P - Compound Probability Lesson

Compound Probability

In this lesson you will learn about compound events, composed of independent and dependent events. You will see how to create probability models that show the sample space for compound events and how to calculate arrangements by using factorials. You will also learn how to set up simulations for experiments and determine outcomes from these simulations.

Think about compound sentences. They are composed of at least two complete sentences. Well, a compound event is composed of two or more separate events. In order to find the probability of a compound event, it is important to understand if it is an independent or dependent event.

Independent events are events in which the outcome of one event does not affect the probability of the other.

• Example: Toss a coin twice - The result of one toss has no effect on the result of the other toss.
• Example: Draw a marble from a bag of assorted marbles, replace the marble, and draw another.

Dependent events are events in which the outcome of one event does affect the probability of the other.

• Example: Draw a marble from a bag of assorted marbles, and then draw another marble from the bag without replacing the first marble.
• Example: One student selects a book from the class library, and then another student selects a book from the remaining books.

The probability of a compound event is the ratio of favorable outcomes to total outcomes in the sample space for which the compound events happen. Remember that you can use sample spaces (tree diagrams, tables, and organized lists) to find the probability of compound events. You may also use the rules below:

• Probability of independent events (Multiply the probabilities): P (A and B) = P (A) * P (B)
• Probability of dependent events: (Calculate the probability of the first event. Calculate the probability that the second event would occur if the first event had already occurred.) In other words, multiply the probability of the first by the probability of the second AFTER the first: P(A and B) = P(A) x P(B after A)

Example

Kelly and her friends are playing a board game. To move her game piece, she must roll the same number on two number cubes (1-6). Represent the sample space and find all of the ways that Kelly could roll the same number.

You can create a Tree Diagram to show the sample space:

Notice that there are 36 outcomes in the sample space and that there are 6 outcomes where the number cubes are the same. Kelly could roll the same number 1 out of 6 times.

You can make a table to show the sample space:

1, 1	1, 2	1, 3	1, 4	1, 5	1, 6
2, 1	2, 2	2, 3	2, 4	2, 5	2, 6
3, 1	3, 2	3, 3	3, 4	3, 5	3, 6
4, 1	4, 2	4, 3	4, 4	4, 5	4, 6
5, 1	5, 2	5, 3	5, 4	5, 5	5, 6
6, 1	6, 2	6, 3	6, 4	6, 5	6, 6

The sample space shows there are 36 outcomes and that 6 of those are doubles.

You can also use the formula for probability of independent events to find the probability. Remember that rolling two number cubes represents two independent events.

$P\left(A\right)\cdot P\left(B\right)=P\left(1\right)\cdot P\left(1\right)=\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$. The probability of rolling doubles is $\frac{1}{36}$. There are 6 numbers on each cube so you add the probabilities together to see the probability of getting the same number. $\frac{6}{36}\:or\:\frac{1}{6}$

Example

The letters in the word mathematics are placed in a container.

If two letters are chosen at random, what is the probability that both will be consonants?

Since the first letter drawn is not replaced, the sample space is different for the second letter, so the events are dependent. Find the probability that the first letter selected is a consonant.

The sample space has 11 outcomes. 7 of those are consonants.

$P\left(first\:consonant\right)=\frac{7}{11}$

If the first letter selected was a consonant, which changes the number of letters in the container to 10 and the total number of consonants to 6. Find the probability that the second letter selected is a consonant.

$P\left(second\:consonant\right)=\frac{6}{10}=\frac{3}{5}$

Now use the formula to find the probability of choosing two consonants.

$P\left(first\:consonant\right)\cdot P\left(second\:consonant\right)=\frac{7}{11}\cdot\frac{3}{5}=\frac{21}{55}$

The probability of selecting two letters that are both consonants is $\frac{21}{55}$.

Before using a strategy to find the probability of compound events, determine whether the event is dependent or independent.

Can you tell whether the following examples are independent or dependent?

1. Drawing a green ball from a bucket and then drawing a yellow ball without replacing the first.
   • Solution: dependent
2. Spinning a 4 on a spinner three times in a row.
   • Solution: independent
3. Flipping a coin and getting heads-up, then flipping the same coin and getting tails-up.
   • Solution: independent
4. Selecting a two green marbles from a bag of assorted color marbles.
   • Solution: dependent

Another type of probability that involves more than one event is a combination. A combination is a grouping of objects or events in which the order does not matter. Since order does not matter, the two arrangements represent the same combination. (As in hamburger and fries is the same as fries and hamburger). One way to find possible combinations is to make a table.

Example

Sherrie asks each of her dinner guests to select one choice from each list:

Meat: chicken, steak, fish

Side: potato or rice

Vegetable: green beans or carrots

What is the probability that a guest will request either rice or beans?

Make a table, tree diagram, or list: (You can abbreviate the words)

Cpg	Cpc	Crg	Crc
Spg	Spc	Srg	Src
Fpg	Fpc	Frg	Frc

**C=chicken S=steak F=fish p=potatoes g =green beans r=rice c=carrots

There is a total of 6 meals that have rice and 6 meals that have green beans. Since we are looking for a probability of either (not both), we add the probabilities together and subtract out when both occur, P(A and B). There are a total of 12 outcomes or different combinations.

P(rice) OR P(green beans) = P(rice) + P(green beans) - P(rice and green beans)

$P\left(\frac{1}{2}\right)+P\left(\frac{1}{2}\right)-P\left(\frac{1}{4}\right)=\frac{3}{4}$

The probability that either rice or green beans will be selected is 75%, or likely, if the guests are choosing from each list.

Let's try another combination:

Example

Jason has 4 shirts and x number of pairs of pants in his closet. He has a total of 8 possible outfits if he chooses one shirt and one pair of pants. How many pairs of pants are in Jason's closet?

You know that 4 shirts and 1 pair of pants is equal to 4 outfits, so you can use the Fundamental Counting Principle which you learned to use in an earlier module, to find out how many are needed to make 8 outfits.  

4 * x = 8  

x = 2  

2 pairs of pants

Fundamental Counting Principle:

Use a table to show the sample space. 

Shirt 1, Pant 1	Shirt 2, Pant 1	Shirt 3, Pant 1	Shirt 4 Pant 1
Shirt 1, Pant 2	Shirt 2, Pant 2	Shirt 3, Pant 2	Shirt 4 Pant 2

He has 2 pairs of pants.

Take a look at this video to see a great example of this type of problem and how to represent it in the sample space.

Compound Probability Practice

Another type of probability that involves more than one event is a permutation. A permutation is an arrangement of objects or events in which the order does matter. Since order does matter, two arrangements that represent the same combination (in a different order) are different events. (As in different ways to arrange the letters MATH).

Example

How many ways can we arrange the letters C.A.T. Use a list to find the possible permutations.

CAT	CTA	TCA
ACT	ATC	TAC

There are 6 ways to arrange these letters.

You can use the Fundamental Counting Principle to find the number of permutations. Remember that the Fundamental Counting Principle states that you can find the total number of outcomes by multiplying the number of outcomes for each separate experiment. Permutations are dependent events, so you will change the number before you multiply.

For example, there are 3 choices for the first letter, which leaves 2 choices for the 2^nd letter, and one choice for the third.

3 * 2 * 1 = 6   (multiply the events together) There are 6 ways to arrange these letters.

Permutations can be written as factorials - 3 factorial is 3! (3 * 2 * 1)

4 factorial is 4! (4 * 3 * 2 * 1)

5 factorial is 5! (5 *4 * 3 * 2 * 1)

A factorial of a whole number is the product of all the whole numbers except zero that are less than or equal to the number.   You can use factorials to find the number of permutations. You can use factorials to find the number of permutations.

Example

Marsha has 7 books to arrange on the bookshelf in her room. How many orders are possible for her to line up those 7 books on her shelf?

Number of permutations: 7!

7 * 6 * 5 * 4 * 3 * 2 * 1 = 5,040

There are 5,040 different ways for Marsha to line up 7 books on the shelf.

You can see why making a table or list is not always the best way to find this type of probability.

Watch the following video to see how this can be set up using colors.

Compound Probability Homework

Now that you have spent some time learning strategies for solving problems involving compound probability, you are ready to complete your Probability: Compound Probability Homework. Download your homework by CLICKING HERE. Links to an external site.

Once you have completed your homework, AND MAKE SURE YOU ATTEMPTED AND WORKED THE PROBLEMS OUT ON YOUR OWN, click here to download your homework key. Links to an external site.

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