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
so the prob that somebody picks the same
= 1 - .000069048
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.
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?
= 360!/360-25)! * 1/360^25
put this in your wolfram calculator:
(360!/(360-25)!) * 1/360^25
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.
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.
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.
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
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
I think I actually computed with 365 first, the calculator wouldn't accept it, then redid it, not thinking.
Here it is with 365 days...
which leads to the same result.
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