Probability of 5 of them having the same birthday while the other 10 having different birthday.

Question

There are 15 students. Assume that there are 365 days in a year, what is the probability that 5 of them have the same birthday on any day of a year (the other 10 students have different birthday)?

$$ P\left(1^{st\ }in\ 5\ to\ have\ same\ birthday\right)=\frac{365}{365}=1 $$ The rest of the 4 students then need to have same birthday as the first student, making probability for each of them, $\frac{1}{365}$. $$ P=\frac{365}{365}\times\frac{1}{365}\times\frac{1}{365}\times \frac{1}{365}\times\frac{1}{365}\times\frac{364}{365}\times \frac{363}{365}\times\frac{362}{365}\times\frac{361}{365}\times\frac{360}{365}\times\frac{359}{365}\times\frac{358}{365}\times\frac{357}{365}\times\frac{356}{365}\times\frac{355}{365} = 4.839 \times 10^{-11} $$

Initially, I am using this way to figure out the probability and I am exploring another way.

$$356^{15} = 2.7189 \times 10^{38} $$

This shows the 365 possibilities for their birthday. There are 11 ways for the birthdays to be selected which decreases after each birthday being selected.

$$ N_{1} = 365 \times 364 \times363\times362\times361\times360\times359\times358\times357\times356\times355 = 1.3157 \times10^{28} $$ $$ N_2 = {{15}\choose{11}} = 1365 $$ $N_2$ is that there are 11 possible combinations in the 15 students.

$$ P = \frac{{N_1}{N_2}}{D} = 6.605 \times 10^{-8} $$

Can someone please explain why is the right way of doing it and why one of them is wrong?

Answer 1

Your first solution is almost correct; you just have to include the choice of which 5 of the 15 students would share the same birthday; i.e. the answer should be

$$\left[ 4.389 \times 10^{-11} \right] \cdot \binom{15}{5} \approx 1.453 \times 10^{-7}$$

The second solution doesn't work because in your computation $365 \times 364 \times \cdots \times 355$, you're assigning an unnecessary ordering to the 10 dates chosen for the people with different birthdays. Also, the second step of calculating $\binom{15}{11}$ doesn't work - which 11 people are being chosen?

Note a simple method to count the number of ways is to break the process into smaller steps:

Step (1): Choose the birth date that five people will share - this can be done in $365$ ways.

Step (2): Choose which five people will share the above date - this can be done in $\binom{15}{5}$ ways.

Step (3): Choose the birth dates for the 10 remaining people - this can be done in $\binom{364}{10}$ ways.

Step (4): Assign the 10 birth dates to the 10 remaining people - this can be done in $10!$ ways.

Thus the required probability is:

$$\frac{365 \binom{15}{5} \binom{364}{10} 10!}{365^{15}} \approx 1.453 \times 10^{-7}$$