MA121: Introduction to Statistics

Definition

The probability of an outcome $e$ in a sample space $S$ is a number $p$ between $o$ and 1 that measures the likelihood that $e$ will occur on a single trial of the corresponding random experiment. The value $p=0$ corresponds to the outcome e being impossible and the value $p=1$ corresponds to the outcome e being certain.

Definition

The probability of an event A is the sum of the probabilities of the individual outcomes of which it is composed. It is denoted $P(A)$.

The following formula expresses the content of the definition of the probability of an event:

If an event $E$ is $E=\left\{e_{1}, e_{2}, \ldots, e_{k}\right\}$, then

$\begin{align*} P(E)=P\left(e_{1}\right)+P\left(e_{2}\right)+\cdots+P\left(e_{k}\right) \end{align*}$

Figure 3.3 "Sample Spaces and Probability" graphically illustrates the definitions.

Figure 3.3 Sample Spaces and Probability

Since the whole sample space $S$ is an event that is certain to occur, the sum of the probabilities of all the outcomes must be the number 1.

In ordinary language probabilities are frequently expressed as percentages. For example, we would say that there is a 70% chance of rain tomorrow, meaning that the probability of rain is 0.70. We will use this practice here, but in all the computational formulas that follow we will use the form 0.70 and not 70%.

EXAMPLE 5

A coin is called "balanced" or "fair" if each side is equally likely to land up. Assign a probability to each outcome in the sample space for the experiment that consists of tossing a single fair coin.

Solution:

With the outcomes labeled $h$ for heads and $t$ for tails, the sample space is the set $S=\{h, t\}$. Since the outcomes have the same probabilities, which must add up to 1, each outcome is assigned probability $1 / 2$.

EXAMPLE 6

A die is called "balanced" or "fair" if each side is equally likely to land on top. Assign a probability to each outcome in the sample space for the experiment that consists of tossing a single fair die. Find the probabilities of the events $E$: "an even number is rolled" and $T$: "a number greater than two is rolled".

Solution:

With outcomes labeled according to the number of dots on the top face of the die, the sample space is the set $S=\{1,2,3,4,5,6\}$. Since there are six equally likely outcomes, which must add up to 1, each is assigned probability $1 / 6$

Since $E=\{2,4,6\}, P(E)=1 / 6+1 / 6+1 / 6=3 / 6=1 / 2$.

Since $T=\{3,4,5,6\}, P(T)=4 / 6=2 / 3$.

EXAMPLE 7

Two fair coins are tossed. Find the probability that the coins match, i.e., either both land heads or both land tails.

Solution:

In Note 3.8 "Example 3 we constructed the sample space $S=\{2 h, 2 t, d\}$ for the situation in which the coins are identical and the sample space $S I=\{h h, h t, t h, t t\}$ for the situation in which the two coins can be told apart.

The theory of probability does not tell us how to assign probabilities to the outcomes, only what to do with them once they are assigned. Specifically, using sample space $S$, matching coins is the event $M=\{2 h, 2 t\}$, which has probability $P(2 h)+P(2 t)$. Using sample space $S$ ', matching coins is the event $M I=\{h h, t t\}$, which has probability $P(h h)+P(t t)$. In the physical world it should make no difference whether the coins are identical or not, and so we would like to assign probabilities to the outcomes so that the numbers $P(M)$ and $P(M \prime)$ are the same and best match what we observe when actual physical experiments are performed with coins that seem to be fair. Actual experience suggests that the outcomes in $S I$ are equally likely, so we assign to each probability $1 / 4$, and then

Similarly, from experience appropriate choices for the outcomes in S are:

which give the same final answer

The previous three examples illustrate how probabilities can be computed simply by counting when the sample space consists of a finite number of equally likely outcomes. In some situations the individual outcomes of any sample space that represents the experiment are unavoidably unequally likely, in which case probabilities cannot be computed merely by counting, but the computational formula given in the definition of the probability of an event must be used.

EXAMPLE 8

The breakdown of the student body in a local high school according to race and ethnicity is 51% white, 27% black, 11% Hispanic, 6% Asian, and 5% for all others. A student is randomly selected from this high school. (To select "randomly" means that every student has the same chance of being selected.) Find the probabilities of the following events:

Solution:

The experiment is the action of randomly selecting a student from the student population of the high school. An obvious sample space is $S=\{w,b,h,a,o\}$. Since 51% of the students are white and all students have the same chance of being selected, $P(w)=0.51$, and similarly for the other outcomes. This information is summarized in the following table:

$\begin{align*} \begin{array}{l|ccccc} \text { Outcome } & w & b & h & a & o \\ \hline \text { Probability } & 0.51 & 0.27 & 0.11 & 0.06 & 0.05 \end{array} \end{align*}$

EXAMPLE 9

The student body in the high school considered in Note 3.18 "Example 8" may be broken down into ten categories as follows: 25% white male, 26% white female, 12% black male, 15% black female, 6% Hispanic male, 5% Hispanic female, 3% Asian male, 3% Asian female, 1% male of other minorities combined, and 4% female of other minorities combined. A student is randomly selected from this high school. Find the probabilities of the following events:

Solution:

Now the sample space is $S=\{wm,bm,hm,am,om,wf,bf,hf,af,of\}$. The information given in the example can be summarized in the following table, called a two-way contingency table:

a. Since $B=\{b m, b f\}, P(B)=P(b m)+P(b f)=0.12+0.15=0.27$

b. Since $M F=\{b f, h f, a f, o f\}$

$\begin{align*} P(M)=P(b f)+P(h f)+P(a f)+P(o f)=0.15+0.05+0.03+0.04=0.27 \end{align*}$

c. Since $F N=\{w f, h f, a f, o f\}$,

$\begin{align*} P(F N)=P(w f)+P(h f)+P(a f)+P(o f)=0.26+0.05+0.03+0.04=0.38 \end{align*}$