Question

Consider an experiment that consists of recording the birthday for each of 20 randomly selected persons. Ignoring leap years, we assume that there are only 365 possible distinct birthdays. Furthermore, we assume that each of the possible sets of birthdays is equi-probable (1)What is the probability that each person in the 20 has a different birthday?(2)Find the minimum number of persons such that the probability of two or more people have a same birthday is at least 50%.(3)Find the minimum number of persons such that the probability of two or more people have a same birthday is at least 95%

Answer 1

Answer:

a)  $p_{20d}$ = 0.588

b) 23

c) 47

Step-by-step explanation:

To find a solution for this question we must consider the following:

If we’d like to know the probability of two or more people having the same birthday we can start by analyzing the cases with 1, 2 and 3 people

For n=1 we only have 1 person, so the probability  $p_{1}$ of sharing a birthday is 0 ($p_{1}$=0)

For n=2 the probability $p_{2}$ can be calculated according to Laplace’s rule. That is, 365 different ways that a person’s birthday coincides, one for every day of the year (favorable result) and 365*365 different ways for the result to happen (possible results), therefore,

$p_{2} = \frac{365}{365^{2} } = \frac{1}{365}$

For n=3 we may calculate the probability $p_{3}$ that at least two of them share their birthday by using the opposite probability P(A)=1-P(B). That means calculating the probability that all three were born on different days using the probability of the intersection of two events, we have:

$p_{3} = 1 - \frac{364}{365}*\frac{363}{365} = 1 - \frac{364*363}{365^{2} }$

So, the second person’s birthday might be on any of the 365 days of the year, but it won’t coincide with the first person on 364 days, same for the third person compared with the first and second person (363).

Let’s make it general for every n:

$p_{n} = 1 - \frac{364}{365}*\frac{363}{365}*\frac{362}{365}*...*\frac{(365-n+1)}{365}$

$p_{n} = \frac{364*363*362*...*(365-n+1)}{365^{n-1} }$

$p_{n} = \frac{365*364*363*...*(365-n+1)}{365^{n} }$

$p_{n} = \frac{365!}{365^{n}*(365-n)! }$

Now, let’s answer the questions!

a) Remember we just calculated the probability for n people having the same birthday by calculating 1 minus the opposite, hence we just need the second part of the first calculation for $p_{n}$, that is:

$p_{20d} = \frac{364}{365}*\frac{363}{365}*\frac{362}{365}*...*\frac{(365-20+1)}{365}$

We replace n=20 and we obtain (you’ll need some excel here, try calculating first the quotients then the products):

$p_{20d}$ = 0.588

So, we have a 58% probability that 20 people chosen randomly have different birthdays.

b) and c) Again, remember all the reasoning above, we actually have the answer in the last calculation for pn:

$p_{n} = \frac{365!}{365^{n}*(365-n)! }$

But here we have to apply some trial and error for 0.50 and 0.95, therefore, use a calculator or Excel to make the calculations replacing n until you find the right n for $p_{n}$=0.50 and $p_{n}$=0.95

b) 0.50 = 365!/(365^n)*(365-n)!

n           $p_{n}$

1              0

2           0,003

3           0,008

….           …

20           0,411

21           0,444

22           0,476

23           0,507

The minimum number of people such that the probability of two or more of them have the same birthday is at least 50% is 23.

c) 0.95 = 365!/(365^n)*(365-n)!

We keep on going with the calculations made for a)

n             $p_{n}$

…                …

43            0,924

44            0,933

45            0,941

46            0,948

47            0,955

The minimum number of people such that the probability of two or more of them have the same birthday is at least 95% is 47.

And we’re done :)