What is the smallest number of people that can be in a room where the probability of two of them having the same birthday is at least 50 percent. (Hint its more than 1, but less than or equal to 31)
2006-08-30 05:59:44 · 11 answers · asked by xxshortstuffcwxx 1 in Science & Mathematics ➔ Mathematics
23, this is a common probability problem, have done in all my probability classes in college. Only thing is we had a set of twins in the major so it always threw it off. but the correct answer is 23.
2006-08-30 07:41:19 · answer #1 · answered by Rach524 2 · 0⤊ 0⤋
The answer is 184 persons. The first person has a birthday. The probability that the second one has the same birthday is 1/365. the probability that the third one has the same birthday of the first or the second person, is 2/365. So the probability that the people number n has the same birthday of one of the past persons is (n-1)/365.
If you want (n-1)/365 be at least 50%, n-1 would be equal to 183. So n=184.
2006-08-30 14:15:49 · answer #2 · answered by Arash 3 · 0⤊ 0⤋
I'm thinking its 20. I do know you have about a 95% chance that in a room of 35 people, 2 will share a birthday. This has to do with summing probabilities. I.E., in a room of 20, you have a 20 in 365 chance you share a birthday, but you have 19 others with equal chances. That helps increase the probability.
2006-08-30 13:08:04 · answer #3 · answered by eric r 2 · 0⤊ 0⤋
You want to find the probability of n people having different birthdays. You are choosing n days from a year without replacement, and there are (365 choose n) ways of this occurring, or 365!/(n!(365-n)!). The probability P of each sequence of days occurring is equal to (1/365)^n times the number of ways n days can be arranged, which is n!. So the probability equals (365 choose n) * P, or [365!/(365-n!)]*(1/365)^n
So we want the minimum value of n such that [(365!/(365-n!)]*(1/365)^n)<.5.
2006-08-30 13:54:35 · answer #4 · answered by john 3 · 0⤊ 0⤋
16
2006-08-30 13:15:16 · answer #5 · answered by Spuddy 2 · 0⤊ 0⤋
21
2006-08-30 13:49:21 · answer #6 · answered by Anonymous · 0⤊ 0⤋
Two. 50/50 chance. (The hint about two people needing to be more than one was a little unnecessary.)
2006-08-30 13:05:41 · answer #7 · answered by James O 1 · 0⤊ 0⤋
maybe 50/50
2006-08-30 13:34:13 · answer #8 · answered by Anonymous · 0⤊ 0⤋
good question!
back when I was in high school, we did this experiments in our class. I suspect the probability is very high! but I forgot the answer.
please tell me, looking forward to it!
2006-08-30 13:43:58 · answer #9 · answered by David F 2 · 0⤊ 1⤋
are you trying to get someone to do your math homework for you?
2006-08-30 13:04:22 · answer #10 · answered by what 3 · 0⤊ 1⤋