What Is the Probability That Students Are Born on Different Days of the Week?

[Zhalfirin88]

Homework Statement

A group of n students decided to find out on what day of the week each of them
was born. Find the probability that all of them were born on different days of the
week if a)n=2, b)n=4, c)n=7, d)n=8.

The Attempt at a Solution

Let's start with n = 2. First, I started out by finding the complement of A, that they are all born on the same day. The total number of combinations is n^r , where r = 7 for 7 days of the week.

Then the probability is 7 choose 2, because there are 7 different possible days, and there are 2 students that were born on the same day. Thus, the probability of A' is 21/128 = .1641, and the probability of A is 1 - .1641 = .8359.

This just doesn't seem right, because if you take n = 8, 8^7 = 2,097,152 but 7 choose 8 doesn't exist, so I'm thinking I'm misinterpreting the formulas or something.

 

[gb7nash]

Then the probability is 7 choose 2, because there are 7 different possible days, and there are 2 students that were born on the same day.

No. Think about it. Listing the ways we can have 2 being born on the same way, we get:

Monday Monday
Tuesday Tuesday
Wednesday Wednesday
Thursday Thursday
Friday Friday
Saturday Saturday
Sunday Sunday

So, there are 7 valid events. What's the total number of events? (It's not 128). Once you have the new probability of A', do what you were doing before.

Unfortunately, you can't use this method for n > 2. Do you see why? What's another way of counting n = 4 and n = 7? By the way, for n = 8, can 8 people be born on different days of the week? Think pigeonhole principle.

 

[Zhalfirin88]

No. Think about it. Listing the ways we can have 2 being born on the same way, we get:

Monday Monday
Tuesday Tuesday
Wednesday Wednesday
Thursday Thursday
Friday Friday
Saturday Saturday
Sunday Sunday

So, there are 7 valid events. What's the total number of events? (It's not 128). Once you have the new probability of A', do what you were doing before.

Unfortunately, you can't use this method for n > 2. Do you see why? What's another way of counting n = 4 and n = 7? By the way, for n = 8, can 8 people be born on different days of the week? Think pigeonhole principle.

So, in this case, the number of samples, r, is the number of students, and the set of objects, n, are the days of the week, thus n^r = 7² = 49 possible ways.

And you listed the favorable events for A', because since we want them to have the same birthday, would be 7 choose 1, correct?

P(A') = 7/49 = 1/7 = 0.142857
P(A) = 1 - 0.142857 = .8571428. And I see why you can't use this for n > 2.

For n = 4, would it not be (7 choose 1)(6 choose 1)...(4 choose 1) divided by 7^3?

For n = 7, there is only 1 favorable outcome, yet there are 7^7 different ways, correct?

Finally, n = 8 is the impossible outcome, yes.

 

[Ray Vickson]

Homework Statement

A group of n students decided to find out on what day of the week each of them
was born. Find the probability that all of them were born on different days of the
week if a)n=2, b)n=4, c)n=7, d)n=8.

The Attempt at a Solution

Let's start with n = 2. First, I started out by finding the complement of A, that they are all born on the same day. The total number of combinations is n^r , where r = 7 for 7 days of the week.

Then the probability is 7 choose 2, because there are 7 different possible days, and there are 2 students that were born on the same day. Thus, the probability of A' is 21/128 = .1641, and the probability of A is 1 - .1641 = .8359.

This just doesn't seem right, because if you take n = 8, 8^7 = 2,097,152 but 7 choose 8 doesn't exist, so I'm thinking I'm misinterpreting the formulas or something.

If the two students are Smith and Jones, look first at Smith. In 6 out of 7 days, the birthday of Jones is different from that of Smith, so P{two different days} = 6/7. For n = 3, P{all different} = (6/7)*(5/7), etc. This just a variant of the "birthday problem"; see, eg.,
http://en.wikipedia.org/wiki/Birthday_problem or
http://mathworld.wolfram.com/BirthdayProblem.html .

RGV