What is the chance that at least two people were born on the same day
of the week if there are 3 people in the room?
I'm wondering if my solution is accurate, as my answer was different than the solution I found:
Probability that there are at least 2 people in the room born on the same day = 1 – (No one was born on the same day) – (Exactly one person was born on the same day)
There are (3 choose 2) different pairs of couples. Each couple has the same birthday as another couple with the chances of 1/7 and different with chances 6/7. Thus:
$$1 – (6/7)^3 – 3(1/7)(6/7)^2 = 0.0553$$
Thanks for any help!
Best Answer
Label the three people $A,B$, and $C$. Suppose that no two were born on the same day of the week. $A$ can be born on any day of the week. The probability that $B$ was born on a different day is $\frac67$. (We are of course assuming that the seven days are equally likely, though I believe that in fact this isn't the case.) Given that $A$ and $B$ were born on different days of the week, the probability that $C$ was born on one of the remaining $5$ days of the week is $\frac57$. Thus, the probability that they were born on three different days of the week is $\frac67\cdot\frac57=\frac{30}{49}$, and the probability that at least two of them were born on the same day of the week is $1-\frac{30}{49}=\frac{19}{49}$.
You can check this as follows. There are $7^3=343$ possible ways of assigning days of the week to $A,B$, and $C$. $7\cdot6\cdot5=210$ of these result in $A,B$, and $C$ being assigned different days of the week, so the remaining $343-210=133$ assignments give at least two of the three people the same day of the week. Since all assignments are (assumed to be) equally likely, the probability of getting one of those is $\frac{133}{343}=\frac{19}{49}$.