If I pick a number between 1 and 99, what is the probability that someone else in my class (30 students) will pick that same number?
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)
etc
for 29 calculations
= (89x88x87x86x...x61)/90^29 = 1 - P(89,28)/90^29
= .000069048
so the prob that somebody picks the same
= 1 - .000069048
= .99993095
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
Pr(NoOnepicks)=(98/99)^29
Put this in your google search engine
(98/99)^29=
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?
Pr:1-pr(nokidshavesame birthday)
pr(no kids)=360/360*359/360*....
= 360!/360-25)! * 1/360^25
put this in your wolfram calculator:
(360!/(360-25)!) * 1/360^25
http://www.wolframalpha.com/input/?i=%28360!%2F%28360-25%29!%29+*+1%2F360^25&dataset=
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
89x88 ...
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...
pr(no one)=.43
which leads to the same result.
To determine the probability that someone else in your class will pick the same number as you, we need to consider the number of possible outcomes and the number of favorable outcomes.
There are 99 possible numbers you can choose from between 1 and 99. Out of those, you select one number.
For each number you choose, there are 30 students in your class who can make their own selection. Each student can, in theory, choose any of the 99 possible numbers.
Therefore, the total number of outcomes is 99 raised to the power of 30 (99^30) because there are 30 independent choices made by your classmates.
The number of favorable outcomes is 1 in each case because only one number was selected by you.
Hence, the probability that someone else in your class will pick the same number as you is 1 divided by 99 raised to the power of 30 (1/99^30).