assume that all 365 days of the year are equally likely and ignore leap years.
2007-01-28 07:43:37 · 9 answers · asked by jay 1 in Science & Mathematics ➔ Mathematics
For at least 2 out of 23, the probability is 50.7%, according to the Wikipedia page on this well-known "birthday paradox".
The page also gives the probabilities for various other sizes of group, and shows how to calculate it.
2007-01-28 07:53:40 · answer #1 · answered by Anonymous · 2⤊ 0⤋
22/365. Out of the 23, one can share a birthday with one of the other 22 persons. There are 365 calendar days in a year.
2007-01-28 07:51:28 · answer #2 · answered by Frankenstein 3 · 0⤊ 2⤋
P = (1 - 364/365 * 363/365 * 362/365 * 361/365 * 360/365 * . . . . . * (365 - 22)/365)
P = (1 - 364!/(342!*365^22)
P = 1 - 0.49270276567601459277458277166297
P ≈ 0.50730
2007-01-28 08:24:11 · answer #3 · answered by Helmut 7 · 1⤊ 0⤋
An easy way to calculate it is to calculate probabilty that none of the 23 share same birthday, and subtract that from 1.
The probabilty that none of them share same birthday is (365*364*363*...*343)/(365^23) = .49 approximately.
SO, its 1-.49 = .51 approximately.
2007-01-28 08:19:23 · answer #4 · answered by yljacktt 5 · 2⤊ 0⤋
69.315% after the first person just add 1/365 + 2/365 + ... + 22/365
2007-01-28 07:51:25 · answer #5 · answered by Anonymous · 0⤊ 2⤋
Well, the birthday ratio is pretty much 1/365
So, if you have 23 people, the ratio is now 23/365
Which is roughly 0.06301. Other wise known as 6.3% - that two of them might share a birthday. I do hope I'm right - I used a calculator.
Answer:
6.3% or roughly 6% or 23/365
2007-01-28 07:52:06 · answer #6 · answered by Anonymous · 0⤊ 3⤋
About 50%.
2007-01-28 07:46:05 · answer #7 · answered by amateur_mathemagician 2 · 0⤊ 1⤋
(23/365)^2
2007-01-28 07:51:22 · answer #8 · answered by Anonymous · 0⤊ 2⤋
Absolutely none, unless half of them were twins and the other half were triplets
2007-01-28 07:48:01 · answer #9 · answered by pop c 2 · 0⤊ 4⤋