Find the probability that among seven persons no two were born on the same day of the week

I Was initially thinking P(born on Monday) = 1/7 and P(Not Born on Monday) = 6/7 and then 1/7 * 6/7 = 6/42, but I don't know if that's the correct approach?

In addition what is P(at least 2 were born on same day) and P(two were born on a Saturday and 2 born on Tuesday) ?

• I recommend asking just one question at a time ... though note that P(at least 2 born on same weekday) = 1-P(all 7 born on different weekdays) Feb 3 '19 at 21:30
• "I Was initially thinking P(born on Monday) = 1/7 and P(Not Born on Monday) = 6/7 and then 1/7 * 6/7 = 6/42" Well, that's a correct answer. But to what question? That's the answer to what is the probability that a specific person was born on Monday and that another specific person was not. Is that the question that was asked? Feb 3 '19 at 22:32

This problem could be a bit more complicated, but the fact that there are 7 people and 7 days of the week makes it simpler. That's because, in this case, each day of the week must be assigned to exactly one person. There are $$7^7$$ ways to assign days of the week to 7 people. How many ways are there to assign days of the week with no repeats? You can think of looking at a calendar week and trying to slot the people in. It's just like ordering 7 books on a shelf. There are 7 choices for who can be born on Sunday, then 6 left for Monday, and so on. Take that number and divide it by $$7^7$$, and you're done.

As @Bram28 says in the comments, this allows you to easily calculate the probability that the opposite happens -- that some two were born on the same day.

I'll leave it to you to figure out the last question.

• The case of $n$ people and $k$ days, with $n\ne k$, is just as simple to solve... only, not with this approach.
– Did
Feb 3 '19 at 21:44
• Actually you can generalize the approach pretty easily. And yes, it is simple. Feb 3 '19 at 21:45

I Was initially thinking P(born on Monday) = 1/7 and P(Not Born on Monday) = 6/7 and then 1/7 * 6/7 = 6/42

OK, let's think about this. How is your P(born on Monday) defined? That is for any random person, right? And so sure, P(Not Born on Monday) = 6/7. But: if you multiple those (which, BTW, would not be $$\frac{6}{42}$$, but $$\frac{6}{49}$$), what does that mean? It is the probability that for some random person, that person is born on a Monday and not on a Monday? But that is assuming these two events are independent .. which clearly they are not. Is it that for two people, the first is born on a Monday, and the second one is not? Those are independent, sure ... but we have $$7$$ people involved, not just $$2$$.

Here is what you need to do: you need to line up all $$7$$ people (call them $$A$$ through $$G$$), and then do something like this: OK, it doesn't matter what day $$A$$ is born, but we need to make sure $$B$$ is born on a different day than $$A$$ was, the probability of which is $$\frac{6}{7}$$. OK, now assuming that $$B$$ is indeed born on a different day than $$A$$, two weekdays are taken, but we need to make sure $$C$$ is born on a different day yet, the probability of which is $$\frac{5}{7}$$. ... see where this is going?

• so then is it (1/7)(6/7)(5/7)(4/7)(3/7)(2/7)(1/7) = 0.0008 ?
– Rob
Feb 3 '19 at 22:16
• @Rob Almost: remember that the first person can be born on any day, it doesn't matter. Feb 3 '19 at 22:25