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Math Help - The Birthday Problem (another probability question)

  1. #1
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    The Birthday Problem (another probability question)

    A room contains n randomly chosen people.

    (a) Assume that a randomly chosen people is equally likely to have been born on any day of the week. The probability that the people in the room were all born on different days of the week is denoted by P.

    (i) Find P in the case n=3.
    (ii) Show that P=120/343 in the case n=4.

    (b) Assume now that a randomly chosen people is equally likely to have been born in any month of the year. Find the smallest value of n such that the probability that the people in the room were all born on different months of the year is less than 0.5.

    (c) Assume now that a randomly chosen people is equally likely to have been born on any of the 365 days in the year. It is given that, for the case n =21, the probability that the people in the room were all botn on different days of the year is 0.55631, correct to 5 places of decimals. Find the smallest value of n such that the probability that at least two of the people were born on the same day of the year exceeds 0.5.


    I thought that for (a) part 1, it would be (3/7) x (2/6) x (1/5) but then I can't show for part 2 so I guess I'm wrong. And I have no clue on (b) and (c) at all! So thanks so much if you could help!
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  2. #2
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    Quote Originally Posted by margaritas View Post
    A room contains n randomly chosen people.

    (a) Assume that a randomly chosen people is equally likely to have been born on any day of the week. The probability that the people in the room were all born on different days of the week is denoted by P.

    (i) Find P in the case n=3.
    (ii) Show that P=120/343 in the case n=4.
    For them all to have different birthdays we have person 1 can have any of 7
    days for their birth dat, the second any of 6 and the third any of 5. There
    are 7^3 ways that the birthdays can be allocated without worrying about
    the days being different so the required probability is:

    [7.6.5]/7^3 ~= 0.6122

    Doing the same with 4 people gives: [7.6.5.4]/7^4 = [6.5.4]/7^3=120/343.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by margaritas View Post
    A room contains n randomly chosen people.

    (b) Assume now that a randomly chosen people is equally likely to have been born in any month of the year. Find the smallest value of n such that the probability that the people in the room were all born on different months of the year is less than 0.5.
    Generalising the method of (a) we have the probability that n people are all
    born in different months is:

    p(n,12) = [12!/(12-n)!]/12^n

    Now we make up a table:

    Code:
       n      p(n,12)
       1         1
       2         0.916667 
       3         0.763889 
       4         0.572917
       5         0.381944
    So the smallest n such that p(n,12)<0.5 is 5.

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by margaritas View Post
    A room contains n randomly chosen people.
    (c) Assume now that a randomly chosen people is equally likely to have been born on any of the 365 days in the year. It is given that, for the case n =21, the probability that the people in the room were all botn on different days of the year is 0.55631, correct to 5 places of decimals. Find the smallest value of n such that the probability that at least two of the people were born on the same day of the year exceeds 0.5.
    Find the smallest value of n such that the probability that at least two of the
    people were born on the same day of the year exceeds 0.5, is the same as
    finding the smallest n such that the probability that they are all born on
    different days is less that or equal 0.5.

    This uses exactly the same method as (b)

    p(n,365) = [365!/(365-n)!]/365^n

    The only problem here is that we cannot handle 365! easily.

    Now I can write some code to compute this:

    Code:
                n       p(n,365)
                1             1 
                2       0.99726 
                3      0.991796 
                4      0.983644 
                5      0.972864 
                6      0.959538 
                7      0.943764 
                8      0.925665 
                9      0.905376 
               10      0.883052 
               11      0.858859 
               12      0.832975 
               13       0.80559 
               14      0.776897 
               15      0.747099 
               16      0.716396 
               17      0.684992 
               18      0.653089 
               19      0.620881 
               20      0.588562 
               21      0.556312 
               22      0.524305 
               23      0.492703 
               24      0.461656 
               25        0.4313 
               26      0.401759 
               27      0.373141 
               28      0.345539 
               29      0.319031 
               30      0.293684
    from this table we see that the required answer is 23.

    We could have got to this result using the given result that

    p(21,365) = 0.55631

    Then

    p(22,365) = p(21,365)*(365-21)/365 = 0.5243

    and:

    p(23,365) = p(22,365)*(365-22)/365 = 0.4927

    RonL
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  5. #5
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    Thanks alot I get (a) but still not too sure about (b) and (c).
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