Examples

**Birthday Problem**. This is a classic example of how probability in
some applications does not coincide with our intuition. Suppose we have a class
of *n* individuals and want to determine the probability that there is
at least one pair of individuals who have the same birthday. To simplify this
problem, we will ignore birthdays that occurred on Feb. 29 during a leap year
and count those as occurring on March 1. The model we will assume for this
problem treats an individual's birthdate as if it was randomly selected from
the set of 365 possible birthdays. Therefore, the experiment in which each
individual selects a birthdate is an experiment with equally likely outcomes.
Therefore, we must count the number of ways to select *n* birthdays from
the population of 365 possible birthdays, and then count the number of ways to
select *n* birthdays with at least one matching pair. It turns out to
be easier to count the number of ways to select *n* birthdays with no
matches. This is equivalent to counting the number of permutations without
replacement of *k* objects selected from a population of 365 objects.
This number is therefore *365!/(365-n)!*. The total number of possible
birthdates for this group is equivalent to the number of permutations with
replacement of *n* birthdates from the population of 365 possible
birthdates. This gives,

A plot of this probability as a function of

**Binomial coefficients**. Note that the number of combinations without
replacement occurs in the binomial series,

Now consider an experiment in which two gamblers play a series of 10 games, the results of which are independent. That is, the event that the first gambler wins (or loses) on game

Any other outcome with exactly 4 wins would just be a rearrangement of the 10 characters in the string, and so would have the same probability. Therefore, the probability that the first gambler wins exactly 4 games is this probability times the number of such outcomes. We can obtain this number by counting the number of combinations of 4 positions taken from the possible 10 positions for

Using the same arguments, we can see that

We can easily extend this to

Finally, note that these probabilities are terms in a binomial series, and that

**Subpopulation selection**. Suppose a committee consists of 40 males
and 20 females and must select a subcommittee of 5 members. It decides to
make this selection randomly. What is the probability that all 5 members of
the subcommittee will be female? What is the probability that at least 2
member of the subcommittee will be male? First note that an outcome of this
experiment is a set of 5 members selected without replacement from the
committee, and this experiment has equally likely outcomes. To answer the
questions, we will first obtain the probability that exactly *k*
members of the subcommittee will be female for .. Note that if
*k* members are female, then *5-k* members will be male. Hence,
the number of outcomes contained in the event that exactly *k* members
are female can be obtained by counting the number of ways to select a subset of
size *k* from the 20 females and multiplying that times the number of
ways to select a subset of size *5-k* from the 40 males. Since order of
selection does not count, this number is then

The number of outcomes in the sample space is the total number of ways to select a subset of size 5 from the 60 committee members, so the probability that exactly

We can now answer the questions.

So, .

2013-12-17