Lesson Objectives
• Learn how to solve a word problem that involves probability

## The Basics of Probability

In this lesson, we want to learn about the basics of probability and how to solve probability word problems. Let's begin the lesson by considering an experiment that has one or more possible outcomes, where each outcome is equally likely to occur. Let's suppose we were to flip a quarter. For this scenario, we can say that there are two equally likely outcomes. Our quarter could land heads up or our quarter could land tails up. The set S of all possible outcomes of a given experiment is called the sample space for the experiment. When we flip our quarter, we can only have an outcome of the quarter landing heads up (h) or landing tails up (t). We can notate this using set notation: $$S=\{h, t\}$$ As an additional example, suppose we repeated this coin toss. In other words, what is the sample space for two coin tosses? Now, we would have four possible outcomes, each equally likely. $$S=\{hh, ht, th, tt\}$$ An event E is any subset of the sample space. We will use the notation P(E) to denote the probability of an event E. When the outcomes in the sample space for an experiment are equally likely, then the probability of the event E occurring can be found using the formula below:

### The Probability of an Event E

$$P(E)=\frac{n(E)}{n(S)}$$ n(E) is the number of outcomes in the event, whereas, n(S) is the number of outcomes in the sample space. Since the number of outcomes in an event must be less than or equal to the number of outcomes in the sample space, the probability of an event must be a number between 0 and 1 inclusive. $$0≤ P(E) ≤ 1$$ 0 tells us the event is impossible or that it can't occur.
1 tells us the event is certain to occur.
Let's look at an example.
Example #1: Solve each word problem.
A nickel is tossed twice. What is the probability of getting a heads each time? $$E=\{hh\}$$ $$S=\{hh, ht, th, tt \}$$ The number of outcomes in the event is just one. The number of outcomes in the sample space is four. $$n(E)=1$$ $$n(S)=4$$ $$P(E)=\frac{n(E)}{n(S)}$$ $$P(E)=\frac{1}{4}$$ The probability of getting two heads on two coin tosses is 1/4 or 25%.

### Independent Events

Independent events are events that are not impacted by previous events. Our previous example of a coin toss is an independent event. Each coin toss does not depend on what happened last time. When we have two or more independent events, we can find probabilities by multiplying. In our last example, we could have shown that the probability of obtaining a result of heads on a given toss is 1/2. In our case, we were looking for the probability of two heads. To calculate this probability, we can multiply the individual probabilities together. $$\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$$

### Probability of Independent Events

If A and B are independent events, the probability that both A and B will occur is: $$P(A \hspace{.2em}\text{and}\hspace{.2em}B)=P(A) \cdot P(B)$$ Let's look at another example.
Example #2: Solve each word problem.
A bag contains 8 red marbles and 8 blue marbles. Another bag contains 8 green marbles and 7 yellow marbles. Mark randomly picks one marble from each bag. What is the probability that one marble is blue and one marble is yellow?
Since the first bag contains 16 total marbles and 8 are blue, the probability of picking 1 blue marble is 8/16 or 1/2. The next bag contains 15 total marbles and 7 are yellow, the probability of picking 1 yellow marble is 7/15. In order to find the probability for both events to occur, we can simply multiply the two probabilities together: $$\frac{1}{2}\cdot \frac{7}{15}=\frac{7}{30}$$

### Dependent Events

In some problems, your scenario will involve dependent events. Now, the events will be impacted by previous events. Let's look at an example.
Example #3: Solve each word problem.
A cooler contains 11 bottles of Gatorade. 7 bottles are lemon-lime flavored, while 4 are orange flavored. Jason randomly grabs a bottle and hands it to Jessica. Then, Jason randomly grabs another bottle for himself. What is the probability that Jason and Jennifer both have lemon-lime flavored Gatorade?
When we begin, the cooler contains 11 bottles of Gatorade, which has 7 bottles that are lemon-lime flavored. This means there is a probability of 7/11 of obtaining a lemon-lime flavored Gatorade on the first grab. After this, we have to adjust. If he grabs a lemon-lime flavored Gatorade, there is 1 less bottle of Gatorade and 1 less bottle of lemon-lime flavored Gatorade. When Jason goes to grab a second Gatorade, there are 6 bottles of lemon-lime flavored Gatorade and 10 total bottles of Gatorade. This means there is a probability of 6/10 or 3/5 of grabbing a lemon-lime flavored Gatorade on the second grab. Now, we can multiply these two probabilities together: $$\frac{7}{11}\cdot \frac{3}{5}=\frac{21}{55}$$ We may also see this expressed as a formula:

### Probability of Dependent Events

$$P(A \hspace{.2em}\text{and}\hspace{.2em}B)=P(A) \cdot P(B|A)$$ Where P(B|A) means the probability of B given that A has occurred.

### Binomial Experiment

A binomial experiment is an experiment that consists of repeated independent trials with either a success or a failure as the outcome in each trial.
p » the probability of success in one trial
1 - p » the probability of failure in one trial
The probability of exactly r successes in n trials is given by: $${n \choose r}p^{r}(1 - p)^{n - r}$$ Let's look at an example.
Example #4: Solve each word problem.
Larry, a basketball player has a 60% chance of making each free throw. What is the probability that Larry will make exactly 3 out of 5 free throws?
Here, p, the probability of a success is .6 as a decimal, r, the number of successes is 3, and n, the number of trials is 5. Let's plug into our formula: $$p=.6$$ $$r=3$$ $$n=5$$ $${n \choose r}p^{r}(1 - p)^{n - r}$$ $${5 \choose 3}.6^{3}(1 - .6)^{5 - 3}$$ $$10 \cdot .216 \cdot .16=.3456$$ There is a 34.56% chance that Larry makes 3 out of 5 free throws.

### Mutually Exclusive Events

Lastly, we will deal with the case where we want to find the probability of event A or event B occurring. If two events A and B are mutually exclusive, they have no outcomes in common. In terms of sets, the intersection of set A and set B is the null set: $$P(A ∩ B)=0$$ In other words, set A and set B are disjoint sets. When this occurs, we can simply add the two probabilities together to find the probability of A or B. $$P(A \hspace{.2em}or \hspace{.2em}B)=P(A) + P(B)$$ Let's look at an example.
Example 5: Solve each word problem. A magazine contains 13 pages. Megan opens to a random page. What is the probability that the page number is 1 or 13?
In this case, these two events are mutually exclusive since she can't open to page 1 and 13 at the same time. One event would be opening to page 1, for which there is a probability of 1/13. The other event would be opening to page 13, for which there is a probability of 1/13. We can find the probability of opening to page 1 or 13 by summing the individual probabilities. $$\frac{1}{13}+ \frac{1}{13}=\frac{2}{13}$$ This tells us there is a 2/13 probability of Megan opening the magazine to either page 1 or 13.
In some cases, the events will not be mutually exclusive. This means it is possible for an event to satisfy both outcomes. When this happens, we need to adjust our formula: $$P(A \hspace{.2em}or \hspace{.2em}B)=P(A) + P(B) - P(A ∩ B)$$ Let's look at an example.
Example 6: Solve each word problem.
Jennifer is on the tennis team and carries a bag full of yellow and green tennis balls to practice. Her bag contains 5 yellow tennis balls that are numbered 1 through 5. Her bag also contains 6 green tennis balls that are numbered 1 through 6. If she randomly picks a tennis ball, what is the probability that it is yellow or has a number less than 5?
Here, we need to consider the fact that she could get a ball that is both yellow and less than 5. This tells us the events are not mutually exclusive. First, let's find the probability of getting a yellow ball, we will call this event A. There are 5 yellow balls in a bag with a total of 11 balls. This means the probability of getting a yellow ball is 5/11. $$P(A)=5/11$$ Next, let's think about the probability of getting a ball with a number that is less than 5. Since there are 11 balls total and 1 yellow that is numbered as a 5 and 2 green (5 and 6) that are numbered at 5 or higher, we can say this event B would have a probability of 8/11. $$P(B)=8/11$$ $$\frac{5}{11}+ \frac{8}{11}=\frac{13}{11}$$ If we add the two probabilities together at this point, we would get the wrong answer. Notice how 5/11 + 8/11 gives us 13/11, which is greater than 1. The problem here is that we have overcounted. We need to subtract out the intersection or the probability of drawing a yellow ball that less than 5. How many yellow balls are less than 5? There are 4 yellow balls that are less than 5, out of a total of 11 balls. Let's now subtract this out. $$\frac{5}{11}+ \frac{8}{11}- \frac{4}{11}=\frac{9}{11}$$ This tells us there is a probability of 9/11 of picking up a ball that is either yellow or has a number less than 5.

#### Skills Check:

Example #1

Solve each word problem.

A bag contains 6 red marbles and 7 blue marbles. You randomly pick a marble and then pick a second marble without returning the marbles to the bag. What is the probability that both marbles are red?

A
$$\frac{5}{26}$$
B
$$\frac{3}{13}$$
C
$$\frac{1}{10}$$
D
$$\frac{5}{9}$$
E
$$\frac{4}{27}$$

Example #2

Solve each word problem.

There are 13 shirts in Olivia's closet, 4 blue shirts, 5 green shirts, and 4 red shirts. She randomly selects one to wear. What is the probability that it is blue or green?

A
$$\frac{9}{17}$$
B
$$\frac{1}{6}$$
C
$$\frac{9}{13}$$
D
$$\frac{2}{3}$$
E
$$\frac{9}{10}$$

Example #3

Solve each word problem.

A test consists of six true/false questions. Beth forgot to study and randomly guesses on every question. What is the probability that Beth answers exactly two questions correctly?

A
$$\frac{2}{3}$$
B
$$\frac{15}{64}$$
C
$$\frac{11}{27}$$
D
$$\frac{9}{61}$$
E
$$\frac{13}{20}$$         