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?
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. Code: P(EventA) = Number of ways EventA can occur divided by the total number of pertinent events. For example what about dice probabilities? Code: 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. Code: 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
