$7$ babies were born in a particular week in a private hospital. What is the probability that three babies were born on the same day of the week?
My answer is the denominator is $7^7$. The numerator will be $6480\times49= 317520$. Is it correct? The reasoning is to first select 3 babies from 7 babies and then the remaining babies have 6 days each, and then multiply it with 7.
Best Answer
Let us introduce seven events $A_1, \ldots, A_7$, where $A_i$ means that some three babies were born at $i$th day. Then we should find the probability of $$ A=A_1\cup A_2\cup\ldots\cup A_7. $$ Note that these events are not disjoint, and $A_1\cap A_2$ means that three babies were born at the first day and another three - at the second day. The intersections of more than two events are impossible. So $$ \mathbb P(A) = \sum_{i=1}^7 \mathbb P(A_i) - \sum_{i<j}\mathbb P(A_i\cap A_j) $$ where the second sum is over all possible pairs of $i,j$ with each pair appeared only once. The first sum containes $7$ identical probabilities $$ \mathbb P(A_1) = \frac{\binom{7}{3}\cdot 6^4}{7^7} $$ and the second sum containes $\binom{7}{2}$ identical probabilities $$ \mathbb P(A_1\cap A_2) = \frac{\binom{7}{3}\cdot\binom{4}{3}\cdot 5}{7^7}. $$ Indeed, firstly we choose three babies to born at the first day, then choose three babies from four to born at the second day, then the rest baby can born at any of the rest five days.
So $$ \mathbb P(A)= 7\cdot \frac{\binom{7}{3}\cdot 6^4}{7^7} - \binom{7}{2} \cdot \frac{\binom{7}{3}\cdot\binom{4}{3}\cdot 5}{7^7} = \frac{302820}{7^7}. $$