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Next: Additional Properties of Probability Up: Class Notes Previous: Combinations without replacement


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,

\begin{displaymath}
P(no\ birthdate\ matches) = \frac{365!/(365-n)!}{365^n}.
\end{displaymath}

A plot of this probability as a function of n is given below. Note that when n=23, there is about a 50% probability of no matches in the group, and when n=50, there is about a 3% chance of no matches in the group.

Image birthday

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

\begin{displaymath}
(a+b)^n = \sum_{k=0}^n {n\choose k}a^kb^{n-k}.
\end{displaymath}

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 r is independent of the event that he wins (or loses) any other game. Suppose that the probability that the first gambler wins a particular game is p, and his probability of winning any other game is the same. Find the probability that the first gambler wins exactly 4 games. To solve this problem, first note that an arbitrary outcome of this experiment can be represented by a string of 10 characters, each of which is either W or L, denoting the outcomes of each game. The event that the first gambler wins 4 games consists of all possible strings in which W occurs 4 times and L occurs 6 times. Each such string can be specified by the 4 positions of W in this string. For example, the outcome WWWWLLLLLL could be specified by the positions, 1234 of W. Since the games are independent, the probability of observing this outcome would be

\begin{displaymath}
P(WWWWLLLLLL) = pppp(1-p)(1-p)(1-p)(1-p)(1-p)(1-p) = p^4(1-p)^6.
\end{displaymath}

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 W in the string. Hence,

\begin{displaymath}
P(4\ wins) = {10\choose 4}p^4(1-p)^6.
\end{displaymath}

Using the same arguments, we can see that

\begin{displaymath}
P(k\ wins\ in\ 10\ games) = {10\choose k}p^k(1-p)^{10-k},\ 0\le k\le 10.
\end{displaymath}

We can easily extend this to n games to obtain

\begin{displaymath}
P(k\ wins\ in\ n\ games) = {n\choose k}p^k(1-p)^{n-k},\ 0\le k\le n.
\end{displaymath}

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

\begin{displaymath}
\sum_{k=0}^n P(k\ wins\ in\ n\ games) = \sum_{k=0}^n {n\choose k}p^k(1-p)^{n-k}
= (p + 1-p)^n = 1.
\end{displaymath}

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 $0\le k\le 5$.. 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

\begin{displaymath}
{20\choose k}{40\choose {5-k}}.
\end{displaymath}

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 k are female is,

\begin{displaymath}
P(k) = \frac{ {20\choose k} {40\choose {5-k}} }{ {60\choose 5} }.
\end{displaymath}

We can now answer the questions.

\begin{eqnarray*}
P(5\ females) &=& P(5) = \frac{{20\choose 5}{40\choose 0}}{{60...
...17\cdot 16}{60\cdot 59\cdot 58\cdot 57\cdot 56} \\
&=& 0.0028.
\end{eqnarray*}

\begin{eqnarray*}
P(at\ least\ 2\ males) &=& P(no\ more\ than\ 3\ females) = P(0) + P(1) + P(2) + P(3) \\
&=& 1 - P(4) - P(5).
\end{eqnarray*}

\begin{eqnarray*}
P(4) &=& \frac{{20\choose 4}{40\choose 1}}{{60\choose 5}} \\
...
... 40\cdot 5}{60\cdot 59\cdot 58\cdot 57\cdot 56} \\
&=& 0.0355.
\end{eqnarray*}

So, $P(at\ least\ 2\ males) = 1 - 0.0355- 0.0028 = 1 - 0.0383 = .9617$.


next up previous
Next: Additional Properties of Probability Up: Class Notes Previous: Combinations without replacement
Larry Ammann
2013-12-17