14 Equally Likely Outcomes, Counting Rules, and Uniform Probability Meaures

14.1 Equally likely outcomes

For a sample space $\Omega$ with finitely many possible outcomes, assuming equally likely outcomes corresponds to a probabiliy measure $\text{P}$ which satisfies

\[ \text{P}(A) = \frac{|A|}{|\Omega|} = \frac{\text{number of outcomes in $A$}}{\text{number of outcomes in $\Omega$}} \qquad{\text{when outcomes are equally likely}} \]

Even if the sample space outcomes are equally likely, the possible values of related random variables are usually not.

Example 14.1

Flip a coin 4 times and record the results in sequence. For example, HTHH indicates T on the second flip and H on the others. Let $X$ be the number of H in the four flips. Assume that the coin is fair and the flips are independent.

Find the probability of the outcome HTHH.
Find the probability of the outcome HHHH.
Make a table of all the possible outcomes; there should be 16. Is it reasonable to assume the outcomes are equally likely? Explain.
Identify the possible values of $X$. Are the values of $X$ equally likely?
Compute and interpret $\text{P}(X = 4)$.
Compute and interpret $\text{P}(X=3)$.
Find the distribution of $X$.
Suppose the coin were biased in favor of landing on H. Would the sample space change? Would the definition of $X$ and its possible values change? Would the 16 outcomes be equally likely? Would the distribution of $X$ change?

Example 14.2

Recall the matching problem with $n=4$: objects labeled 1, 2, 3, 4, are placed at random in spots labeled 1, 2, 3, 4, with spot 1 the correct spot for object 1, etc. Recall the sample space from Example @ref(exm:matching-outcome). Let the random variable $X$ count the number of objects that are put back in the correct spot. (Hint: recall Table @ref(tab:matching-indicator-tab).) Let $\text{P}$ denote the probability measure corresponding to the assumption that the objects are equally likely to be placed in any spot, so that the 24 possible placements are equally.

Find $\text{P}(X=0)$.
Find the distribution of $X$.

14.2 Some counting rules

Computing probabilities in the equally likely case reduces to just counting outcomes.

Example 14.3

I’m serving ice cream to my kids. They can choose to have a bowl or a cone with a single scoop from one of four different flavors.

How many different ways could I serve the ice cream? (For example, peppermint in a cone, birthday cake in a bowl, etc)
Now suppose they can either add rainbow or chocolate sprinkles or not. How many different ways could I serve the ice cream? (For example, peppermint in a cone with chocolate sprinkles, birthday cake in a bowl with rainbow sprinkles, etc.)
Now suppose the kids who requested bowls could choose whether to have whipped cream on top. Is the number of different ways I could serve the ice cream equal to the answer to the previous part multiplied by two?

*Multiplication principle for counting.** Suppose that stage 1 of a process can be completed in any one of $n_1$ ways. Further, suppose that for each way of completing the stage 1, stage 2 can be completed in any one of $n_2$ ways. Then the two-stage process can be completed in any one of $n_1\times n_2$ ways.
This rule extends naturally to a $\ell$-stage process, which can then be completed in any one of $n_1\times n_2\times n_3\times\cdots\times n_\ell$ ways.
In the multiplication principle it is not important whether there is a “first” or “second” stage. What is important is that there are distinct stages, each with its own number of “choices”.

Example 14.4 Suppose the board of directors of a corporation has identified 5 candidates — Ariana, Beyonce, Cardi, Drake, Elvis — for three executive positions: chief executive officer (CEO), chief financial officer (CFO), and chief operating officer (COO). In the interest of fairness, the board assigns 3 of the 5 candidates to the positions completely at random. No individual can hold more than one of the positions.

When calculating probabilities below, consider the sample space of all possible executive teams.

How many executive teams are possible?
What is the probability that Ariana is CEO, Beyonce is CFO, and Cardi is COO?
What is the probability that Ariana is CEO and Beyonce is CFO?
What is the probability that Ariana is CEO?

Number of ordered arrangements. The number of ordered arrangements of $k$ items, selected without replacement from a set of $n$ distinct items is

\[ n(n-1)(n-2)\cdots(n-k+1) = \frac{n!}{(n-k)!} \]

Recall the factorial notation: $m!=m(m-1)(m-2)\cdots (3)(2)(1)$. For example, $5!=5\times4\times3\times2\times1=120$. By definition, 0!=1.

Example 14.5

Your boss is forming a committee of 3 people for a new project team, and 5 people — Ariana, Beyonce, Cardi, Drake, Elvis— have volunteered to be on the committee. In the interest of fairness, 3 of the 5 people will be selected uniformly at random to form the committee.

How is this situation different from the executive team example?
How many possible committees consist of Ariana, Beyonce, Cardi? How many executive teams consisted of Ariana, Beyonce, Cardi?
How many different possible committees of 3 people can be formed from the 5 volunteers?

The following is the relationship between “ordered” and “unordered” counting. { \[\begin{align*} \left(\text{number of \emph{ordered} selections of $k$ from $n$}\right) & = \left(\text{number of \emph{unordered} selections of $k$ from $n$}\right)\\ &\quad \times\left(\text{number of ways of arranging the $k$ items in order}\right). \end{align*}\] }
Number of permutations. The number of ways of arranging $k$ items in order is

\[ k\times (k-1)\times (k-2)\times\cdots\times 3\times 2\times1 = k! \]

Number of combinations. The number of ways to choose $k$ items without replacement from a group of $n$ distinct items where order does not matter, denoted $\binom{n}{k}$, is

\[ \binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} = \frac{n!}{k!(n-k)!} \]

The quantity on the right is just a compact way of representing the quantity in the middle. But since factorials can be very large, it’s best to use the quantity in the middle to compute. In R: choose(n, k). In Python: scipy.special.comb(n, k)
The symbol $\binom{n}{k}$ is by definition equal to the quantity in the middle above. It is read as “$n$ choose $k$” and is referred to as a binomial coefficient.

Example 14.6

Find the probability that the committee consists of Ariana, Beyonce, and Cardi.
Find the probability that Ariana and Beyonce are on the committee.
Find the probability that Ariana is on the committee.

Example 14.7

Continuing Example 14.1.

Use counting rules to find a formula for $\text{P}(X = 3)$.
Use counting rules to find a formula for $\text{P}(X = 2)$.
Use counting rules to find a formula for $\text{P}(X = x)$ for each possible value of $x$.
Now suppose the coin is flipped $n$ times. Continue to assume the coin is fair and the flips are independent. Let $X$ count the number of H in $n$ flips. Use counting rules to find a formula for $\text{P}(X = x)$ for each possible value of $x$.

14.3 Uniform probability measures

The continuous analog of equally likely outcomes is a uniform probability measure.
When the sample space is uncountable, size is measured continuously (length, area, volume) rather that discretely (counting).

\[ \text{P}(A) = \frac{|A|}{|\Omega|} = \frac{\text{size of } A}{\text{size of } \Omega} \qquad \text{if $\text{P}$ is a uniform probability measure} \]

Example 14.8

In the meeting problem, assume that Regina arrives at a time chosen uniformly at random between noon and 1. If we measure measure time in minutes after noon, we can model Regina’s arrival with the sample space $[0, 60]$ and a uniform probability measure.

Find the probability that Regina arrives before 12:15.
Find the probability that Regina arrives after 12:45.
Find the probability that Regina arrives between 12:15 and 12:45.
Find the probability that Regina arrives between 12:15:00 and 12:16:00.
Find the probability that Regina arrives between 12:15:00 and 12:15:01.
Find the probability that Regina arrives at the exact time 12:15:00 (with infinite precision).

For a continuous sample space, the probability of any particular outcome is 0.
For a continuous random variable, the probability it takes any particular value is 0.

Example 14.9

Katniss throws a dart at a circular dartboard with radius 1 foot. Suppose that Katniss’s dart lands at a uniformly random location on the dartboard (and she never misses the dartboard).

Compute the probability that Katniss’s dart lands within 1 inch of the center of the dartboard.
Compute the probability that Katniss’s dart lands more than 1 inch but less than 2 inches away from the center of the dartboard.
Compute the probability that Katniss’s dart lands within 1 inch of the outside edge of the dartboard (but on the dartboard).
Continue the previous parts for the remaining 1 inch increments from the center to the edge of the dartboard.
Let $R$ be the distance (inches) from the location of the dart to the center of the dartboard. Find $\text{P}(R < 1)$.
Find $\text{P}(1 < R < 2)$.
Sketch a plot of the distribution of $R$. Does $R$ follow a Uniform distribution?
Find and interpret the 50th percentile of $R$.
Find and interpret the 25th percentile of $R$.
Find and interpret the 75th percentile of $R$.
Sketch a spinner corresponding to the distribution of $R$.