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Math Help - Help me please: difficult problems

  1. #1
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    Help me please: difficult problems

    Hi everyone
    I got an exam tomorrow and I just found this website I shoulda found earlier. Please help me if you know how to do these problems:

    1. If 3 people are picked from a group of 4 married couples, what is the probability of NOT including a pair of spouses?

    2. Find a formula for the probability that EXACTLY one of the events A, B, or C occurs. Then, make up a problem and solve the problem using the formula.

    3. Find the probability of getting three of a kind in a regular poker (excluding 2 jokers so there is 52 pokers)

    a) 3 of a kind means KKK A 2
    My answer is [(13C1)*(4C3) + 2*[(12C1)*(4C1)]] / (52C5)

    b) full house: [(13C1)*(4C3) + (12C1)*(4C2)] / (52C5)

    c) 4 of a kind: [(13C1)*(4C4) + (12C1)*(4C1)] / (52C5)

    What do you guys think?
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  2. #2
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    Quote Originally Posted by bondvista View Post
    Hi everyone
    I got an exam tomorrow and I just found this website I shoulda found earlier. Please help me if you know how to do these problems:

    1. If 3 people are picked from a group of 4 married couples, what is the probability of NOT including a pair of spouses?

    2. Find a formula for the probability that EXACTLY one of the events A, B, or C occurs. Then, make up a problem and solve the problem using the formula.

    3. Find the probability of getting three of a kind in a regular poker (excluding 2 jokers so there is 52 pokers)

    a) 3 of a kind means KKK A 2
    My answer is [(13C1)*(4C3) + 2*[(12C1)*(4C1)]] / (52C5)

    b) full house: [(13C1)*(4C3) + (12C1)*(4C2)] / (52C5)

    c) 4 of a kind: [(13C1)*(4C4) + (12C1)*(4C1)] / (52C5)

    What do you guys think?
    Hi bondvista,

    3 of a kind:
    \frac{\binom{13}{1} \binom{4}{3} \binom{12}{2} \binom{4}{1}^2} {\binom{52}{5}}

    Full house:
    \frac{\binom{13}{1} \binom{4}{3} \binom{12}{1} \binom{4}{2}} {\binom{52}{5}}

    (You seem to be adding in some cases above where you should be multiplying.)
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  3. #3
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    what about the first 2 problems please? Thank you!
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  4. #4
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    Quote Originally Posted by bondvista View Post
    what about the first 2 problems please? Thank you!
    On #1. There are 24 ways to choose a married couple to be among the three chosen.
    So take that away from the total: {8 \choose 3}.

    For #2. P(A \cap B^c  \cap C^c ) + P(A^c  \cap B \cap C^c ) + P(A^c  \cap B^c  \cap C)
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  5. #5
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    Quote Originally Posted by Plato View Post
    On #1. There are 24 ways to choose a married couple to be among the three chosen.
    So take that away from the total: {8 \choose 3}.

    For #2. P(A \cap B^c  \cap C^c ) + P(A^c  \cap B \cap C^c ) + P(A^c  \cap B^c  \cap C)
    24 ways. Is this how u got it, 4C1*3C1*2=24?
    Thank you guys so much!
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