what is the probability of 2 out of 23 people having the same birthday? answer key
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This is usually stated to be "what is the probability of AT LEAST 2 out of 23 people having the same birthday?"
= 1 - prob(all have different birthday dates)
Start with any student, the prob that he will have a birthday that year
= 365/365 = 1
then the prob that the first 2 students will have different birthdays
= (365/365)(364/365)
the prob that the first 3 students will have different birthdays
= (365/365)(364/365)(363/365)
....
the prob that the 23 students will have different birthdays
= (365/365)(364/365)(363/365)...(343/365)
= (365!/342!)/365^23
Your standard calculator cannot handle such large numbers, but if we
do 23 divisions, we get ...
= .4927
prob(that at least 2 of 23 will have the same birthday)
= 1 -.4927 = .5073
Very patiently check my calculations to get .4927
I only did it once, might have hit a wrong key somewhere.
The concept is correct.
we could try this using a high level "calculator" such as Wolfram
let k = (365! - 342!)/365^23
ln k = ln [ (365! - 342!)/365^23 ]
= ln 365! - ln 342! - 23ln 365
Wolfram gave me:
ln 365! = 1792.33
ln 343! = 1657.34
23ln365 = 135.698
ln k = ln 365! - ln 342! - 23ln 365 = 1792.33-1657.34-135.698 = -.708
k = .4926 , hey, almost dead-on both ways
prob(that at least 2 of 23 will have the same birthday)
= 1 -.4926 = .5074 , off by .0001 from the other answer
How about just say: 50.7%
Wow, that was fun!!!
To calculate the probability of 2 out of 23 people having the same birthday, we can use the concept of the birthday paradox.
First, let's calculate the probability of two people NOT sharing a birthday. The probability that the second person has a different birthday than the first is 364/365, since there are 365 days in a year and the first person has already taken one of those days.
Now, for a group of three people, the probability that the third person has a different birthday than the first two is 363/365, since there are now only 363 remaining days in the year.
We can continue this pattern for each additional person. The probability that the k-th person has a different birthday than the previous (k-1) people is (365 - (k-1))/365.
Therefore, for a group of 23 people, the probability that no two people share a birthday is:
(364/365) * (363/365) * (362/365) * ... * (343/365)
To find the probability that at least two people share a birthday, we subtract the probability of nobody sharing a birthday from 1:
1 - [(364/365) * (363/365) * (362/365) * ... * (343/365)]
Calculating this expression gives us the probability of 2 out of 23 people having the same birthday.
Answer key:
The probability of 2 out of 23 people having the same birthday is approximately 0.5073, or 50.73%.
Note that this calculation assumes that all 365 days of the year are equally likely to be someone's birthday and that the birthdays of the individuals are independent of each other.