<< Chapter < Page Chapter >> Page >
This chapter covers principles of probability. After completing this chapter students should be able to: write sample spaces; determine whether two events are mutually exclusive; use the addition rule; calculate probabilities using tree diagrams and combinations; solve problems involving conditional probability; determine whether two events are independent.

Chapter overview

In this chapter, you will learn to:

  1. Write sample spaces.
  2. Determine whether two events are mutually exclusive.
  3. Use the Addition Rule.
  4. Calculate probabilities using both tree diagrams and combinations.
  5. Do problems involving conditional probability.
  6. Determine whether two events are independent.

Sample spaces and probability

If two coins are tossed, what is the probability that both coins will fall heads? The problem seems simple enough, but it is not uncommon to hear the incorrect answer 1 / 3 size 12{1/3} {} . A student may incorrectly reason that if two coins are tossed there are three possibilities, one head, two heads, or no heads. Therefore, the probability of two heads is one out of three. The answer is wrong because if we toss two coins there are four possibilities and not three. For clarity, assume that one coin is a penny and the other a nickel. Then we have the following four possibilities.

HH HT TH TT

The possibility HT, for example, indicates a head on the penny and a tail on the nickel, while TH represents a tail on the penny and a head on the nickel.

It is for this reason, we emphasize the need for understanding sample spaces.

An act of flipping coins, rolling dice, drawing cards, or surveying people are referred to as an experiment .

Sample Spaces
A sample space of an experiment is the set of all possible outcomes.

If a die is rolled, write a sample space.

A die has six faces each having an equally likely chance of appearing. Therefore, the set of all possible outcomes S size 12{S} {} is

1,2,3,4,5,6 size 12{ left lbrace 1,2,3,4,5,6 right rbrace } {} .

Got questions? Get instant answers now!
Got questions? Get instant answers now!

A family has three children. Write a sample space.

The sample space consists of eight possibilities.

BBB , BBG , BGB , BGG , GBB , GBG , GGB , GGG size 12{ left lbrace ital "BBB", ital "BBG", ital "BGB", ital "BGG", ital "GBB", ital "GBG", ital "GGB", ital "GGG" right rbrace } {}

The possibility BGB size 12{ ital "BGB"} {} , for example, indicates that the first born is a boy, the second born a girl, and the third a boy.

We illustrate these possibilities with a tree diagram.

The tree diagram illustrates the different possibilities for the gender of three children in a family.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

Two dice are rolled. Write the sample space.

We assume one of the dice is red, and the other green. We have the following 36 possibilities.

Green
Red 1 2 3 4 5 6
1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

The entry (2, 5), for example, indicates that the red die shows a two, and the green a 5.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

Now that we understand the concept of a sample space, we will define probability.

Probability

For a sample space S size 12{S} {} , and an outcome A size 12{A} {} of S size 12{S} {} , the following two properties are satisfied.
  1. If A size 12{A} {} is an outcome of a sample space, then the probability of A size 12{A} {} , denoted by P A size 12{P left (A right )} {} , is between 0 and 1, inclusive.
    0 P A 1 size 12{0<= P left (A right )<= 1} {}
  2. The sum of the probabilities of all the outcomes in S size 12{S} {} equals 1.

If two dice, one red and one green, are rolled, find the probability that the red die shows a 3 and the green shows a six.

Since two dice are rolled, there are 36 possibilities. The probability of each outcome, listed in [link] , is equally likely.

Since (3, 6) is one such outcome, the probability of obtaining (3, 6) is 1 / 36 size 12{1/"36"} {} .

Got questions? Get instant answers now!
Got questions? Get instant answers now!

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Applied finite mathematics' conversation and receive update notifications?

Ask