Respuesta :
at least 24 people must be in the party to guarantee that at least 3 people have the same date of birth, date of birth, date of birth.
To warrant that at least three people have the same date, month, and year of birth, we need to find the minimum number of people such that there is a probability that at least three people have the same date, month, and year. there is. Birth year she is 1 or more. Non-leap years have 365 days, and leap years have her 366 days. The birth year range includes two leap years, so there are 3652 + 3662 = 2,922 possible births. This means that the probability that a particular person's birthday is unique is (2922-1)÷2922 = 1÷2922.
Now we can use the birthday puzzle to calculate the probability that at least three people have the same birthday. The probability that no one has the same birthday is (1 - 1÷2922)^n, where n is the number of people in your party. At least he has 1-(1-1÷2922)^n probability that 3 of them have the same birthday. You can set this probability to 1 or more and solve for n.
1 - (1 - 1÷2922)^n >= 1
(1 - 1÷2922)^n <= 0
n >= log(0) ÷ log(1 - 1÷2922)
The logarithm of 0 is not defined, so you can use a small value on the right hand side instead. B. 10^-6. This will tell you:
n >= log(10^-6) ÷ log(1 - 1/2922)
n >= log(10^-6) ÷ log(2921/2922)
n >= log(10^-6) ÷ log(2921) - log(10^-6) / log(2922)
n >= log(10^-6) × log(2922) ÷ log(2921) - log(10^-6)
n >= log(10^-6) × log(2922) ÷ log(2921) - log(10^-6)
n >= 23.
Hence, at least 24 people must be in the party to guarantee that at least 3 people have the same date of birth, date of birth, date of birth.
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