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There are $15$ students. Assume that there are $365$ days in a year.

What is the probability that $3, 5$ and $7$ of them have the same birthday on any three days of a year, with these three groups of students have a different birthday?

$$ P\left(3\right)=\frac{365}{365}\times\frac{1}{365}\times\frac{1}{365}=\frac{1}{133225} $$ $$ P\left(5\right)=\frac{364}{365}\times\frac{1}{365}\times\frac{1}{365}\times\frac{1}{365}\times\frac{1}{365}=5.6187\times{10}^{-11} $$ $$ P\left(7\right)=\frac{363}{365}\times\frac{1}{365}\times\frac{1}{365}\times\frac{1}{365}\times\frac{1}{365}\times\frac{1}{365}\times\frac{1}{365}=4.206\times{10}^{-16} $$

should I change the denominator for $P(5)$ to $364$ and $P(7)$ to $363$ as after each group the available days for birthday decreases?

Can someone please explain it to me?

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  • $\begingroup$ Hint: your formula for $P(3)$ is wrong for two reasons. (1) If you assign #'s to the people, (e.g. people are p-1, p-2, ...p-3) then you have calculated the probability that p-1, p-2, p-3 all have the same birthday. You have to consider that there could be three others (for example) in the group of 15 that have the same birthday. (2) You have to very carefully distinguish between exactly 3 people having the same birthday, and at least 3 people having the same birthday. $\endgroup$
    – user2661923
    Nov 5, 2020 at 19:22
  • $\begingroup$ @user2661923 maybe the OP is looking for something like $P(3) \cdot P(5) \cdot P(7)$ $\endgroup$
    – Tortar
    Nov 5, 2020 at 20:37

1 Answer 1

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Tortar's comment makes the very fair point that perhaps the OP wants to compute the following probability.

  • There are three distinct pertinent days in the 365 day year, day-1, day-2, and day-3.

  • Of the 15 people, exactly 3 of the people have day-1 as their birthday.

  • Of the 15 people, exactly 5 of the people have day-2 as their birthday.

  • Of the 15 people, exactly 7 of the people have day-3 as their birthday.

In my opinion, the easiest way to attack this problem is to express the probability as

$$\frac{N\text{(umerator)}}{D\text{(enominator)}}$$

with

$$D = (365)^{(15)}.$$

$N$, which represents the # of distinct ways that the scenario can occur, will be computed as the product of factors.

The # of ways that the 3 days can be selected is

$$N_1 = 365 \times 364 \times 363.$$

The # of ways that the 15 people can be partitioned into groups of 3, 5, and 7 is

$$N_2 = \binom{15}{3} \times \binom{12}{5}.$$

Final answer is

$$\frac{N_1 \times N_2}{D}.$$

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  • $\begingroup$ Understood. So does this means my way of doing that is completely wrong? If wrong, it is wrong in which way? I am clear about your way. $\endgroup$
    – Hames
    Nov 6, 2020 at 3:11
  • $\begingroup$ does the denominator 365^15 means the number of possible ways for 15 of them to have birthday? $\endgroup$
    – Hames
    Nov 6, 2020 at 3:15
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    $\begingroup$ @Hames The denominator of $(365)^{(15)}$ represents that each of the 15 people have 365 different possibilities for their birthday. If you examine your answer, you will see that this corresponds to your having the number $(365)$ appear as a denominator 15 times. If you also compare the numerators of our two answers, you will see that you also had the $(365 \times 364 \times 363)$ factor. This means that the only mistake that you made is that there are $\binom{15}{3} \times \binom{12}{5}$ ways of partitioning the 15 people into groups of 3, 5, and 7. $\endgroup$
    – user2661923
    Nov 6, 2020 at 3:52
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    $\begingroup$ @Hames Although technically, you could have salvaged your way by figuring out about the needed factor of $\binom{15}{3} \times \binom{12}{5}$, your way is not generally recommended. Typically, in a problem like this, the easy road is to assume that (for example), the denominator is $(365)^{(15)}$ and use that as the starting point of your attack. $\endgroup$
    – user2661923
    Nov 6, 2020 at 3:59
  • $\begingroup$ I see! Now i understood. Thank you very much!! $\endgroup$
    – Hames
    Nov 6, 2020 at 4:56

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