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### The Birthday Problem; Queueing at a Bank

```Date: 08/26/99 at 09:52:06
From: Sophia
Subject: Probability

I am having an awful time figuring out probability! Can you please
help?

1. In a gathering of 23 people, what is the probability that at least
two people have the same birthday?  Assume a 365-day year.

I figure that I have (365)^23 possible outcomes. But the problem I am
having trouble with is the numerator. I don't know how to say that _at
least_ two people have the same birthday. Any help?

2. Forty customers at a bank are waiting for the doors to open,
whereupon they will each choose, at random, one of five tellers (and
after choosing they do not change lines). If no more customers after
the forty arrive, then...

a) What is the probability that teller #2 has no customers?

I _think_ I may have this one. I think the total possible outcomes is
(5)^40, and the possibility that teller #2 is not selected by everyone
is (4/5)^40. So I think the answer is: [(4/5)^40]/[(5)^40]. Am I
right?

But I am stuck and really confused by the others...

b) What is the most likely number of customers that teller #3 is to
serve?

I am guessing that I need to find out which number between 0-40 is
most likely to occur in a teller line. But I don't know how to do
that.

c) What is the probability that every teller has at least one
customer?

d) What is the probability that exactly two tellers do not have any
customers?

I know it seems like I haven't made much progress, and I guess I
haven't. But I have really been racking my brain on this. I just need
a good explanation of how to think about these problems. I would be so
grateful for any help!

Thanks,
Sophia

```

```Date: 08/26/99 at 14:41:38
From: Doctor Anthony
Subject: Re: Probability

>1. In a gathering of 23 people, what is the probability that at least
>   two people have the same birthday?  Assume a 365-day year.

They chose 23 people because that is the number above which there is a
better than 50% probability of at least two sharing a birthday.

If no two share a birthday then the first person can choose a birthday
in 365 ways, the second person in 364 ways, the third in 363 ways, and
so on. If there are n people in the room, the probability that no two
share a birthday is

365 x 364 x 363 x ... x (365-n +1)     P(365,n)
----------------------------------  =  --------
365^n                     365^n

So the probability that two at least share a birthday is:

P(365,n)
1 - --------
365^n
P(365,n)
If this must be >= 0.5 then we require  -------- <= 0.5
365^n

(Here, P(365,n) is the number of permutations of n things that can be

There is no easy way to solve the equation for n, but with a
calculator which gives permutations and combinations, trial and error
will quickly establish that n must be around 23.

P(365,23)
---------  = 0.4927
365^23

and 1 - 0.4927 = 0.5073

and so 23 people are sufficient to make the probability of a shared
birthday greater than 50%.

>2. Forty customers at a bank are waiting for the doors to open...
>a) What is the probability that teller #2 has no customers?

You can think of this as binomial probability. For any teller the
probability of 'success' with a particular customer is 1/5. So n = 40,
p = 1/5, and q = 4/5

P(0) = (4/5)^40 = 0.0001329

>b) What is the most likely number of customers that teller #3 is to
>   serve?

The mean number at any teller is

np = 40 x 1/5 = 8

and the most likely number at any teller MUST lie within 1 of the
mean. So we test P(7), P(8), P(9) and find which of these is greatest.

P(7) = C(40,7)(1/5)^7 (4/5)^33 = 0.15125
P(8) = C(40,8)(1/5)^8 (4/5)^32 = 0.15598
P(9) = C(40,9)(1/5)^9 (4/5)^31 = 0.13865

and so P(8) is the greatest probability and we conclude that the most
likely number of customers at ANY position is 8.

>c) What is the probability that every teller has at least one
>   customer?

This is equivalent to distributing 40 numbered balls into 5 numbered
urns such that no urn is empty, and we require the coefficient of
x^40/40! in the expansion of

[e^x - 1]^5 = e^5x - 5e^4x + 10e^3x - 10e^2x + 5e^x - 1

and the coefficient of x^40/40! is

(x^40/40!).[5^40 - 5.(4^40) + 10.(3^40) - 10.(2^40) + 5]

= (x^40/40!).[9.08890251 x 10^27]

9.08890251 x 10^27
and the required probability = -------------------
5^40

=  0.999335

>d) What is the probability that exactly two tellers do not have any
>   customers?

You can choose 2 tellers from 5 in C(5,2) = 10 ways

You then place 40 balls at random into 3 cells so that no cell is
empty. Again we must deal with equiprobable events and the number of
ways is given by the coefficient of x^40/40! in the expansion of

[e^x - 1]^3  = e^3x - 3.e^2x + 3.e^x - 1

The term in x^40/40! is

(x^40/40!).[3^40 - 3(2^40) + 3]

= (x^40/40!).[1.2157662 x 10^19]

10 x 1.2157662 x 10^19
and the required probability =  -----------------------
5^40

=  1.33675 x 10^-8

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

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