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Math Help - [SOLVED] Gender distribution. Wrong (official) solution?

  1. #1
    Member courteous's Avatar
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    Question [SOLVED] Gender distribution. Wrong (official) solution?

    Couple are planning four kids. What is more likely: both genders equal (2 boys, 2 girls) or 3 of one gender and 1 of the other gender?
    P(\text{2 boys, 2 girls})=\frac{6}{16} I agree with that.

    But is P(\text{3 boys, 1 girl or vice-versa}) really \frac{8}{16}? I'd say it is rather \frac{{4\choose 1}}{16}=\frac{4}{16}.

    How could they get 8 in the numerator?!

    Is there already any [tex] keyboard shortcut? I vote for FN/CTRL+M.
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  2. #2
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    Isn't the combination of 3 one gender and 1 the other gender 4C1*3C3? However that only takes into account ONE gender combo.
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  3. #3
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    In this one, there ARE only 16. Enumerate them.

    4 vs 0
    BBBB
    GGGG

    3 vs 1
    BBBG
    BBGB
    BGBB
    GGBG
    GGGB
    BGGG
    GBBB
    GBGG

    2 vs 2
    GBBG
    GBGB
    GGBB
    BGBG
    BGGB
    BBGG

    Of course, had you pondered one important thing, you would have kept trying. After counting 2 vs 2, you could have counted 4 vs 0 and subtracted, since there are only three types. This would have discouraged your result.

    Further, a little binomial expansion would reveal it.

    (B+G)^{4}\;=\;B^{4}+4B^{3}G+6B^{2}G^{2}+4BG^{3}+G^  {4}

    It's pretty obvious from that.

    The REAL lesson is that there are many ways to proceed. If you know only one way, you will not have a backup plan or an alternative approach.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by courteous View Post
    P(\text{2 boys, 2 girls})=\frac{6}{16} I agree with that.

    But is P(\text{3 boys, 1 girl or vice-versa}) really \frac{8}{16}? I'd say it is rather \frac{{4\choose 1}}{16}=\frac{4}{16}.

    How could they get 8 in the numerator?!

    Is there already any [tex] keyboard shortcut? I vote for FN/CTRL+M.
    The number of girls has a binomial distribution B(4,0.5)

    So probability of exactly two girls is:

    p_1=b(2;4;0.5)=\frac{4!}{2!2!}0.5^4=\frac{3}{8}

    Probability of three of one gender and the one of the other gender is equal to the sum of the probability of exactly one girl and of exactly three girls:

    p_2=b(1;4,0.5)+b(3;4,0.5)=1/2

    CB
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  5. #5
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    Hello, courteous!

    A couple is planning on four chldren.
    What is more likely: 2 boys and 2 girls, or 3 of one gender, 1 of the other?
    If you're not convinced, we can LIST the possible outcomes.

    . . \begin{array}{c} \text{Two of each} \\ \hline BBGG \\BGBG \\ BGGB \\ GBBG \\ GGBB \\ GBGB \end{array} \qquad \text{ 6 ways}


    . . \begin{array}{c}\text{Three and one} \\ \hline<br />
BBBG \\ BBGB \\ BGBB \\ GBBB \\ BGGG \\ GBGG \\ GGBG \\ GGGB \end{array} \qquad \text{ 8 ways}

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  6. #6
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    Quote Originally Posted by courteous View Post
    P(\text{2 boys, 2 girls})=\frac{6}{16} I agree with that.

    But is P(\text{3 boys, 1 girl or vice-versa}) really \frac{8}{16}? I'd say it is rather \frac{{4\choose 1}}{16}=\frac{4}{16}.

    How could they get 8 in the numerator?!

    Is there already any [tex] keyboard shortcut? I vote for FN/CTRL+M.
    Hi courteous,

    2 boys, 2 girls

    arrange 4, with 2 sets of 2 the same \Rightarrow\ P(2 girls)=\frac{\left(\frac{4!}{2!2!}\right)}{4!}

    3 boys, 1 girl

    arrange 4, with a set of 3 the same \Rightarrow\ P(3 boys)=\frac{\left(\frac{4!}{3!}\right)}{4!}

    3 girls, 1 boy.... same calculation.

    You neglected to count both cases.
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  7. #7
    Member courteous's Avatar
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    Red face

    I should have counted the other gender "combination" (instead of trying to dispute their solution). Thank you all for your help, I was surprised to see 5 replies. I also like TKHunny's quick binomial way of showing it.
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