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If I pick a number between 1 and 99, what is the probability that someone else in my class (30 students) will pick that same number?

  • MATH -

    proceed with the following argument
    Whatever number you pick, the prob that the
    second person picks a different number is 89/90
    the prob that the 2nd and 3rd person will pick a different number = (89/90)(88/90)
    the prob that the 2nd, the 3rd and the 4th pick different numbers = (89/90)(88/90)(87/90)
    for 29 calculations
    = (89x88x87x86x...x61)/90^29 = 1 - P(89,28)/90^29
    = .000069048

    so the prob that somebody picks the same
    = 1 - .000069048
    = .99993095

  • go with bob MATH -

    Go with bobpursley's solution

    I found the probabilty that at least two people in the class will have picked the same number, not just the number that you picked.

  • MATH -

    Ok, there are 29 other students.
    the probability that someone will pick your number is 1-noOnewillpick


    Put this in your google search engine

    then subtract it from one.

    I get about .25, or a quarter of the time, someone will pick you number.

    There is an old question, in a class of 25 kids, what is the probability that two kids in the class celebrates their birthday on the same day?

    Pr:1-pr(nokidshavesame birthday)

    pr(no kids)=360/360*359/360*....
    = 360!/360-25)! * 1/360^25

    put this in your wolfram calculator:
    (360!/(360-25)!) * 1/360^25!%2F%28360-25%29!%29+*+1%2F360^25&dataset=

    I get about .43

    which means the probility of two kids celebrating their birthdays on the same day in a clss of 25 is 1-.43=.57, better than even.

  • MATH -

    Reiny: I read your post, then glanced at mine, said oh my, erased my post, then said oh my again, and reposted. Such a life when one is tired.

  • Michael - MATH -

    Don't even look at my solution, I really messed up

    Somehow I read it as 90 not 99
    so all the calculations are bogus.

  • at bob -MATH -

    And I did the same thing, glad you were able to get yours back.
    I also don't even know why I started my series as
    89x88 ...
    when it should have been 98x97x...
    looks like I read the 99 as 90

    I usually make the font bigger, my reading glasses are a bit too weak.

    BTW, aren't there 365 days in a year ??
    look at your birthday problem solution, lol

  • MATH -

    I think I actually computed with 365 first, the calculator wouldn't accept it, then redid it, not thinking.

  • MATH -

    Here it is with 365 days...
    pr(no one)=.43
    which leads to the same result.

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