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Thread: 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.

    . . $\displaystyle (a)\;\frac{1080}{7^5}\qquad(b)\;\frac{2160}{7^5}\q quad(c)\;\frac{540}{7^4}\qquad(d)\;\text{None of these}$
    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: .$\displaystyle {7\choose2} = 21$ ways to pick the two people.
    . . Call them $\displaystyle A$ and $\displaystyle B.$

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


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

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

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  3. #3
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    Thanks a lot Soroban.
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