Originally Posted by

**margaritas** A room contains *n* randomly chosen people.

(c) Assume now that a randomly chosen people is equally likely to have been born on any of the 365 days in the year. It is given that, for the case n =21, the probability that the people in the room were all botn on different days of the year is 0.55631, correct to 5 places of decimals. Find the smallest value of n such that the probability that at least two of the people were born on the same day of the year exceeds 0.5.

Find the smallest value of n such that the probability that at least two of the

people were born on the same day of the year exceeds 0.5, is the same as

finding the smallest n such that the probability that they are all born on

different days is less that or equal 0.5.

This uses exactly the same method as (b)

p(n,365) = [365!/(365-n)!]/365^n

The only problem here is that we cannot handle 365! easily.

Now I can write some code to compute this:

Code:

n p(n,365)
1 1
2 0.99726
3 0.991796
4 0.983644
5 0.972864
6 0.959538
7 0.943764
8 0.925665
9 0.905376
10 0.883052
11 0.858859
12 0.832975
13 0.80559
14 0.776897
15 0.747099
16 0.716396
17 0.684992
18 0.653089
19 0.620881
20 0.588562
21 0.556312
22 0.524305
23 0.492703
24 0.461656
25 0.4313
26 0.401759
27 0.373141
28 0.345539
29 0.319031
30 0.293684

from this table we see that the required answer is 23.

We could have got to this result using the given result that

p(21,365) = 0.55631

Then

p(22,365) = p(21,365)*(365-21)/365 = 0.5243

and:

p(23,365) = p(22,365)*(365-22)/365 = 0.4927

RonL