How rare is this? it happened to me

Discussion in 'Physics & Math' started by ali_baba, Nov 28, 2013.

  1. ali_baba Registered Member

    In high school, my day was split into 8 classes. I had 5 of those 8 classes with this one friend, what are the odds of that.
    200 students in my class
    20 students per classroom
    8 classes per day
    what are the odds of 2 people being in the same class 5 out of the 8 times?

    The answer I got is 1/2138573647911200638050300 or ~4*10^-25

    Is that correct and what is an event that has a similar probability of occurring?
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  3. francescakat Registered Member

    Maybe you both have the same test scores or last name.
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  5. Dinosaur Rational Skeptic Valued Senior Member

    Ali Baba: Probability is one of the simplest disciplines in theory & one of the most difficult to deal with practice. Simple in principle due to the following.
    P(EventA) = Number of ways EventA can occur divided by the total number of pertinent events.
    For example what about dice probabilities?
    P(Snake Eyes) = 1/36 or .027 777 778
    P(Seven) = 6/36 or .16666667
    Now for the question asked: I disagree with your estimate of 4 * 10[sup]-25[/sup], but note my caveat at the end of this post.

    First: The question is a bit ambiguous. I will assume that students are assigned to classes randomly. I will also assume that you want the probability of both of you being in exactly 5 classes, rather than both being in 5 or more classes.

    BTW: Terms like random & randomly are somewhat ambiguous.

    There are C(200,20) ways 200 people can be assigned to a class of 20 people.
    There are C(200,18) ways a class can include two specific people (Id est: You & your friend).

    For a specific class, P(Both) = C(200,18) / C(200,20)
    P(Both) = [200! / 182!*18!] / [200! / 180!*20!]
    P(Both) = 200!*180!*20! / 200!*182!*18!
    P(Both) = 20*19 / 182*181
    P(Both) = .011 535 426 Call this P
    P(Not Both) = .988 464 574 Call this Q

    Now consider expanding (P + Q)[sup]8[/sup], which represents the probabilities of both being in no class together to both being in all eight classes together.
    The term of interest is the one with P[sup]5[/sup] * Q[sup]3[/sup]
    This term has a coefficient of C(8,5) = 56

    Hence the answer is 56 *.011 535 426[sup]5[/sup] *.988 464 574[sup]3[/sup] = .000 000 011

    My HP50G calculates (P + Q)[sup]8[/sup] as.
    0[b]:[/b] .911 357 711
    1[b]:[/b] .085 084 684
    2[b]:[/b] .003 475 297
    3[b]:[/b] .000 081 114
    4[b]:[/b] .000 001 188
    5[b]:[/b] .000 000 011
    Others: zilch
    In the above, the integer specifies the number of classes in which both are present & the decimal fraction specifies the probablity of that event.

    I trust my HP50G more than my memory of how to make up the correct calculations for the 50G & my ability to make neither keying errors nor typo’s
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