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Homework Help: Conditional Probability: probability of having a girl.

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data
    I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?

    2. Relevant equations

    3. The attempt at a solution
    I think it's 1/3, because the possibilities are: Boy Boy, Boy Girl, Girl Boy, Girl Girl. There are three possibilities with at least one girl. Out of these three, only one has two girls.

    But my friend says order doesn't matter, so BG is the same as GB. So the probability is 1/2.

    Which is right?
  2. jcsd
  3. Oct 25, 2011 #2
    it doesn't matter
    Last edited: Oct 25, 2011
  4. Oct 25, 2011 #3


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    My vote's 1/3. This is a different question to 'My eldest child of two is a girl, what is the probability that they are both girls', which is 1/2. In your case, the piece of information provided is not 'ordered' but, instead, excludes the BB possibility.
  5. Oct 25, 2011 #4
    Using Bayes theorem, the probability of event A, given event B, (with events independent) is the probability of A intersect B divided by probability of B.

    So, the probability of the kid being a girl, given that there is already a boy, is the probability of a girl intersected with a boy divided by the probability of a boy.

    To find A intersect B for independent events, just multiply the probability of each. (1/2)(1/2). So, (1/4) divided by (1/2).
  6. Oct 25, 2011 #5
    It's the probability that it's a girl, given that at least one is a girl - nothing involving a boy. Does that change things?
  7. Oct 26, 2011 #6
    I think you're right. The outcome of one boy and one girl should be twice as likely as the other 2 outcomes, since this is a http://en.wikipedia.org/wiki/Binomial_distribution" [Broken] with n=2 and p=1/2.

    More formally, let [itex] p = 1/2[/itex] be the probability a girl is born and on our universe U define the random variables X = number of girls, Y = number of boys. Define the events [itex] C = \{X = 2 \} [/itex] and [itex] D = \{X \geq 1 \}[/itex]. Certainly [itex] C \subseteq D [/itex] so [itex] C \cap D = C [/itex]. Since [itex]p = 1/2 [/itex] and the two births are independent events, then [itex]\mathbf{P}(C) = (1/2)^2 = 1/4 [/itex]. The probability of the complement of D, [itex] \mathbf{P}(U \setminus D) [/itex] is the same as [itex]\mathbf{P}\{X=0\}[/itex], which is the probability of 2 boys. Similarly, since [itex]1-p = 1/2 [/itex] and the two births are independent events, then [itex] \mathbf{P}(U \setminus D) = (1/2)^2 = 1/4[/itex]. So [itex] \mathbf{P}(D) = 1 - \mathbf{P}(U \setminus D) = 3/4[/itex]. Finally
    \mathbf{P}(C|D) = \frac{\mathbf{P}(C \cap D)}{\mathbf{P}(D)} = \frac{\mathbf{P}(C)}{\mathbf{P}(D)} = \frac{1/4}{3/4} = 1/3 \, .

    I think your proof using the reduced sample space is correct as well--and shorter, too!
    Last edited by a moderator: May 5, 2017
  8. Oct 26, 2011 #7


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    Yep, 1/3. For a less technical explanation, if it helps anyone:

    Imagine surveying a large number of families with 2 children. In half of them, the older child will be a boy. Within that half, half again (or a quarter of the total) will have the younger child be a boy, and the rest (another quarter of the total) will have the younger child be a girl. The same goes for the remaining half of the families whose older child is a girl. So you have four possibilities: BB, BG, GB, GG (in order of age), each representing a quarter of the total population.

    Now, the statement "given that at least one of them is a girl" excludes one of those possibilities, BB. There are three equally likely possibilities left (BG, GB, GG), of which exactly one has both children being girls. So the probability is 1/3.
  9. Oct 26, 2011 #8


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    Yes, that was the the means I convinced myself of my analysis.

    I think the comment about Bayes has actually taken the conclusion the wrong way. What we learn in the statement 'there is a girl' is a piece of information that teaches us to exclude 1/4 of the population set of 2-child families.

    @AN; You have concluded Bayes says 'To have a boy is twice as likely as to have a girl'. Therefore probability of two girls is 1/3 and a boy and a girl is 2/3, because 2/3 is twice 1/3!!! (Because we've excluded the BB option from the P=1 outcome.)
    Last edited: Oct 26, 2011
  10. Oct 26, 2011 #9
    Oops, I was talking about girls and boys. That doesn't change the answer though. The Bayes theorem is sound. The answer is 1/2.
  11. Oct 26, 2011 #10


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    But what's the question? '1/2' is not the answer to the op, but is the answer to 'what is the probability of GG compared with not GG' (conditional on 'there is a girl').
  12. Oct 26, 2011 #11

    Ray Vickson

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    This is the same problem as: we toss a (fair) coin twice and get at least one head; what is the probability we get another head as well? The results LOOK a bit paradoxical when written out as follows: events are C = {two heads}, A = {first toss is heads}, B = {second toss is heads}. We have P{C|A} = P{C|B} = 1/2, but P{C|A or B} = 1/3. That is your problem in a nutshell. This looks paradoxical because the statement that we get at least one head is the same as saying that A or B occur, and each of these separately implies a 1/2 probability of C; however, when taken together they imply a 1/3 probability of C!

  13. Oct 26, 2011 #12

    I like Serena

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    Hi everyone! :smile:

    Here's my view on the matter.

    There is a difference between saying:
    "given the first (or youngest) kid is a girl" (chance 2/4)
    and "given that at least one kid is a girl" (chance 3/4).

    In the first case the chance would be 1/2 as Arcana claims.
    We can also write this as:
    [tex]P(\text{2 girls | the youngest kid is a girl}) = {\text{number of favorable outcomes} \over \text{number of possible outcomes}} = {1 \over 2}[/tex]

    In the second case it is:
    [tex]P(\text{2 girls | at least 1 girl}) = {\text{number of favorable outcomes} \over \text{number of possible outcomes}} = {1 \over 3}[/tex]

    So I believe the reasoning of pearapple is sound for this problem and the chance is 1/3.
  14. Oct 26, 2011 #13


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    Since the others have explained how to get the correct answer, I'll point out what mistake your friend is making:

    You have four possible outcomes: BB, BG, GB, and GG. BG and GB are distinct outcomes. The order matters when you're enumerating outcomes.

    When you combine them to form events, then the order may or may not matter depending on how the event is defined. The event "exactly 1 girl" would be {BG, GB}. Whether the girl comes first or last doesn't really matter.

    Note that even though the order doesn't matter in determining what outcomes are included in an event, you still have to add up the probabilities for each outcome to get the probability of the event. That is, P({BG, GB}) = P({BG})+P({GB}) = 1/2. It's incorrect to say: order doesn't matter, so BG=GB and P({BG, GB})=P({BG})=1/4.

    The event "at least one girl" is {BG, GB, GG}, which corresponds to a probability of 3/4. If you incorrectly reduce this to {BG, GG}, as your friend has done, you get a probability of 1/2. If you take the probability GG, which is 1/4, and divide it by 3/4, you get the correct answer of 1/3. If instead you divide it by the incorrect probability of 1/2, you get 1/2, the answer your friend got.
  15. Oct 27, 2011 #14


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    Its amazing that even such a simple example has an answer that is not immediately obvious.
  16. Oct 27, 2011 #15
    The answer to this question is ambiguous because the question itself is ambiguous.

    I believe this falls under a similar category of the monty hall game show. Lets say you start off with two doors, and behind each door there could either be a boy or a girl. If the host says that behind one of the doors there is a girl, then you know the chances of there being a girl behind both doors is one in three. (since he didn't specify which door). If he tells you that there is a girl behind the door on the left, then you know that the chances of there being two girls are one in two.

    So you and your friend have seperate interpretations of the question. The question boils down to if you know which kid is a girl, or you just know that there is A girl.

    If you have your 4 scenarios, GG, BG, GB, and BB, him telling you that there is at least one girl only means you get to cross off BB. Him telling you that there is a girl on the left allows you to cross of BG And BB
  17. Oct 27, 2011 #16


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    I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?

    Is different to:

    I have two kids. I know one of them, which is a girl. What is the probability that they are both girls?

    I suppose pearapple's friend misinterpreted it as the second question, because he assumed that if the father knew he had at least one girl, then it would have meant that he met one of his kids, which was a girl. But this is not necessarily true.

    I guess in language, the two statements are very similar, but you have to be careful about what you're actually saying.
  18. Oct 27, 2011 #17


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    Pretty much what others are saying: there are two different questions that can be confused here:
    1) A person has two children, one of which is a girl. What is the probability that the other child is a girl?

    2) A person has two children, the older of which (or the younger or the one we happen to know- anything that distinguishes one child from another) is a girl. What is the probability that the other child is a girl?

    With two children, and assuming that "boy" and "girl" for any given child are equally likely, There are four equally likely possibilities: BB, BG, GB, GG where, of course, "B" means "boy", "G" means "girl" and the order depends upon which was born first, or was the one we knew, etc.

    With problem (1) above, knowing that one of the children is a girl removes "BB" from the list, leaving "BG", "GB", and "GG". In only one of those is the other child also a girl. Probability of two girls, 1/3.

    With problem (2) above, knowing that the older child (or younger child or whichever is distinguished in some way) we remove both "BB" and "BG". We are left with "GB" and "GG" of which 1 has both girls. Probability of two girls, 1/2.
  19. Oct 27, 2011 #18

    I like Serena

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    Uhh :uhh:. Aren't those the same? :confused:
  20. Oct 27, 2011 #19
    Is there anything that you don't understand specifically about the above explanation?
  21. Oct 27, 2011 #20

    I like Serena

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    Your explanation is correct.
    It seems to me that BruceW's explanation is not correct. :redface:
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