knock off of the "birthday problem"

byrne

New member
Joined
Nov 28, 2011
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5
hello!
I have tried solving this problem, I realize it is extremely similar to the 'birthday problem,' but I still cannot figure out which formulas to use.

QUESTION: "In a class of 36 students, what is the probability that at least 2 people have the same birthday? Assume that there are 365 days in a calendar year. (You can do a search for the birthday problem and discover that in a group of 23 people, the probability is 0.507 that at least two have the same birthday.)"

So far, I have this:

36 students ---- 2 same birthday --- 365 days.
P(n) = 1 - P(n)

Probability that it's not: 36/2 = 36 * 35/2 = 630 pairs.

NOW, I'm lost after this. :(

I'd really appreciate the help, thank you in advance!!!
 
Are you insane? Why would you calculate this directly?

p(at least 2) = 1 - pr(0) - p(1)

There's a whole lot less to calculate on the right hand side.
 
Could you please just respond with a more helpful description rather than "are you insane?"
Thank you, I appreciate it.
 
Could you please just respond with a more helpful description rather than "are you insane?"
Thank you, I appreciate it.

Pretty sure that's why the other two sentences are there. The opening was just to get your attention. It appears to have worked as intended.

Pick a player. Your choice has a birthday. This is more important than it sounds.

Now, go through the rest of the party, one at a time.

Person #2. p(same birthday as #1) = 1/365. p(not the same) = 364/365
Person #3. p(same birthday as NEITHER #1 nor #2) = 363/365
Person #4. p(same birthday as none of #1, #2, or #3) = 362/365
keep this going until you have considerd everyone.

p(no match) = \(\displaystyle \prod_{n=1}^{36-1}\frac{365-n}{365}\)
 
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I don't understand what the formulas in the other two sentences mean; no need to be derogatory. Could you please explain why and how you use these formulas?
Thanks, I do appreciate it!
 
Who's doing anything derogatory? Anyway...

If you have two birthdays to miss, assuming any day is as probable as any other, 2/365 is the probability that you will miss the two. Now you have three for the next candidate to miss.
 
Wow, I think I am actually understanding this one a bit, which means a lot for me in math. Thank you!!
 
Watch out for crazy notation, too. That bar over the second 'p' is pretty important. \(\displaystyle P(n) = 1 - \overline{P}(n)\)
 
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