## 7.3: Properties of Probability Distributions

(Based on Section 7.3 in Finite Mathematics and Finite Mathematics and Applied Calculus)

If you look back over the tutorials for estimated probability and theoretical probability, you will notice the following common properties.

• The probability of each outcome in the sample space is a real number between 0 and 1 (inclusive).
• The probabilities of all the outcomes add to 1.
• We can compute the probability of any event E by adding the probabilities of all the outcomes in E.

Since the different kinds of probability share these same properties, we will refer to estimated and theoretical probability collectively as just probability. What they have in common is the idea of a probability distribution. Note that everything we say about probability distributions applies equally well to estimated and theoretical probability.

Sample Space, Probability, and Probability Distribution

An finite sample space is just any old finite set

S = {s1, s2, ... , sn}.

A probability distribution is an assignment of a number P(si) to each outcome si in a sample space S ={s1, s2, ... , sn} such that

 (a) 0 P(si) 1 The probability of each outcome is between 0 and 1 inclusive. (b) P(s1) + P(s2) + ... + P(sn) = 1. The probabilities of all the events add to 1.

P(si) is called the probability of si.

A few more facts and things to note:

• To get the probability of an event (recall that an event is a set of outcomes) all we do is add the probabilities of the individual outcomes.
• If P(E) = 0, we call E an impossible event. The event is always impossible, since something must happen.
• Everything we say about probability from now on applies equally well to estimated and theoretical probability.

Examples

1. All the examples of estimated and theoretical probability distributions we have considered are examples of probability distributions.

2. A new one: Take S = {H, T} and make the assignments

P(H) = .2, P(T) = .8.
Since these numbers are between 0 and 1, and add to 1, they specify a probability distribution.

3. Take S = {1, 2, 3, 4, 5, 6}. The following table gives a probability distribution for S.

 Outcome 1 2 3 4 5 6 Probability 0.3 0.3 0 0.1 0.2 0.1

Using the table, we get

 P({1, 6}) = .3 + .1 = .4 P({2, 3, 4}) = P({1, 5}) =

Q Fill in the missing probabilities, given that P(6) = 3P(1).

 Outcome 1 2 3 4 5 6 Probability 0.125 0.125 0.125 0.125

Q The probability that an even number will come up is .

Probability of Unions, Intersections, and Complements

First, a warmup on unions and mutually exclusive events to get started... (To review unions and intersections of events, go to the tutorial for Section 7.1.)

If E is the event that the price of your stocks will increase by at least 50% and F is the event that the price of your stocks will stay the same, then EF is:

 the event that the price of your stocks will stay between their present value and 50% above their present value the event that the price of your stocks will either stay the same or increase by at least 50% the event that that the price of your stocks will neither stay the same nor increase by at least 50% the impossible event that the price of your stocks will simultaneously stay the same and increase by at least 50%

You select an exercise at random from Finite Mathematics . Which of the following are mutually exclusive pairs of events?

 E: the exericse you select is an even-numbered one F: the exercise you select is numbered betwen 1 and 10 E: the exercise you select is either an application or a communication and reasoning exercise F: the exercise you select is neither an application nor a communication and reasoning exercise E: the exercise you select is an even-numbered one F: the exercise you select is numbered either 11 or 13 E: the exercise you select is either an application or a communication and reasoning exercise F: the exercise you select is an application

We now see how to calculate the probability of a union of two or more events.

Principle

Example
Addition Principle for Mutually Exclusive Events

If E and F are mutually exclusive events, then

P(EF) = P(E) + P(F).

In words: The probability that either E or F occurs is the probability that E occurs, plus the probability that F occurs.

Cast two dice and add the numbers facing up.
E: the sum is 5
F: the sum is even.

Since these are mutually exclusive events, the probability that the sum is either 5 or even is

 P(EF) = P(E) + P(F) = 19 + 12 = 1118

Q How did you get P(E) = 1/9?
A Press here to see how we got that.

If E and F are not mutually exclusive events, then we must use the more general formula

P(EF) = P(E) + P(F) - P(EF).

(For an explanation of where these formulas come from, consult Section 7.3 in Finite Mathematics, or Finite Mathematics and Applied Calculus. )

Cast a single die and note the number facing up. Take

E: the outcome is 1, 2 or 3; P(E) = 1/2
F: the outcome is even; P(E) = 1/2

We will also need:

EF: the outcome is 2

Since these are not mutually exclusive, we use

 P(EF) = P(E) + P(F) - P(EF). = 12 + 12 - 16 = 56

In a survey of voters, 30% of the respondents reported supporting the Republican candidate in the last presidential election, and 35% reported supporting the Democrat candidate. (Voting machines made it impossible to "spoil the ballot" by voting for both candidates.) Based on this data, the experimental probability that a voter will vote for a presidential candidate of either major political party is:

 65% 0.05 .35 .65 0.63

In a survey of voters who were eligible to vote in both of the last two presidential elections, 20% reported voting in both, 25% reported voting in the first, while 35% did not vote at all. What percentage of voters voted in the second presidential election?

 20% 60% 30% 40% 55%

Further Principles of Probability

Here is an expanded table which lists all the principles of probability. You will recognize the first and third from above.

Principle

Example
Addition Principle for Mutually Exclusive Events

If E and F are mutually exclusive events, then

P(EF) = P(E) + P(F)

In words: The probability that either E or F occurs is the probablity that E occurs, plus the probability that F occurs.

Cast two dice and add the numbers facing up.
E: the sum is 5
F: the sum is even.

Since these are mutually exclusive events, the probability that the sum is either 5 or even is

 P(EF) = P(E) + P(F) = 19 + 12 = 1118

Addition Principle for Any Number of Mutually Exclusive Events

If E1, E2, . . . , Es are all mutually exclusive events, then

P(E1E2...Es) = P(E1) + P(E2) + ... + P(Es)

In words: The probability that some Ei occurs is the sum of the probabilities that the individual Ei occur.

Cast two dice and add the numbers facing up.

E1: the sum is 5
E2: the sum is 7.
E3: the sum is even

Since these are mutually exclusive events, the probability that the sum is either 5, 7, or even is

 P(E1E2E3) = P(E1) + P(E2) + P(E3) = 19 + 16 + 12 = 79

If E and F are not mutually exclusive events, then we must use the more general formula

P(EF) = P(E) + P(F)-P(EF)

Cast a single die and note the number facing up. Take

E: the outcome is 1, 2 or 3; P(E) = 1/2
F: the outcome is even; P(E) = 1/2

We will also need:

EF: the outcome is 2; P(EF) = 1/6

Since these are not mutually exclusive, we use

 P(EF) = P(E) + P(F) - P(EF). = 12 + 12 - 16 = 56

Probability of the Whole Sample Space and the Impossible Event

P(S) = 1
P() = 0

In words: The probability that something happens is 1; the probability of the impossible event is 0.

In the experiment immediately above,

S = {1, 2, 3, 4, 5, 6};
P(S) = 1

(the outcome is always a number in the range 1-6.)

Complement

P(E') = 1 - P(E)

In words: The probability of E not happening is 1 minus the probability of E

In the experiment immediately above, take

E: the event that the outcome is 6.
then
 P(the outcome is not 6) = 1-P(E) = 1 - 16 = 56

We now put all of these principles to use.

90% of the Jupiter colonists have been vaccinated against Jovian Influenza X. 3% of this group gets the flu anyway. 10% of the total population of colonists gets this flu. Take

E: the event that a colonist is vaccinated and gets the flu
F: the event that a colonist is unvaccinated and gets the flu.
Then P(E) and P(F) are:
 P(E) = .03;P(F) = .07 P(E) = .42; P(F) = .01 P(E) = .934; P(F) = .031 P(E) = .027;P(F) = .973 P(E) = .027;P(F) = .073

In the above situation, what percentage of population are unvaccinated and still do not get the flu?
[Hint: Consider the three following 3 events:

E: the event that a colonist was vaccinated
F: the event that a colonist was unvaccinated and gets the flu (same as F above)
G: the event that a colonist was unvaccinated and does not get the flu.]
 0.048 0.073 .027 0.1 0.043

Finally, what percent of the colonists has been exposed to the virus either by vaccination or by getting the flu?

 0.01 0.073 .927 0.027 0.973

For more practice, try some of the questions in the chapter quiz (Warning: it covers the whole of Chapter 7) by pressing the button on the sidebar. Then try the exercises in Section 7.3 of Finite Mathematics and Finite Mathematics and Applied Calculus

Last Updated: June, 2006