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Probability and Probability Models
Mathematicians tend to avoid the whole debate, and talk instead about abstract probability, or probability distributions, based purely on the properties of relative frequency. Specific probability distributions can then be used as models in real-life situations, such as flipping a coin or tossing a die, to predict (or model) relative frequency.

Probability and Probability Models
Probability Distribution; Probability A (finite) probability distribution is an assignment of a number P(si), the probability of si, to each outcome of a finite sample space S = {s1, s2, , sn}. The probabilities must satisfy 1. 0 ≤ P(si) ≤ 1 and 2. P(s1) + P(s2) P(sn) = 1.

Probability and Probability Models
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 empty event ∅ is always impossible, since something must happen.

Probability and Probability Models
Quick Examples 1. Let us 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.

Probability and Probability Models
It follows that P({1, 6}) = = .4 P({2, 3}) = = .3 P(3) = 0. {3} is an impossible event.

Probability and 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 think of relative frequency as estimated probability, we can think of modeled probability as theoretical probability.

Probability and Probability Models
Quick Examples 1. Fair Coin Model: 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.

Probability and Probability Models
Figure 4 suggests that the relative frequency of heads approaches .5 as the number of coin tosses gets large, so the fair coin model predicts the relative frequency for a large number of coin tosses quite well. Figure 4

Probability and Probability Models
2. Unfair Coin Model: Take S = {H, T} and 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.

Probability and Probability Models
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. This model predicts for example, that the relative frequency of throwing a 5 approaches 1/6 as the number of times you roll the die gets large.

Probability and Probability Models
4. Roll a pair of fair dice (recall that there are a total of outcomes if the dice are distinguishable). Then an appropriate model of the experiment has with each outcome being assigned a probability of 1/36.

Probability and Probability Models
5. In the experiment in Quick Example 4, take E to be the event that the sum of the numbers that face up is 5, so E = {(1, 4), (2, 3), (3, 2), (4, 1)}. By the properties of probability distributions,

Probability and Probability Models
Notice that, in all of the Quick Examples above except for the unfair coin, all the outcomes are equally likely, and each outcome s has a probability of More generally, in the last Quick Example 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):

Probability and Probability Models
Probability Model 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 Visualizing Probability for Equally Likely Outcomes

Probability and Probability Models
Quick Example 1. Toss a fair coin three times, so S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. The probability that we throw exactly two heads is There are eight equally likely outcomes and E = {HHT, HTH, THH}.

Example 1 – Sales of Hybrid Vehicles
A total of 1.9 million hybrid vehicles had been sold in the United States through October of Of these, 955,000 were Toyota Prii, 205,000 were Honda Civics, 170,000 were Toyota Camrys, 105,000 were Ford Escapes, and the rest were other makes. a. What is the probability that a randomly selected hybrid vehicle sold in the United States was either a Toyota Prius or a Honda Civic? b. What is the probability that a randomly selected hybrid vehicle sold in the United States was not a Toyota Camry?

Example 1(a) – Solution The experiment suggested by the question consists of randomly choosing a hybrid vehicle sold in the United States and determining its make. We are interested in the event E that the hybrid vehicle was either a Toyota Prius or a Honda Civic. So, S = the set of hybrid vehicles sold; n(S) = 1,900,000

Example 1(a) – Solution cont’d E = the set of Toyota Prii and Honda Civics sold; n(E) = 955, ,000 Are the outcomes equally likely in this experiment? Yes, because we are as likely to choose one vehicle as another. Thus, = 1,160,000.

Example 1(b) – Solution cont’d Let the event F consist of those hybrid vehicles sold that were not Toyota Camrys. n(F) = 1,900,000 – 170,000 Hence, = 1,730,000

Probability of Unions, Intersections, and Complements

Probability of Unions, Intersections, and Complements
So far, all we know about computing the probability of an event E is that P(E) is the sum of the probabilities of the individual outcomes in E. Suppose, though, that we do not know the probabilities of the individual outcomes in E but we do know that E = A  B, where we happen to know P(A) and P(B).

Probability of Unions, Intersections, and Complements
How do we compute the probability of A  B? We might be tempted to say that P(A  B) is P(A) + P(B), but let us look at an example using the probability distribution in Quick Example 5: For A let us take the event {2, 4, 5}, and for B let us take {2, 4, 6}. A  B is then the event {2, 4, 5, 6}.

Probability of Unions, Intersections, and Complements
We know that we can find the probabilities P(A), P(B), and P(A  B) by adding the probabilities of all the outcomes in these events, so P(A) = P({2, 4, 5}) = = .6 P(B) = P({2, 4, 6}) = = .5, and P(A  B) = P({2, 4, 5, 6}) = = .7. Our first guess was wrong: P(A  B)  P(A) + P(B).

Probability of Unions, Intersections, and Complements
Notice, however, that the outcomes in A  B are counted twice in computing P(A) + P(B), but only once in computing P(A  B): P(A) + P(B) = P({2, 4, 5}) + P({2, 4, 6}) = ( ) + ( ) = 1.1 Whereas P(A  B) = P({2, 4, 5, 6}) = = .7. A  B = {2, 4} P(A  B) counted twice P(A  B) counted once

Probability of Unions, Intersections, and Complements
Thus, if we take P(A) + P(B) and then subtract the surplus P(A  B), we get P(A  B). In symbols, P(A  B) = P(A) + P(B) – P(A  B) .7 = – .4. (see Figure 6). Figure 6

Probability of Unions, Intersections, and Complements
We call this formula the addition principle. One more thing: Notice that our original guess P(A  B) = P(A) + P(B) would have worked if we had chosen A and B with no outcomes in common; that is, if A  B = ∅. When A  B = ∅, recall that we say that A and B are mutually exclusive.

Probability of Unions, Intersections, and Complements
Addition Principle If A and B are any two events, then P(A  B) = P(A) + P(B) – P(A  B). Visualizing the Addition Principle In the figure, the area of the union is obtained by adding the areas of A and B and then subtracting the overlap (because it is counted twice when we add the areas).

Probability of Unions, Intersections, and Complements
Addition Principle for Mutually Exclusive Events If A  B = ∅, we say that A and B are mutually exclusive, and we have P(A  B) = P(A) + P(B). Visualizing the Addition Principle for Mutually Exclusive Events If A and B do not overlap, then the area of the union is obtained by adding the areas of A and B. Because P(A  B) = 0

Probability of Unions, Intersections, and Complements
This holds true also for more than two events: If A1, A2, , An are mutually exclusive events (that is, the intersection of every pair of them is empty), then P(A1  A2   An) = P(A1) + P(A2) P(An). Addition principle for many mutually exclusive events

Probability of Unions, Intersections, and Complements
Quick Example There is a 10% chance of rain (R) tomorrow, a 20% chance of high winds (W), and a 5% chance of both. The probability of either rain or high winds (or both) is P(R  W) = P(R) + P(W) – P(R  W) = – .05 = .25.

Example 4 – School and Work
A survey conducted by the Bureau of Labor Statistics found that 68% of the high school graduating class of 2010 went on to college the following year, while 42% of the class was working. Furthermore, 92% were either in college or working, or both. a. What percentage went on to college and work at the same time? b. What percentage went on to college but not work?

Example 4 – Solution We can think of the experiment of choosing a member of the high school graduating class of 2010 at random. The sample space is the set of all these graduates. a. We are given information about two events: A: A graduate went on to college; P(A) = .68. B: A graduate went on to work; P(B) = .42. We are also told that P(A  B) = .92. We are asked for the probability that a graduate went on to both college and work, P(A  B).

Example 4 – Solution cont’d To find P(A  B), we take advantage of the fact that the formula P(A  B) = P(A) + P(B) – P(A  B) can be used to calculate any one of the four quantities that appear in it as long as we know the other three. Substituting the quantities we know, we get .92 = – P(A  B)

Example 4 – Solution so P(A  B) = .68 + .42 – .92 = .18.
cont’d so P(A  B) = – .92 = .18. Thus, 18% of the graduates went on to college and work at the same time.

Example 4 – Solution cont’d b. We are asked for the probability of a new event: C: A graduate went on to college but not work. C is the part of A outside of A  B, so C  (A  B) = A, and C and A  B are mutually exclusive. (See Figure 7.) Figure 7

Example 4 – Solution Thus, applying the addition principle, we have
cont’d Thus, applying the addition principle, we have P(C) + P(A  B) = P(A). From part (a), we know that P(A  B) = .18, so P(C) = .68 giving P(C) = .50. In other words, 50% of the graduates went on to college but not work.

Probability of Unions, Intersections, and Complements
More Principles of Probability Distributions The following rules hold for any sample space S and any event A: P(S) = 1 P(∅) = 0 P(A) = 1 – P(A). The probability of something happening is 1. The probability of nothing happening is 0. The probability of A not happening is 1 minus the probability of A.

Probability of Unions, Intersections, and Complements
Note We can also write the third equation as P(A) = 1 – P(A) or P(A) + P(A) = 1.

Probability of Unions, Intersections, and Complements
Visualizing the Rule for Complements Think of A as the portion of S outside of A. Adding the two areas gives the area of all of S, equal to 1. Sample Space S

Probability of Unions, Intersections, and Complements
Quick Example There is a 10% chance of rain (R) tomorrow. Therefore, the probability that it will not rain is P(R ) = 1 – P(R) = 1 – .10 = .90.