7.3: Probability and Probability Models

This tutorial: Part A: Basic Concepts
Next tutorial: Part B: Probability of Unions, Intersections, and Complements

If you look back over the tutorial for relative frequency you will see that relative frequency satisfies certain basic properties. The mathematical notion of "probability" generalizes this idea:
Probability Distribution; Probability

A (finite) probability distribution is an assignment of a number P(s_i), the probability of s_i, to each outcome s_i of a finite sample space S = \{\.s_1, s_2, ..., s_n\.\}. The probabilities must satisfy

    1. 0 ≤ P(s_i) ≤ 1 The probability of each outcome is between 0 and 1 inclusive.
    2. P(s_1) + P(s_2) + ... + P(s_n) = 1 The probabilities of all the outcomes add up to 1.

We find the probability of an event E, written P(E), by adding up the probabilities of the outcomes in E. If P(E) = 0, we call E an impossible event. The event is impossible, since there are no outcomes to add.

Examples:

1. All the examples of relative frequency distributions discussed in the preceding tutorial are examples of probability distributions.

2. Take S = \{H, T\} and make the assignments P(H) = .5, P(T) = .5. Because these numbers are between 0 and 1 and add to 1, they specify a probability distribution.

3. There is no reason that P(H) and P(T) have to be .5: We can choose any nonnegative values for P(H) and P(T), as long as they add up to 1, such as P(H) = .2, P(T) = .8.

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

Outcome123456
Probability .3 .3   0   .1 .2 .1

Using the table, we get

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

Note: Since relative frequency has the above properties, all the properties of probability we will derive apply equally well to relative frequency.

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

Outcome 123 456
Probability  .125    .125    .125    .125  
     

The probability that an even number will come up is .

     

Instead of actually performing an experiment many times to come up with a probability distribution, we can often obtain one by analyzing the experiment and predicting what the relative frequencies should be:
Probability Models

A probability model for a particular experiment is a probability distribution that predicts the relative frequency of each outcome if the experiment is performed a large number of times. Just as we occasionally refer to relative frequency as estimated probability, we occasionally refer to modeled probability as theoretical probability.

Examples:

1. Fair Coin Model (See Example 2 above): Flip a fair coin and observe the side that faces up. Because we expect that heads is as likely to come up as tails, we model this experiment with the probability distribution specified by S = \{H, T\}, P(H) = .5, P(T) = .5.

2. Unfair Coin Model (See Example 3 above): Take S = \{H, T\}, P(H) = .2, P(T) = .8. We can think of this distribution as a model for the experiment of flipping an unfair coin that is four times as likely to land with tails uppermost than heads.

3. Fair Die Model Roll a fair die and observe the uppermost number. Because we expect to roll each specific number one sixth of the time, we model the experiment with the probability distribution specified by S = \{\.1, 2, 3, 4, 5, 6\.\}, P(1) = 1/6, P(2) = 1/6, ..., P(6) = 1/6.

4. Two Dice Model Roll a pair of fair dice (recall that there are a total of 36 outcomes if the dice are distinguishable). Then an appropriate model of the experiment has

with each outcome being assigned a probability of 1/36.

5. Look at the preceding example, and take E to be the event that the sum of the numbers that face up is 5, so .

By the properties of probability distributions, P(E) =

     

In Example 5 above we saw that adding the probabilities of the individual outcomes in an event E amounted to computing the ratio (Number of favorable outcomes)/(Total number of outcomes);

This procedure works when the outcomes are all equally likely. That is, where each outcome is as likely to occur as any other.

Probability Models for Equally Likely Outcomes

In an experiment in which all outcomes are equally likely, we model the experiment by taking the probability of an event E to be

    P(E) =
    Number of favorable outcomes

    Total number of outcomes
    =
    n(E)

    n(S)
    The "favorable outcomes" are the outcomes in E.

Examples

1. Q If I toss a coin three times, what is the (modeled) probability of throwing exactly two heads?
    A To use the formula above we need to know n(S) and n(E) :

    S = \{\. HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \.\}, so n(S) = 8.
    E = Set of outcomes with two heads = \{\. HHT, HTH, THH \.\}, so n(E) = 3.
    So,
    P(E) = \frac{n(E)}{n(S)} = \frac{3}{8}.

2. Toss a coin three times. The probability of rolling at most two heads is

    P(E) =        

3. Q If I roll a pair of fair dice, what is the probability of rolling a double (both dice show the same number)?
    A Here we use the distinguishable dice model (whose outcomes are equally likely) seen above:
    n(S) = 36
    E = \{\.(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\.\}
    So,
    P(E) = \frac{n(E)}{n(S)} = \frac{6}{36} = \frac{1}{6}.

4. Randomly choose a person from a class of 40, in which 6 have red hair. If E is the event that a randomly selected person in the class has red hair, then

    P(E) =        

 

We said above that modeled probability predicts the relative frequency if the experiment is performed a large number of times. Here is a simulation to check this experimentally:
Dice Simulation
In this experiment we throw two dice repeatedly and compute the relative frequency that the sum of the numbers is 5. We saw above that the probability is 1/9 ≈ .1111. Press any of the "Throw Dice" buttons in succession to see the graph of relative frequency versus the number of throws N. As N gets large, the relative frequency should approach the modeled probability of 1/9.
 
N (Sample size) Sum (Red + Green)
fr(E) (Frequency)       P(E) (relative frequency)

The next quis is somewhat similar to Example 1 in Section 7.3 of and : You are dealt a hand of five cards from a standard deck of playing cards. The number of possible hands is 2,598,960. The number of possible hands consisting entirely of red cards (diamonds and hearts) is 65,780, and equal to the number of possible hands consisting entirely of black cards (clubs and spades). The number of possible hands consisting entirely of diamonds is 1287. The probability of being dealt a hand that does not consist entirely of red cards is approximately:

The probability of being dealt a hand of 5 cards of the same color is approximately: If you are dealt a hand of five red cards from a standard deck of playing cards, the probability of having a hand consisting entirely of diamonds is approximately:

You toss a fair coin 3 times. Let

Fill in the following probability distribution table and press "Check."
Event E_0E_1E_2E_3
Probability
     
The probability that either one or two heads will come up is .
   

A Note on Populations and Samples

Q Last year, 450 of my 500 customers purchased one or more of my on-line video clips. Thus, the probability that a randomly selected customer purchased an on-line video clip is 450/500 = .9. Is this estimated probability (that is, relative frequency) or "actual" probability (that is, modeled probability)?
A We are given all the information we need to calculate the modeled probability, since we know the actual proportion of customers who purchased the video clips. On the other hand, if the scenario was

then we would not know the actual proportion, since there may be more customers than were surveyed. Thus, the probability would be estimated (relative frequency) being based on a sample, rather than the entire population.

For more practice, try some of the questions in the chapter review exercises (Warning: it covers the whole of Chapter 7). Also try the exercises dealing with estimated probability in Section 7.3 of or . Otherwise, press "Next Topic" on the left to continue.

Last Updated: May, 2009
Copyright © 2009 Stefan Waner

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