In this chapter, you will learn to:
Write sample spaces.
Determine whether two events are mutually exclusive.
Use the Addition Rule.
Calculate probabilities using both tree diagrams and combinations.
Do problems involving conditional probability.
Determine whether two events are independent.
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. 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.
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 is
{1,2,3,4,5,6}.
A family has three children. Write a sample space.
The sample space consists of eight possibilities.
The possibility
, 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.
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.
Now that we understand the concept of a sample space, we will define probability.
For a sample space
S, and an outcome
A of
S, the following two properties are satisfied.
If A is an outcome of a sample space, then the probability of A, denoted by P(A), is between 0 and 1, inclusive.
The sum of the probabilities of all the outcomes in 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 Example 13.3, is equally likely.
Since (3, 6) is one such outcome, the probability of obtaining (3, 6) is 1/36.
The example we just considered consisted of only one outcome of the sample space. We are often interested in finding probabilities of several outcomes represented by an event.
An event is a subset of a sample space. If an event consists of only one outcome, it is called a simple event.
If two dice are rolled, find the probability that the sum of the faces of the dice is 7.
Let E represent the event that the sum of the faces of two dice is 7.
Since the possible cases for the sum to be 7 are: (1, 6), (2,5), (3, 4), (4, 3), (5, 2), and (6, 1).
and the probability of the event E,
P(E)=6/36 or 1/6.
A jar contains 3 red, 4 white, and 3 blue marbles. If a marble is chosen at random, what is the probability that the marble is a red marble or a blue marble?
We assume the marbles are r1, r2, r3, w1, w2, w3, w4, b1, b2, b3. Let the event C represent that the marble is red or blue.
The sample space S={r1,r2,r3,w1,w2,w3,w4,b1,b2,b3}
And the event C={r1,r2,r3,b1,b2,b3}
Therefore, the probability of C,
P(C)=6/10 or 3/5.
A jar contains three marbles numbered 1, 2, and 3. If two marbles are drawn, what is the probability that the sum of the numbers is 4?
Since two marbles are drawn, the sample space consists of the following six possibilities.
Let the event F represent that the sum of the numbers is four. Then
Therefore, the probability of F is
P(F)=2/6 or 1/3.
A jar contains three marbles numbered 1, 2, and 3. If two marbles are drawn, what is the probability that the sum of the numbers is at least 4?
The sample space, as in Example 13.7, consists of the following six possibilities.
Let the event A represent that the sum of the numbers is at least four. Then
Therefore, the probability of F is
P(F)=4/6 or 2/3.
In the Section 11.1, we learned to find the union, intersection, and complement of a set. We will now use these set operations to describe events.
The union of two events E and F, E∪F, is the set of outcomes that are in E or in F or in both.
The intersection of two events E and F, E∩F, is the set of outcomes that are in both E and F.
The complement of an event E, denoted by Ec, is the set of outcomes in the sample space S that are not in E. It is worth noting that P(EC)=1−P(E). This follows from the fact that if the sample space has n elements and E has k elements, then Ec has n−k elements. Therefore,
.
Of particular interest to us are the events whose outcomes do not overlap. We call these events mutually exclusive.
Two events E and F are said to be mutually exclusive if they do not intersect. That is, E∩F=∅.
Next we'll determine whether a given pair of events are mutually exclusive.
A card is drawn from a standard deck. Determine whether the pair of events given below is mutually exclusive.