# How rare is this? it happened to me

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

1. ### ali_babaRegistered Member

Messages:
1
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?

3. ### francescakatRegistered Member

Messages:
62
Maybe you both have the same test scores or last name.

5. ### DinosaurRational SkepticValued Senior Member

Messages:
4,885
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