Chapter 6 Probability

# 6.1 Concepts of Probability

Learning Objectives

By the end of this section, you will be able to:

- Identify a simple event, a compound event, and the sample space of an experiment
- Compute probabilities using the basic probability formula
- Identify complementary events and compute their probabilities
- Compute the odds of an event
- Identify independent events and compute their probabilities
- Compute the probability of an “or” statement

If you roll a die, pick a card from a deck of playing cards, or randomly select a person and observe their hair color, you are executing an experiment or procedure. In probability, we look at the likelihood of different outcomes. We begin with some terminology.

# Events and Outcomes

The result of an experiment is called an **outcome**.

An **event** is any particular outcome or group of outcomes.

A **simple event **is an event that cannot be broken down further.

A simple event consists of exactly one outcome.

The **sample space** is the set of all possible simple events.

Example 1

If we roll a standard 6-sided die, describe the sample space and some simple events.

The sample space is the set of all possible simple events: {1, 2, 3, 4, 5, 6}

Some examples of simple events:

We roll a 1.

We roll a 5.

Some compound events:

We roll a number bigger than 4.

We roll an even number.

# Basic Probability

Given that all outcomes are equally likely, we can compute the probability of an event [latex]E[/latex] using this formula:

[latex]P(E) =[/latex] [latex]\frac{\text { number of outcomes corresponding to event E}}{\text { total number of equally likely outcomes}}[/latex]

Example 2

If we roll a 6-sided die, calculate:

a) [latex]P(\text{rolling a 1})[/latex]

Recall that the sample space is {1, 2, 3, 4, 5, 6}. There is one outcome corresponding to “rolling a 1,” so the probability is [latex]\frac{1}{6}[/latex].

b) [latex]P(\text{rolling a number bigger than 4})[/latex]

There are two outcomes bigger than a 4, so the probability is [latex]\frac{2}{6} = \frac{1}{3}[/latex] .

Probabilities are essentially fractions and can be reduced to lower terms like fractions.

Example 3

Let’s say you have a bag with 20 figs, 14 ripe and 6 not-quite-ripe. If you pick a fig at random, what is the probability that it will be ripe?

There are 20 possible figs that could be picked, so the number of possible outcomes is 20. Of these 20 possible outcomes, 14 are favorable (ripe), so the probability that the fig will be ripe is [latex]\frac{14}{20} = \frac{7}{10}[/latex].

There is one potential complication to this example, however. It must be assumed that the probability of picking any of the figs is the same as the probability of picking any other. This wouldn’t be true if (let us imagine) the ripe figs are smaller than the not-quite-ripe ones. (The not-quite-ripe figs would come to hand more readily when you sampled from the bag.) Let us keep in mind, therefore, that when we assess probabilities in terms of the ratio of favorable to all potential cases, **we rely heavily on the assumption of equal probability for all outcomes**.

Exercise 1

At some random moment, you look at your clock and note the minutes reading.

a) What is the probability that the minutes reading is 15?

b) What is the probability that the minutes reading is 15 or less?

**Solution**

There are 60 possible readings, from 00 to 59.

a) [latex]\frac{1}{60}[/latex]

b) [latex]\frac{16}{60}[/latex] (counting 00 through 15)

## Cards

A standard deck of 52 playing cards consists of four **suits** (hearts, spades, diamonds, and clubs). Spades and clubs are black, while hearts and diamonds are red. Each suit contains 13 cards, each of a different **rank**: an Ace (which in many games functions as both a low card and a high card), cards numbered 2 through 10, a Jack, a Queen, and a King.

Example 4

Compute the probability of randomly drawing one card from a deck and getting an Ace.

There are 52 cards in the deck and 4 Aces, so [latex]P(\text{ace}) =[/latex] [latex]\frac{4}{52} = \frac{1}{13} \approx[/latex] 0.0769 .

We can also think of probabilities as percentages: There is a 7.69% chance that a randomly selected card will be an Ace.

Notice that the smallest possible probability is 0—if there are no outcomes that correspond with the event. The largest possible probability is 1—if all possible outcomes correspond with the event.

# Certain and Impossible Events

An impossible event has a probability of 0.

A certain event has a probability of 1.

The probability of any event must be [latex]0 \leq P(E) \leq 1[/latex].

As you’re working through this chapter, *if you compute a probability and get an answer that is negative or greater than 1, you have made a mistake and should check your work*.

# Complementary Events

Now let us examine the probability that an event does **not** happen. As in the previous section, consider the situation of rolling a six-sided die and first compute the probability of rolling a six: the answer is [latex]P(\text{six}) =[/latex] *[latex]\frac{1}{6}[/latex]*. Now consider the probability that we do *not* roll a six: there are 5 outcomes that are **not** a six, so the answer is [latex]P(\text{not a six}) =[/latex] [latex]\frac{5}{6}[/latex]. Notice that [latex]P(\text{six}) + P(\text{not a six}) =[/latex] [latex]\frac{1}{6} + \frac{5}{6} = \frac{6}{6} = 1[/latex].

This is not a coincidence. Consider a generic situation with [latex]n[/latex] possible outcomes and an event [latex]E[/latex] that corresponds to [latex]m[/latex] of these outcomes. Then the remaining [latex]n-m[/latex] outcomes correspond to [latex]E[/latex] not happening, thus

[latex]P(\text{not E}) =[/latex] [latex]\frac{n-m}{n} = \frac{n}{n} - \frac{m}{n} = 1 - \frac{m}{n}[/latex] [latex]= 1 - P(E)[/latex].

## Complement of an Event

The **complement** of an event is the event “[latex]E[/latex] doesn’t happen.”

The notation [latex]E'[/latex] is used for the complement of event [latex]E[/latex].

We can compute the probability of the complement using [latex]P(E') = 1 - P(E)[/latex] .

Notice also that [latex]P(E) = 1 - P(E')[/latex] .

Example 5

If you pull a random card from a deck of playing cards, what is the probability it is not a heart?

There are 13 hearts in the deck, so [latex]P(\text{heart}) =[/latex] [latex]\frac{13}{52} = \frac{1}{4}[/latex].

The probability of *not* drawing a heart is the complement:

[latex]P(\text{not heart}) = 1 - P(\text{heart}) = 1 -[/latex] [latex]\frac{1}{4} = \frac{3}{4}[/latex] .

Sometimes you will see probabilities expressed as **odds**.

# Odds

Odds of an event are typically expressed in the form [latex]A:B[/latex]*. *It is the ratio of

(Number of outcomes corresponding to event [latex]E[/latex]):(Number of outcomes corresponding to event not [latex]E[/latex])

Example 6

If you pull a random card from a deck of playing cards, what are the odds that it is an Ace?

There are 4 Aces in the deck and 48 cards that are not Aces, so the odds would be:

4:48, or 1:12

Notice how this is different than a probability—with probabilities we use the number of successes out of the total number of possible outcomes, while with odds we use the number of successes compared to the number of failures. As another example, if we flipped a coin, the odds of getting a heads would be 1:1.

# Probability of Two Independent Events

Example 7

Suppose we flipped a coin and rolled a die and wanted to know the probability of getting a head on the coin and a 6 on the die.

We could list all possible outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.

Notice there are [latex]2 \cdot 6 = 12[/latex] total outcomes. Out of these, only 1 is the desired outcome, so the probability is [latex]\frac{1}{12}[/latex] .

The prior example was looking at two independent events.

## Independent Events

Events A and B are **independent events** if the probability of Event B occurring is the same whether or not Event A occurs.

Example 8

Are these events independent?

a) A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second toss is a head.

The probability that a head comes up on the second toss is [latex]\frac{1}{2}[/latex] regardless of whether or not a head came up on the first toss, so these events are independent.

b) The two events (1) “It will rain tomorrow in Shreveport” and (2) “It will rain tomorrow in Bossier City” (a city near Shreveport).

These events are not independent because it is more likely that it will rain in Bossier City on days it rains in Shreveport than on days it does not.

c) You draw a card from a deck, then draw a second card without replacing the first.

The probability of the second card being red depends on whether the first card is red or not, so these events are not independent.

When two events are independent, the probability of both occurring is the product of the probabilities of the individual events.

[latex]P(A \text{ and } B)[/latex] for **independent events:**

If events [latex]A[/latex] and [latex]B[/latex] are independent, then the probability of both [latex]A[/latex] and [latex]B[/latex] occurring is

[latex]P(A \text{ and } B) = P(A) \cdot P(B)[/latex]

where [latex]P(A \text{ and } B)[/latex] is the probability of events [latex]A[/latex] and [latex]B[/latex] both occurring, [latex]P(A)[/latex] is the probability of event [latex]A[/latex] occurring, and [latex]P(B)[/latex] is the probability of event [latex]B[/latex] occurring.

If you look back at the coin and die example from earlier, you can see how the number of outcomes of the first event multiplied by the number of outcomes in the second event equals the total number of possible outcomes in the combined event.

Example 9

In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you randomly reach in and pull out a pair of socks and a tee shirt, what is the probability both are white?

The probability of choosing a white pair of socks is [latex]\frac{6}{10}[/latex].

The probability of choosing a white tee shirt is [latex]\frac{3}{7}[/latex].

The probability of both being white is [latex]\frac{6}{10} \cdot \frac{3}{7} = \frac{18}{70} = \frac{9}{35}[/latex].

Example 10

The manufacturing process for a certain product has a 0.2% defect rate, meaning 2 products out of 1,000 are defective on average. If two items are pulled randomly off the assembly line, what’s the probability both are defective?

The probability of each being defective is independent, so the probability of both defective is [latex]\frac{2}{1000} \cdot \frac{2}{1000} = \frac{4}{1,000,000} = \frac{1}{250,000}[/latex]

Exercise 2

A card is pulled from a deck of cards and noted. The card is then replaced, the deck is shuffled, and a second card is removed and noted. What is the probability that both cards are Aces?

**Solution**

Since the second draw is made after replacing the first card, these events are independent. The probability of an ace on each draw is [latex]\frac{4}{52}=\frac{1}{13}[/latex], so the probability of an Ace on both draws is [latex]\frac{1}{13} \cdot \frac{1}{13} = \frac{1}{169}[/latex].

The previous examples looked at the probability of *both* events occurring. Now we will look at the probability of *either* event occurring.

Example 11

Suppose we flipped a coin and rolled a die and wanted to know the probability of getting a head on the coin *or* a 6 on the die.

Here, there are still 12 possible outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

By simply counting, we can see that 7 of the outcomes have a head on the coin *or* a 6 on the die *or* both—we use *or* inclusively here (these 7 outcomes are H1, H2, H3, H4, H5, H6, T6), so the probability is [latex]\frac{7}{12}[/latex] . How could we have found this from the individual probabilities?

As we would expect, [latex]\frac{1}{2}[/latex] of these outcomes have a head, and [latex]\frac{1}{6}[/latex] of these outcomes have a 6 on the die. If we add these, [latex]\frac{1}{2} + \frac{1}{6} = \frac{6}{12} +\frac{2}{12} = \frac{8}{12}[/latex], which is not the correct probability. Looking at the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head *and* rolling a 6 is [latex]\frac{1}{12}[/latex].

If we subtract out the probability that we double counted, we have the correct probability: [latex]\frac{8}{12} - \frac{1}{12} = \frac{7}{12}[/latex].

[latex]P(A \text{ or } B)[/latex]

The probability of either [latex]A[/latex] or [latex]B[/latex] occurring (or both) is [latex]P(A \text{ or } B) = P(A) + P(B) – P(A \text{ and } B)[/latex].

Example 12

Suppose we draw one card from a standard deck. What is the probability that we get a Queen or a King?

There are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is [latex]\frac{8}{52}[/latex]

Note that in this case, there are no cards that are both a Queen and a King, so P(King and Queen) = 0 . Using our probability rule, we could have said:

[latex]P(\text{King or Queen}) = P(\text{King}) + P(\text{Queen}) - P(\text{King and Queen}) =[/latex][latex]\frac{4}{52} + \frac{4}{52} - 0 = \frac{8}{52}[/latex]

In the last example, the events were **mutually exclusive**, so [latex]P(A \text{ or } B) = P(A) + P(B)[/latex].

Events are mutually exclusive if [latex]P(A \text{ and } B) = 0[/latex].

Example 13

Suppose we draw one card from a standard deck. What is the probability that we get a red card or a King?

Half the cards are red, so [latex]P(\text{red}) =[/latex] [latex]\frac{26}{52}[/latex] .

There are four kings, so [latex]P(\text{King}) =[/latex] [latex]\frac{4}{52}[/latex] .

There are two red kings, so [latex]P(\text{red and King}) =[/latex] [latex]\frac{2}{52}[/latex] .

We can then calculate

[latex]P(\text{red or King}) = P(\text{red}) + P(\text{King}) - P(\text{red and King}) =[/latex][latex]\frac{26}{52} + \frac{4}{52} - \frac{2}{52} = \frac{28}{52}[/latex].

Exercise 3

In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you reach in and randomly grab a pair of socks and a tee shirt, what is the probability at least one is white?

**Solution**

[latex]P(\text{white sock or white tee}) =[/latex] [latex]\frac{6}{10} + \frac{3}{7} - \frac{9}{35} = \frac{27}{35}[/latex]

Example 14

The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year and the color of their car.

[latex]\begin{array} {|c|c|c|c|} \hline \text{}& \text{Speeding} &\text{No speeding} & \text{Total}\\ &\text{ticket}& \text{ticket}\\ \hline \text{Red car} & 15 & 135 & 150\\ \hline \text{Not red car} & 45 & 470 & 515\\ \hline \text{Total} & 60 & 605 & 665\\ \hline \end{array}[/latex]

Find the probability that a randomly chosen person:

a) Has a red car *and* got a speeding ticket.

We can see that 15 people of the 665 surveyed had both a red car and got a speeding ticket, so the probability is [latex]P(\text{ had a red car and got a speeding ticket}) = [/latex][latex]\frac{15}{665} \approx[/latex] 0.0226 .

Notice that having a red car and getting a speeding ticket are not independent events, so the probability of both of them occurring is not simply the product of probabilities of each one occurring.

b) Has a red car *or* got a speeding ticket.

We could answer this question by simply adding up the numbers:

[latex]15 \text{ people with red cars and speeding tickets} \\ + 135 \text{ with red cars but no ticket} \\ + 45 \text{ with a ticket but no red car} \\ = 195 \text{ people}[/latex]

So the probability is [latex]\frac{195}{665} \approx{0.2932}[/latex].

We also could have found this probability by:

[latex]P(\text{had a red car or got a speeding ticket}) = P( \text{had a red car}) + P( \text{got a speeding ticket}) - P( \text{had a red car and got a speeding ticket}) =[/latex][latex]\frac{150}{665} + \frac{60}{665} - \frac{15}{665} = \frac{195}{665}[/latex].

The result of an experiment.

Any particular outcome or group of outcomes of a probability experiment.

An event that can only have a single outcome.

The set of all possible simple events.