  11. Oct 13, 2009 #10
    Doh, I calculated all FF possibilities and thought I was finished with the problem. Sorry. Yes so I just worked out the other case P(both are girls | one is a girl) to be essentially the same, I think. Divide all FF possibilities over sum of final outcomes with at least one F. I hope this makes sense.
  12. Oct 13, 2009 #11


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    yeah i think that makes sense - so writing it out

    P(boths girls, given oldest girl) = P(FF)/(P(FF) + P(FM))

    P(boths girls, given 1 girl) = P(FF)/(P(FF) + P(FM)+P(MF))
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