Probability that at least one student will share a birthday with another in a class of 25?
+23246 (I'm going to assume that the probability of being born on any day is the same as the probability of being born on any other day and I'm going to ignore the difficulties of twin, triplets, etc. and the difficulty of having to deal with the possibility of one or more having been born on February 29.)
Let's see what the probability that all 25 will have different birthdays (which is the oppositie of what we want) and then subtract this from 1.000 to get the probability that we want.
So that all the students have different birthdays:
The first person can have a birthday on any day of the year. This person has 365 choices out of all 365 choices to have a factor of 365/365.
The second person can have a birthday on any day but the day of the first person; this person has 364 choices out of all 365 choices to have a factor of 364/365.
The third person can have a birthday on any day but the two days already taken; this person has 363 choices out of all 365 choices to have a factor of 363/365.
Notice that the numerator keeps dropping by one each time while the denominator remains at 365.
This pattern continues; the 25th person has a factor of 341/365. Yes; it is 341/365, not 340/365; check it out!
To find the probability that all 25 persons have different birthdays, find this product:
(365/365) x (364/365) x (363/365) x (362/365) x ... x (343/365) x (342/365) x (341/365)
Since this is the opposite of what we want, subtract that answer from 1.0000.
+23246 (I'm going to assume that the probability of being born on any day is the same as the probability of being born on any other day and I'm going to ignore the difficulties of twin, triplets, etc. and the difficulty of having to deal with the possibility of one or more having been born on February 29.)
Let's see what the probability that all 25 will have different birthdays (which is the oppositie of what we want) and then subtract this from 1.000 to get the probability that we want.
So that all the students have different birthdays:
The first person can have a birthday on any day of the year. This person has 365 choices out of all 365 choices to have a factor of 365/365.
The second person can have a birthday on any day but the day of the first person; this person has 364 choices out of all 365 choices to have a factor of 364/365.
The third person can have a birthday on any day but the two days already taken; this person has 363 choices out of all 365 choices to have a factor of 363/365.
Notice that the numerator keeps dropping by one each time while the denominator remains at 365.
This pattern continues; the 25th person has a factor of 341/365. Yes; it is 341/365, not 340/365; check it out!
To find the probability that all 25 persons have different birthdays, find this product:
(365/365) x (364/365) x (363/365) x (362/365) x ... x (343/365) x (342/365) x (341/365)
Since this is the opposite of what we want, subtract that answer from 1.0000.
