Classic Birthday Problem Find the probability that among 25 randomly selected pe
ID: 3235934 • Letter: C
Question
Classic Birthday Problem Find the probability that among 25 randomly selected people at least 2 have the same birthday. To solve this problem you have to use a simulation. A simulation of a procedure is a process that behaves the same way as the actual procedure so that similar results are produced. For the above classic birthday problem, a simulation begins by representing birthdays by integers from 1 through 365, where 1 represents a birthday of January 1, 2 represents January 2, 3 represents January 3, and so on. We can simulate 25 birthdays by using a calculator or computer to generate 25 random numbers (with repetition allowed) between 1 and 365. Those numbers can then be sorted, so it becomes easy to examine the list to determine whether any 2 of the simulated birth days are the same. (After sorting, equal numbers are adjacent.) We can repeat the process as many times as we like, until we are satisfied that we have a good estimate of the probability. To do the simulation, you can use either the TI-83/84 Plus graphing calculator or StatCrunch. Instructions for both of these are on the next page. TI-83/84 Plus: Press MATH, select PRB, then choose 5:randInt. Press ENTER. Enter the minimum of 1, the maximum of 365, and 25 for the number of values, all separated by commas, as in randInt (1, 365, 25). Press ENTER. You can then store the data in list L1. To do this, press STO then 2^ND then L1 To see the numbers listed in in L1, press STAT then ENTER You can then sort L1 by pressing STAT and selecting 2:SortA. Press ENTER. You will see Sort A (on the screen. Press 2^ND then L1 then ENTER. You will see Sort A (L1 and Done on the screen. Go to see L1 by pressing STAT then ENTER. L1 is now sorted.Explanation / Answer
what is the probability that no two people will share a birthday?
Again, the first person can have any birthday. The second person's birthday has to be different. There are 364 different days to choose from, so the chance that two people have different birthdays is 364/365. That leaves 363 birthdays out of 365 open for the third person.
To find the probability that both the second person and the third person will have different birthdays, we have to multiply:
(365/365) * (364/365) * (363/365) = 132 132/133 225,
which is about 99.18%.
If we want to know the probability that four people will all have different birthdays, we multiply again:
(364/365) * (363/365) * (362/365) = 47 831 784/ 48 627 125,
or about 98.36%.
We can keep on going the same way as long as we want. A formula for the probability that n people have different birthdays is
If you know permutation notation, you can write this formula as
That's the same as
here n=25
It follows that if there are 25 people in a room the probability that they will all have different birthdays is
365!/((365-25)!*365^25)
=365!/(340!*365^25)
=0.43129
hence required probability = 1-0.43129= 0.5687