$7$ 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.

  • $\begingroup$ How did you come to those conclusions? If you show your working, then people will be more willing to help you out. $\endgroup$ – lioness99a Feb 14 at 10:21
  • $\begingroup$ Thanks for the suggestion. $\endgroup$ – Conorlash Feb 14 at 12:28
  • $\begingroup$ No, the numerator is not correct. If there appears that another three babies will be born at the same day, you counted this variant twice. $\endgroup$ – NCh Feb 14 at 15:25
  • $\begingroup$ Please elaborate. $\endgroup$ – Conorlash Feb 14 at 15:48
  • $\begingroup$ Already writing, wait just a minute $\endgroup$ – NCh Feb 14 at 15:49

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 $$ \mathbb P(A)= 7\cdot \frac{\binom{7}{3}\cdot 6^4}{7^7} - \binom{7}{2} \cdot \frac{\binom{7}{3}\cdot\binom{4}{3}\cdot 5}{7^7} = \frac{302820}{7^7}. $$

  • $\begingroup$ Thanks. Now I got it. $\endgroup$ – Conorlash Feb 14 at 17:08
  • $\begingroup$ @Conorlash After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. $\endgroup$ – NCh Feb 15 at 1:08
  • $\begingroup$ Okay. I solved it by counting 5 different scenarios, and I am getting the same result. $\endgroup$ – Conorlash Feb 15 at 6:56
  • $\begingroup$ Yes, it is also the possible way to count the total number of arrangements in five variants $3+4+0+0+0+0+0$, $3+2+2+0+0+0+0$, $3+3+1+0+0+0+0$, $3+2+1+1+0+0+0$ and $3+1+1+1+1+0+0+0$. $\endgroup$ – NCh Feb 15 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.