7.5 Basic Concepts of Probability

Introducing Probability

Uncertainty is, almost by definition, a nebulous concept. In order to put enough constraints on it that we can mathematically study it, we will focus on uncertainty strictly in the context of experiments. Recall that experiments are processes whose outcomes are unknown; the sample space for the experiment is the collection of all those possible outcomes. When we want to talk about the likelihood of particular outcomes, we sometimes group outcomes together; for example, in the Monopoly example at the beginning of this section, we were interested in the roll of 2 dice that might fall as a 4, 5, or 7. A grouping of outcomes that we’re interested in is called an event. In other words, an event is a subset of the sample space of an experiment; it often consists of the outcomes of interest to the experimenter.

Once we have defined the event that interests us, we can try to assess the likelihood of that event. We do that by assigning a number to each event ($E$) called the probability of that event ($P(E)$). The probability of an event is a number between 0 and 1 (inclusive). If the probability of an event is 0, then the event is impossible. On the other hand, an event with probability 1 is certain to occur. In general, the higher the probability of an event, the more likely it is that the event will occur.

Three Ways to Assign Probabilities

The probabilities of events that are certain or impossible are easy to assign; they’re just 1 or 0, respectively. What do we do about those in-between cases, for events that might or might not occur? There are three methods to assign probabilities that we can choose from. We’ll discuss them here, in order of reliability.

Method 1: Theoretical Probability

The theoretical method gives the most reliable results, but it cannot always be used. If the sample space of an experiment consists of equally likely outcomes, then the theoretical probability of an event is defined to be the ratio of the number of outcomes in the event to the number of outcomes in the sample space.

FORMULA

For an experiment whose sample space $S$ consists of equally likely outcomes, the theoretical probability of the event $E$ is the ratio

$$P(E)=\frac{n(E)}{n(S)}\text{,}$$

where $n(E)$ and $n(S)$ denote the number of outcomes in the event and in the sample space, respectively.

Example 7.17

Computing Theoretical Probabilities

Recall that a standard deck of cards consists of 52 unique cards which are labeled with a rank (the whole numbers from 2 to 10, plus J, Q, K, and A) and a suit ($♣$, $♢$, $♡$, or $♠$). A standard deck is thoroughly shuffled, and you draw one card at random (so every card has an equal chance of being drawn). Find the theoretical probability of each of these events:

1. The card is $10♠$.
2. The card is a $♡$.
3. The card is a king (K).

Solution

There are 52 cards in the deck, so the sample space for each of these experiments has 52 elements. That will be the denominator for each of our probabilities.

1. There is only one $10♠$ in the deck, so this event only has one outcome in it. Thus, $P(10♠)=\frac{1}{52}$.
2. There are 13 $♡\text{s}$ in the deck, so $P(♡)=\frac{13}{52}=\frac{1}{4}$.
3. There are 4 cards of each rank in the deck, so $P(\text{K})=\frac{4}{52}=\frac{1}{13}$.

Your Turn 7.17

You are about to roll a fair (meaning that each face has an equal chance of landing up) 6-sided die, whose faces are labeled with the numbers 1 through 6. Find the theoretical probabilities of each outcome.

1.

You roll a 4.

2.

You roll a number greater than 2.

3.

You roll an odd number.

Checkpoint

It is critical that you make sure that every outcome in a sample space is equally likely before you compute theoretical probabilities!

Example 7.18

Using Tables to Find Theoretical Probabilities

In the Basic Concepts of Probability, we were considering a Monopoly game where, if your sister rolled a sum of 4, 5, or 7 with 2 standard dice, you would win the game. What is the probability of this event? Use tables to determine your answer.

Solution

We should think of this experiment as occurring in two stages: (1) one die roll, then (2) another die roll. Even though these two stages will usually occur simultaneously in practice, since they’re independent, it’s okay to treat them separately.

Step 1: Since we have two independent stages, let’s create a table (Figure 7.27), which is probably the most efficient method for determining the sample space.

Figure 7.27

Now, each of the 36 ordered pairs in the table represent an equally likely outcome.

Step 2: To make our analysis easier, let’s replace each ordered pair with the sum (Figure 7.28).

Figure 7.28

Step 3: Since the event we’re interested in is the one consisting of rolls of 4, 5, or 7. Let’s shade those in (Figure 7.29).

Figure 7.29

Our event contains 13 outcomes, so the probability that your sister rolls a losing number is $\frac{13}{36}$.

Your Turn 7.18

1.

If you roll a pair of 4-sided dice with faces labeled with the numbers 1 through 4, what is the probability of rolling a sum of 6 or 7?

Example 7.19

Using Tree Diagrams to Compute Theoretical Probability

If you flip a fair coin 3 times, what is the probability of each event? Use a tree diagram to determine your answer

1. You flip exactly 2 heads.
2. You flip 2 consecutive heads at some point in the 3 flips.
3. All 3 flips show the same result.

Solution

Let’s build a tree to identify the sample space (Figure 7.30).

Figure 7.30

The sample space is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, which has 8 elements.

1. Flipping exactly 2 heads occurs three times (HHT, HTH, THH), so the probability is $\frac{3}{8}$.
2. Flipping 2 consecutive heads at some point in the experiment happens 3 times: HHH, HHT, THH. So, the probability is $\frac{3}{8}$.
3. There are 2 outcomes that all show the same result: HHH and TTT. So, the probability is $\frac{2}{8}=\frac{1}{4}$.

Your Turn 7.19

You have a modified deck of cards containing only 3$\mathrm{♡<!--\; ♡\; -->}$, 4$\mathrm{♡<!--\; ♡\; -->}$, and 5$\mathrm{♡<!--\; ♡\; -->}$. You draw 2 two cards without replacing them (where order matters). What is the probability of each event?

1.

The first card drawn is 3$\mathrm{♡<!--\; ♡\; -->}$.

2.

The first card drawn has a lower number than the second card.

3.

One of the cards drawn is 4$\mathrm{♡<!--\; ♡\; -->}$.

People in Mathematics

Gerolamo Cardano

The first known text that provided a systematic approach to probabilities was written in 1564 by Gerolamo Cardano (1501–1576). Cardano was a physician whose illegitimate birth closed many doors that would have otherwise been open to someone with a medical degree in 16th-century Italy. As a result, Cardano often turned to gambling to help ends meet. He was a remarkable mathematician, and he used his knowledge to gain an edge when playing at cards or dice. His 1564 work, titled Liber de ludo aleae (which translates as Book on Games of Chance), summarized everything he knew about probability. Of course, if that book fell into the hands of those he played against, his advantage would disappear. That’s why he never allowed it to be published in his lifetime (it was eventually published in 1663). Cardano made other contributions to mathematics; he was the first person to publish the third degree analogue of the Quadratic Formula (though he didn’t discover it himself), and he popularized the use of negative numbers.

New Probabilities from Old: Complements

One of the goals of the rest of this chapter is learning how to break down complicated probability calculations into easier probability calculations. We’ll look at the first of the tools we can use to accomplish this goal in this section; the rest will come later.

Given an event $E$, the complement of $E$ (denoted $E\prime$) is the collection of all of the outcomes that are not in $E$. (This is language that is taken from set theory, which you can learn more about elsewhere in this text.) Since every outcome in the sample space either is or is not in $E$, it follows that $n(E)+n(E\prime )=n(S)$. So, if the outcomes in $S$ are equally likely, we can compute theoretical probabilities $P(E)=\frac{n(E)}{n(S)}$ and $P(E\prime )=\frac{n(E\prime )}{n(S)}$. Then, adding these last two equations, we get $$$$

$$\begin{array}{ccc}\hfill P(E)+P(E\prime )& \hfill =\hfill & \frac{n(E)}{n(S)}+\frac{n(E\prime )}{n(S)}\hfill \\ \hfill & \hfill =\hfill & \frac{n(E)+n(E\prime )}{n(S)}\hfill \\ \hfill & \hfill =\hfill & \frac{n(S)}{n(S)}\hfill \\ \hfill & \hfill =\hfill & 1\hfill \end{array}$$

Thus, if we subtract $P(E\prime )$ from both sides, we can conclude that $P(E)=1-P(E\prime )$. Though we performed this calculation under the assumption that the outcomes in $S$ are all equally likely, the last equation is true in every situation.

How is this helpful? Sometimes it is easier to compute the probability that an event won’t happen than it is to compute the probability that it will. To apply this principle, it’s helpful to review some tricks for dealing with inequalities. If an event is defined in terms of an inequality, the complement will be defined in terms of the opposite inequality: Both the direction and the inclusivity will be reversed, as shown in the table below.