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Thread: Married couple raffle question.

  1. #1
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    Married couple raffle question.

    Hey, I don't think I am doing this question correctly, my solutions just seem to simple.

    Question:
    There are 8 opposite-sex married couples at a party. Two people are chosen at random to win a door prize.
    a) What is the probability that the 2 people will be married to each other?
    b) What is the probability that the 2 people will be of the same sex?
    c) If 6 people are chosen, what is the probability that they are 3 married couples?

    Solution:
    a) $\displaystyle n(s) = 8C2 = 120 $
    $\displaystyle n(married couples) = 8 $
    So, that means that it's 8/120, or 1/15.

    b) $\displaystyle (\frac {8}{16})(\frac {7}{15}) $
    $\displaystyle = \frac {7}{30} $

    c) $\displaystyle n(s) = 16C6 = 8008 $
    There are 8 married couples, so 8C3.
    $\displaystyle \frac {56}{8008} $
    $\displaystyle \frac {1}{143} $


    I am unsure if any of these are correct, as I am not used to doing these sorts of problems with things like married couples.
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  2. #2
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    Hello, Kakariki!

    The problems are simple, but require some careful thinking.
    And you did fine!


    There are 8 opposite-sex married couples at a party.
    Two people are chosen at random to win a door prize.

    a) What is the probability that the 2 people will be married to each other?

    Solution:

    $\displaystyle n(s) \:=\: _8C_2 \:=\: 120 $

    $\displaystyle n(\text{married couples}) \:=\: 8 $

    So, that means that it's: .$\displaystyle \frac{8}{120} \:=\:\frac{1}{15}$ . . Right!


    b) What is the probability that the 2 people will be of the same sex?
    The first person can be any of the 16 people: .$\displaystyle \frac{16}{16} \:=\:1$

    The second must be one of the other 7 people of the same sex: .$\displaystyle \frac{7}{15}$

    Therefore: .$\displaystyle P(\text{same sex}) \;=\;\frac{7}{15}$

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Another approach . . .

    There are 120 possible outcomes.


    $\displaystyle \text{We want: }\:\begin{Bmatrix} \text{2 men:} & _8C_2 \:=\:28\text{ ways} \\

    \text{or} \\ \text{2 women: } & _8C_2\:=\:28\text{ ways} \end{Bmatrix} $

    Therefore: .$\displaystyle P(\text{same sex}) \;=\;\frac{56}{120} \;=\;\frac{7}{15}$




    c) If 6 people are chosen, what is the probability that they are 3 married couples?

    $\displaystyle n(s) \:=\:_{16}C_6 \:=\:8008$

    There are 8 married couples, so: .$\displaystyle _8C_3 \:=\:56$

    Therefore: .$\displaystyle \frac{56}{8008} \:=\:\frac{1}{143}$ . . Yes!
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