# Thread: The Birthday Problem (another probability question)

1. ## The Birthday Problem (another probability question)

A room contains n randomly chosen people.

(a) Assume that a randomly chosen people is equally likely to have been born on any day of the week. The probability that the people in the room were all born on different days of the week is denoted by P.

(i) Find P in the case n=3.
(ii) Show that P=120/343 in the case n=4.

(b) Assume now that a randomly chosen people is equally likely to have been born in any month of the year. Find the smallest value of n such that the probability that the people in the room were all born on different months of the year is less than 0.5.

(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.

I thought that for (a) part 1, it would be (3/7) x (2/6) x (1/5) but then I can't show for part 2 so I guess I'm wrong. And I have no clue on (b) and (c) at all! So thanks so much if you could help!

2. Originally Posted by margaritas
A room contains n randomly chosen people.

(a) Assume that a randomly chosen people is equally likely to have been born on any day of the week. The probability that the people in the room were all born on different days of the week is denoted by P.

(i) Find P in the case n=3.
(ii) Show that P=120/343 in the case n=4.
For them all to have different birthdays we have person 1 can have any of 7
days for their birth dat, the second any of 6 and the third any of 5. There
are 7^3 ways that the birthdays can be allocated without worrying about
the days being different so the required probability is:

[7.6.5]/7^3 ~= 0.6122

Doing the same with 4 people gives: [7.6.5.4]/7^4 = [6.5.4]/7^3=120/343.

RonL

3. Originally Posted by margaritas
A room contains n randomly chosen people.

(b) Assume now that a randomly chosen people is equally likely to have been born in any month of the year. Find the smallest value of n such that the probability that the people in the room were all born on different months of the year is less than 0.5.
Generalising the method of (a) we have the probability that n people are all
born in different months is:

p(n,12) = [12!/(12-n)!]/12^n

Now we make up a table:

Code:
   n      p(n,12)
1         1
2         0.916667
3         0.763889
4         0.572917
5         0.381944
So the smallest n such that p(n,12)<0.5 is 5.

RonL

4. 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

5. Thanks alot I get (a) but still not too sure about (b) and (c).