7 People born any day of the week

Question

I tried to apply the stars and bars theorem so that I can arrange the $3$ people that are left in the $5$ days between monday and sunday and that gave me $35.$

I know that the total ways to arrange 7 people in the 7 days of the week is $7^7.$ I don't know where's the fallacy in this logic.

If people can be born with the same probability any day of the week, what is the probability that in a random group of seven people two were born on Monday and two on Sunday?

Answer 1

There are a couple of mistakes here. First, you seem to be neglecting the number of ways to choose the two people born on Sunday and the two born on Monday. Second, stars and bars is not applicable, because the people are distinguishable. The number of ways for the remaining $3$ people to be born on Tuesday through Saturday is $5^3$, just like you figured out that the number of ways for everyone is $7^7.$

Answer 2

$P(m)=\frac{1}{7}$ (for Monday), $P(s)=\frac17$ as well (for Sunday), the probability of another day is $P(o)=\frac57$. And $7$ people means $7$ independent trials, we can choose the required Monday, Sunday and other birthdays among them in $\binom{7}{2,2,3}$ ways, and each such run has chance $(\frac{1}{7})^2 (\frac{1}{7})^2 (\frac{5}{7})^3$ so we get probability

$$\binom{7}{2,2,3} (\frac{1}{7})^2 (\frac{1}{7})^2 (\frac{5}{7})^3 = \frac{7 \times 6 \times 5 \times 125}{7^7}$$

in total. (i.e. about $0.031874$, a good 3 %)

Answer 3

The people are all different so total number of ways to choose $2$ from $7$ to have their bday on Sunday is ${7\choose 2}$ The total ways to choose $2$ from remaining $5$ to have bday on Monday are ${5\choose 2}$. The remaining three each have any of the $5$ days on which they can have their b'days. Thus total ways for them are $5^3$ hence probability is $\frac{{7\choose 2}{5\choose 2}5^3}{7^7}=0.03187$. In stars and bars you have to assume that each person is special ie it's not like arranging similar stars in the gaps.