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Math Help - Probability of a girl?

  1. #1
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    Exclamation Probability of a girl?

    Please help me understand conditional probability.

    Q: Find the probability of a couple having a baby girl when their fourth child is born, given that the first 3 children are girls.

    A: I think the answer is 0.5. Because regardless of the situation the probability will always be 0.5 as each event is independent.

    How can I solve this problem in terms of P(B|A)=P(A and B)/P(A)
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  2. #2
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    Hello, crazydeo!

    Good question!


    Q: Find the probability of a couple's fourth child is a girl,
    given that their first 3 children are girls.

    A: I think the answer is 0.5, because regardless of the situation,
    the probability will always be 0.5, as each event is independent.

    How can I solve this problem in terms of: . P(B|A)\:=\:\frac{P(A \wedge B)}{P(A)}

    We want: . P(\text{4th is girl }|\text{ first 3 are girls}) \;=\;\frac{P(\text{[4th is girl] }\wedge\text{ [first 3 are girls]})}{P\text{(first 3 are girls})}


    The numerator is: . P(\text{all 4 are girls}) \;=\;\left(\frac{1}{2}\right)^4 \:=\:\frac{1}{16}

    The denominator is: . P(\text{first 3 are girls}) \:=\:\left(\frac{1}{2}\right)^3 \:=\:\frac{1}{8}


    Therefore: . P(\text{4th is a girl }|\text{ first 3 are girls}) \;=\;\frac{\frac{1}{16}}{\frac{1}{8}} \;=\;\frac{1}{2}

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  3. #3
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    Just a comment on the problem--

    Both crazydeo and Soroban assume that the genders of the children are (statistically) independent. I think this is very likely true, but it is not automatically so. One can imagine biological mechanisms which would result in non-independence.

    (According to this web page

    Houghton Mifflin Science for Families: Cricket Connections

    there was an article in "Chance" which concluded there is no compelling evidence for non-independence.)
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