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Thread: Family of Three Children - Conditional Probability

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    Family of Three Children - Conditional Probability

    From families with three children, a family is selected at random and found to have
    a boy. What is the probability that the boy has (a) an older brother and a younger
    sister; (b) an older brother; (c) a brother and a sister? Assume that in a three-child
    family all gender distributions have equal probabilities.
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    Re: Family of Three Children - Conditional Probability

    Quote Originally Posted by joshuaa View Post
    From families with three children, a family is selected at random and found to have
    a boy. What is the probability that the boy has (a) an older brother and a younger
    sister; (b) an older brother; (c) a brother and a sister? Assume that in a three-child
    family all gender distributions have equal probabilities.
    Here is the probability space: $\begin{array}{*{20}{c}}B&B&B\\B&B&G\\B&G&B\\B&G&G \\G&B&B\\G&B&G\\G&G&B\\G&G&G\end{array}$ You are given that the family has a boy.
    That creates a subspace. Now we just count.
    a) is $\dfrac{1}{7}$. Please explain that.
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    Re: Family of Three Children - Conditional Probability

    First, the book answer for (a) is 1/14

    -------------------------
    But let me try to tell you why you got 1/7

    we have 8 outcomes in the sample space
    the probability of selecting a boy is 7/8 since only G G G does not have a boy

    B B G and B G B and G B B has two boys and a girl
    so the probability of getting two boys and a girl is 3/8

    but if order is important only B B G will be the outcome of a boy has an older brother and a sister
    so the probability is 1/8

    Therefore, the answer for (a) is (1/8) / (7/8) = 1/7

    Is the answer of the book wrong?
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    Re: Family of Three Children - Conditional Probability

    Quote Originally Posted by joshuaa View Post
    First, the book answer for (a) is 1/14
    we have 8 outcomes in the sample space
    the probability of selecting a boy is 7/8 since only G G G does not have a boy
    B B G and B G B and G B B has two boys and a girl
    so the probability of getting two boys and a girl is 3/8
    but if order is important only B B G will be the outcome of a boy has an older brother and a sister
    so the probability is 1/8
    Therefore, the answer for (a) is (1/8) / (7/8) = 1/7
    Is the answer of the book wrong?
    If someone tell you that a family has three children and at least one is a boy, what is the probability of BBG? clearly order does matter.
    What is the probability of two boys and a girl? (order does not matter)
    Yes, I argue the answer is 1/7 as written. I would like to hear what that author says.
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    Re: Family of Three Children - Conditional Probability

    Quote Originally Posted by joshuaa View Post
    First, the book answer for (a) is 1/14

    -------------------------
    But let me try to tell you why you got 1/7

    we have 8 outcomes in the sample space
    the probability of selecting a boy is 7/8 since only G G G does not have a boy

    B B G and B G B and G B B has two boys and a girl
    so the probability of getting two boys and a girl is 3/8

    but if order is important only B B G will be the outcome of a boy has an older brother and a sister
    so the probability is 1/8

    Therefore, the answer for (a) is (1/8) / (7/8) = 1/7

    Is the answer of the book wrong?
    I think it's $\dfrac{1}{14}$ because getting GBB could be either the "boy" having an older brother as we want, or
    it could be that the "boy" is the older brother and thus has a younger brother.

    Both are equiprobable and they total to $\dfrac{1}{7}$ in probability so they are both $\dfrac{1}{14}$ in probability.
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    Re: Family of Three Children - Conditional Probability

    Quote Originally Posted by joshuaa View Post
    From families with three children, a family is selected at random and found to have
    a boy. What is the probability that the boy has (a) an older brother and a younger
    sister; (b) an older brother; (c) a brother and a sister? Assume that in a three-child
    family all gender distributions have equal probabilities.

    Quote Originally Posted by romsek View Post
    I think it's $\dfrac{1}{14}$ because getting GBB could be either the "boy" having an older brother as we want, or
    it could be that the "boy" is the older brother and thus has a younger brother.
    Both are equiprobable and they total to $\dfrac{1}{7}$ in probability so they are both $\dfrac{1}{14}$ in probability.
    @Romsek, it clearly says THAT boy has an older brother and younger sister. So the elementary event is BBG. We cannot count GBB because it violates the given.
    Last edited by Plato; Mar 20th 2019 at 03:17 PM.
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    Re: Family of Three Children - Conditional Probability

    Quote Originally Posted by Plato View Post
    @Romsek, it clearly says THAT boy has an older brother and younger sister. So the elementary event is BBG. We cannot count GBB because it violates the given.
    No. It asks what is the probability that the boy has an older brother and younger sister.
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    Re: Family of Three Children - Conditional Probability

    Quote Originally Posted by romsek View Post
    GBB could be either the "boy" having an older brother as we want, or
    it could be that the "boy" is the older brother and thus has a younger brother.
    Both are equiprobable and they total to $\dfrac{1}{7}$ in probability so they are both $\dfrac{1}{14}$ in probability.
    I agree GBB are equiprobable, but they violate the event as Plato said

    Why not saying the Event BBG could be either the "boy" having an older brother as we want, or
    it could be that the "boy" is the older brother and thus has a younger brother.

    then 1/14 answer makes more sense
    Last edited by joshuaa; Mar 20th 2019 at 05:21 PM.
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    Re: Family of Three Children - Conditional Probability

    Quote Originally Posted by joshuaa View Post
    I agree GBB are equiprobable, but they violate the event as Plato said

    Why not saying the Event BBG could be either the "boy" having an older brother as we want, or
    it could be that the "boy" is the older brother and thus has a younger brother.

    then 1/14 answer makes more sense
    That's what I meant.

    we know that BBG has probability $\dfrac 1 7$

    BBG can be either our boy has an older brother, or that he is the older brother.

    Both with equal probability, and summing to $\dfrac 1 7$, i.e. $\dfrac{1}{14}$

    So $P[\text{our boy has older brother and younger sister}] = \dfrac{1}{14}$
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    Re: Family of Three Children - Conditional Probability

    Quote Originally Posted by joshuaa View Post
    From families with three children, a family is selected at random and found to have
    a boy. What is the probability that has (a) an older brother and a younger
    sister; (b) an older brother; (c) a brother and a sister? Assume that in a three-child
    family all gender distributions have equal probabilities.
    If I were tasked with editing this question, here is what I would submit.
    In a family of with three children what is the probability that a boy who has:
    A) an older brother and a younger sister?
    B) an older brother?
    C) a brother and a sister?

    The advantage in that wording is that any hint of a conditional is removed.
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    Re: Family of Three Children - Conditional Probability

    Quote Originally Posted by romsek View Post
    That's what I meant.

    we know that BBG has probability $\dfrac 1 7$

    BBG can be either our boy has an older brother, or that he is the older brother.

    Both with equal probability, and summing to $\dfrac 1 7$, i.e. $\dfrac{1}{14}$

    So $P[\text{our boy has older brother and younger sister}] = \dfrac{1}{14}$
    Please tell us where does the $14$ come from?
    $\dfrac{1}{7}+\dfrac{1}{7}=\dfrac{2}{7}$ OR $\dfrac{1}{7}\cdot\dfrac{1}{7}=\dfrac{1}{49}$

    The count in the probability space does not change. So where is it?
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    Re: Family of Three Children - Conditional Probability

    Quote Originally Posted by Plato View Post
    Please tell us where does the $14$ come from?
    $\dfrac{1}{7}+\dfrac{1}{7}=\dfrac{2}{7}$ OR $\dfrac{1}{7}\cdot\dfrac{1}{7}=\dfrac{1}{49}$

    The count in the probability space does not change. So where is it?
    it comes from $2p = \dfrac 1 7 \Rightarrow p = \dfrac{1}{14}$
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    Re: Family of Three Children - Conditional Probability

    for part (b), I will follow the same style of equiprobability as romsek did

    we want
    B B B
    B B G
    B G B
    G B B

    let us assume (B) is the older brother, then
    Probability of (B) B B = B (B) B = B B (B) = 1/21
    Probability of (B) B G = B (B) G = 1/14
    Probability of (B) G B = B G (B) = 1/14
    Probability of G (B) B = G B (B) = 1/14

    from all of those I will exclude the outcomes when the older boy comes before the boy

    then I will remain with
    B (B) B
    B B (B)
    B (B) G
    B G (B)
    G B (B)

    summing their probabilities respectively = 1/21 + 1/21 + 1/14 + 1/14 + 1/14 = 13/42 (which is the same as the book answer)

    But why here the order of the boy must be before the older boy while we don't care about the order of the girl?

    for part (c)
    the answer is 3/7 and to get that answer, why order does not matter? Even I have to take G B (B), G (B) B, ...etc But it says a boy and a girl. I thought the girl should always be the 3rd. But if I take only the outcomes when the girl is 3rd, the answer will be wrong.
    Last edited by joshuaa; Mar 20th 2019 at 07:57 PM.
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    Re: Family of Three Children - Conditional Probability

    Quote Originally Posted by joshuaa View Post
    for part (b), I will follow the same style of equiprobability as romsek did

    we want
    B B B
    B B G
    B G B
    G B B

    let us assume (B) is the older brother, then
    Probability of (B) B B = B (B) B = B B (B) = 1/21
    Probability of (B) B G = B (B) G = 1/14
    Probability of (B) G B = B G (B) = 1/14
    Probability of G (B) B = G B (B) = 1/14

    from all of those I will exclude the outcomes when the older boy comes before the boy

    then I will remain with
    B (B) B
    B B (B)
    B (B) G
    B G (B)
    G B (B)

    summing their probabilities respectively = 1/21 + 1/21 + 1/14 + 1/14 + 1/14 = 13/42 (which is the same as the book answer)

    But why here the order of the boy must be before the older boy while we don't care about the order of the girl?
    This is all exactly as I would have (and did) do it. You care about the boys because the problem asked for the
    probability of an older brother. Not an older sister.

    (c) is correct. Any combination of 2 B's and 1 G satisfies this. There are 3 of them.
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    Re: Family of Three Children - Conditional Probability

    Yeah the problem (b) cares about boys, especially the older brother and (B) B B event has boys. Why do we have to exclude it?
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