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Math Help - Person and week problem.

  1. #1
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    Person and week problem.

    Q - In a room there are 7 persons.The chances that the two of them born on the same day of week.

    a) 1080/7^5 b) 2160/7^5 c) 540/7^4 d) None of these

    Any hint would be greatly appreciated.

    Thanks,
    Ashish
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  2. #2
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    Hello, Ashish!

    In a room there are 7 persons.
    Find the probability that two of them are born on the same day of week.

    . . (a)\;\frac{1080}{7^5}\qquad(b)\;\frac{2160}{7^5}\q  quad(c)\;\frac{540}{7^4}\qquad(d)\;\text{None of these}
    I got "None of these" until I read the problem again . . .

    I believe the problem means "exactly two have the same weekday of birth".


    There are: . {7\choose2} = 21 ways to pick the two people.
    . . Call them A and B.

    A can have any day for his birthday: \frac{7}{7} = 1
    . . B must have the same day: . \frac{1}{7}
    . . . . C must have one of the remaining six days: . \frac{6}{7}
    . . . . . . D must have one of the remaining five days: . \frac{5}{7}
    . . . . . . . . E must have one of the remaining four days: . \frac{4}{7}
    . . . . . . . . . . F must have one of the remaining three days: . \frac{3}{7}
    . . . . . . . . . . . . . G must have one of the remaining two days: . \frac{2}{7}


    Therefore, the probability is: . 21\cdot1\cdot\frac{1}{7}\cdot\frac{6}{7}\cdot\frac  {5}{7}\cdot\frac{4}{7}\cdot\frac{3}{7}\cdot\frac{2  }{7}

    . . which simplifies to: . \frac{2160}{7^5} . . . answer (c)

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  3. #3
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    Thanks a lot Soroban.
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