Question

what is the probability that at least two of them have the same birthday (month and day)? (assume that each day is equally likely to be a student's birthday, that there are no sets of twins, and that there are 365 days in the year. do not include leap years).

Answer 1

To calculate the probability that at least two students have the same birthday, use the complement rule. The probability can be calculated using the formula 1 - (365/365) * (364/365) * (363/365) * ... * (365 - n + 1)/365, where n is the number of students.

Step-by-step explanation:

To calculate the probability that at least two students have the same birthday (month and day), we can use the complement rule. The complement rule states that the probability of an event happening is equal to 1 minus the probability of the event not happening. In this case, the event not happening is that all students have different birthdays.

There are 365 possible birthdays in a year (assuming no leap years), so for the first student, the probability of having a unique birthday is 365/365. For the second student, the probability of having a unique birthday is 364/365, since one day has been taken by the first student. For the third student, the probability of having a unique birthday is 363/365, and so on.

Using the complement rule, the probability that at least two students have the same birthday is 1 - (365/365) * (364/365) * (363/365) * ... * (365 - n + 1)/365, where n is the number of students.

For example, if there are 8 students, the probability is 1 - (365/365) * (364/365) * (363/365) * (362/365) * (361/365) * (360/365) * (359/365) * (358/365) = 0.994.

Answer 2

probability is 1.

Step-by-step explanation:

In a lot of probability problems when asking for “at least two” is usually easier to see the opposite case, that is, the “none of them” case…

P(“at least one … X”) = 1 - P(“none of them … X”)

I will we exclude 29th of February, that is, considering none of the people was born on the leap day of a leap year.

First, let’s see what happens for 2 people. We can name one of those 2 people as “first” and the other as “second”. Whatever the birthday of the “first” one is, the other has 364 out of 365 days in the year to have a different birthday.

p(2) = 1–364/365 = 1/365

For 3 people, the “first” has a birthday, the “second” has a probability of 364/365 to have a different birthday and the third has a probability of 363/365.

p(3) = 1 - 364/365 * 363/365= (365^2 - 364*363)/365^2

For 4 people:

p(4) = 1 - 364/365 * 363/365 * 362/365

…

For n people, if n is less than 366:

p(n) = 1 - (364! / [365 - n]! )/365^(n-1)

p(n)=1−364!(365−n)!⋅365n−1

Obviously, if n is 366 or more, since we excluded the leap day there must be always 2 people with the same birthday (pigeonhole principle), and then in this cases the probability is 1.