Which of the following refers to the ratio of the number of ways the event can occur to the number of outcomes?

9.1 Fundamental Concepts

The probability of an event occurring is intuitively understood to be the likelihood or chance of it occurring. In the very simplest cases, the probability of a particular event A occurring from an experiment is obtained from the number of ways that A can occur divided by the total number of possible outcomes. For example, the probability of obtaining

from a single roll of a fair1 dice is obtained from the following observations,

the possible outcomes are

the number of ways of obtaining

Therefore, probability of obtaining

Example 9.1

Calculate the probability of the following outcomes from a single roll of a fair dice.

Either

Any one of

Solution

In each case, the total number of possible outcomes is six.

a.

The total number of ways of obtaining

is one. The required probability is then 16.b.

The total number of ways of obtaining either

or

is two. The required probability is then 26 =13.c.

The total number of ways of obtaining any one of

,

,

,

,

, or

is six. The required probability is therefore 66=1.

Example 9.2

Calculate the probability of obtaining the following total scores from simultaneously rolling two fair dice.

1

12

6

7

Solution

We need to determine the set of all possible outcomes. These are listed in Table 9.1 and we see that there are 36.

Table 9.1. Possible outcomes of simultaneously rolling two fair dice

a.

A total of 1 cannot be obtained. That is, the number of ways of obtaining a total of 1 is 0. The required probability is therefore 0.

b.

A total of 12 can be obtained from only

. That is, there is only one way. The required probability is therefore 136.c.

A total of 6 can be obtained from any of

,

,

,

, or

. That is, there are six ways. The required probability is therefore 636=16 .d.

A total of 7 can be obtained from any of

,

,

,

,

or

. That is, there are six ways. The required probability is therefore 636=16 .

Example 9.3

Consider a bag containing a number of balls colored either red, blue, green, or yellow, denoted Ⓡ, Ⓑ, Ⓖ, or Ⓨ, respectively. In particular, there are 2 × Ⓡ, 1 × Ⓑ, 1 × Ⓖ, and 1 × Ⓨ. Calculate the probability that the draw of a single ball will be the following.

Ⓨ

Ⓡ

either Ⓡ or Ⓑ

a black ball

Solution

In each case, the total number of possible outcomes is five. That is, one can draw either Ⓡ, Ⓡ, Ⓑ, Ⓖ, or Ⓨ.

a.

The total number of ways of drawing Ⓨ is one. The required probability is then 15.

b.

The total number of ways of drawing Ⓡ is two. The required probability is then 25.

c.

The total number of ways of drawing either Ⓡ or Ⓑ is three. The required probability is then 35.

d.

The total number of ways of drawing

is zero. The required probability is then 0.

It should be clear from the above examples that the probability of an outcome is 1 if that outcome is certain, and 0 if it that outcome is impossible. For example, in Example 9.1 (c), the probability of obtaining either

We intuitively see that probabilities are real numbers on the interval [0,1]. Probabilities are typically stated as either fractions, as in the previous examples, or decimals.

Now let us now consider multiple experiments. For example, rolling a dice twice and asking about the probability they we obtain

First, let us consider rolling a fair dice. The probability that we obtain

Now consider drawing two balls from the bag described in Example 9.3. There are two variations of this experiment,

1.

draw a ball, note its color and return it to the bag before drawing another,

draw a ball, note its color but do not return it to the bag before drawing another.

Since variation 1 involves replacement of the first ball, the outcome of the second draw has no bearing on the result of the first. The two draws in variation 1 are therefore independent.

Variation 2, however, is different. Since the first ball is not replaced, the outcome of the second draw is dependent on the outcome of the first. For example, if Ⓨ is drawn first from a bad containing ⓇⓇⒼⓎⒷ, it is impossible to draw it for a second time. This is explored in the following example.

Example 9.4

Using the bag of balls in Example 9.3, calculate the probability of obtaining the following outcomes from two draws. State whether the draws are independent or dependent.

Ⓖ is drawn then replaced,

Ⓨ is drawn.

Ⓖ is drawn, but not replaced,

Ⓨ is drawn.

The independence or otherwise of multiple draws is a crucial consideration when calculating the “overall” probability of drawing a particular set of sequential outcomes. We return these ideas in Section 9.4.

Concept Questions

1.

The probability of an event is a value between __ and __, the odds of the event are between __ and __, and the ln(odds) are between __ and __.

In one situation, Poisson regression and logistic regression can substitute for each other. Describe that situation.

Your professor comments, “what appears as an interaction when a profile plot is made for the probabilities may not appear as an interaction when the ln(odds) are plotted.” Use an example with some probabilities you make up to illustrate the professor's meaning.

Neither logistic regression nor Poisson regression produce an estimate of the error variance. Why?

Suppose that in Example 13.4 the number of workers had been expressed in millions, that is, 2.850803 rather than 2,850,803. How would the estimated regression coefficients change?

Concept Questions

1.

The probability of an event is a value between __ and __, the odds of the event are between __ and __, and the ln(odds) are between __ and __.

For each situation below, say whether the choice of a regression-like method is most likely logistic, Poisson, nonlinear with an S curve, or nonlinear with a unimodal curve.

y is the height of children followed from ages 6 to 18 years.

y is whether or not a child receives a measles vaccine by age 6.

y is the number of times a person is hospitalized between the ages of 18 and 50.

y is the concentration in the blood of an antibiotic, followed from time of injection and for many hours thereafter.

y is the intensity of sunlight falling on a solar panel, plotted against time of day.

Your professor comments, “what appears as an interaction when a profile plot is made for the probabilities may not appear as an interaction when the ln(odds) are plotted.” Use an example with some probabilities you make up to illustrate the professor’s meaning.

Neither logistic regression nor Poisson regression produce an estimate of the error variance. Why?

Suppose that in Example 13.4 the number of workers had been expressed in millions, that is, 2.850803 rather than 2,850,803. How would the estimated regression coefficients change?

2.2.1 Definitions and Concepts

For example, the toss of a fair coin (gambling activities are popular examples for studying probability) is an experiment.

In the toss of a coin, a head would be one outcome, a tail the other. In the measles study, one outcome would be “yes,” the other “no.”

In Example 2.2, determining whether an individual has had measles is an experiment. The information on outcomes for this experiment may be obtained in a variety of ways, including the use of health certificates, medical records, a questionnaire, or perhaps a blood test.

In the measles study, an event may be defined as “one member of the couple has had measles.” This event could occur if the husband has and the wife has not had measles, or if the husband has not and the wife has. An event may also be the result of more than one replicate of an experiment. For example, asking the couple may be considered as a combination of two replicates: (1) asking if the wife has had measles and (2) asking if the husband has had measles.

We will represent outcomes and events by capital letters. Letting A be the outcome “an individual of childbearing age has had measles,” then, based on the national health study, we write the probability of A occurring as

P(A)=0.20.

Note that any probability has the property

0≤P(A)≤1.

This is, of course, a result of the definition of probability as a relative frequency.

Note that two individual observations are mutually exclusive. The sum of the probabilities of all the mutually exclusive events in an experiment must be one. This is apparent because the sum of all the relative frequencies in a problem must be one.

Thus, not having had measles is the complement of having had measles. The complement of outcome A is represented by A′. Because A and A′ are mutually exclusive, and because A and A′ are all the events that can occur in any experiment, the probabilities of A and A′ sum to one:

P(A′)=1−P(A).

Thus the probability of an individual not having had measles is

P(no measles)=1−0.2=0.8.

One Event Given Another: Reflecting Current Information

When you revise the probability of an event to reflect information that another event has occurred, the result is the conditional probability of the first event given the other event. (All of the ordinary probabilities you have learned about so far can be called unconditional probabilities, if necessary, to avoid confusion.) Here are some examples of conditional probabilities:

Suppose the home team has a 70% chance of winning the big game. Now introduce new information in terms of the event “the team is ahead at halftime.” Depending on how this event turns out, the probability of winning should be revised. The probability of winning given that we are ahead at halftime would be larger—say, 85%. This 85% is the conditional probability of the event “win” given the event “ahead at halftime.” The probability of winning given that we are behind at halftime would be less than the 70% overall chance of winning—say, 35%. This 35% is the conditional probability of “win” given “behind at halftime.”

Success of a new business project is influenced by many factors. To describe their effects, you could discuss the conditional probability of success given various factors, such as a favorable or unfavorable economic climate and actions by competitors. An expanding economy would increase the chances of success; that is, the conditional probability of success given an expanding economy would be larger than the (unconditional) probability of success.

8.9.2 First Passage Time via the Reflection Principle

The computation of the probability of an event associated with a random walk is essentially the counting of the number of paths that define that event. These probabilities can often be derived from the reflection principle, which states as follows:

Reflection principle: Let k,v>0 . Any path from A=(a,k) to N=(n,v) that touches or crosses the x-axis in between corresponds to a path from A′=(a,−k) to N=(n,v).

Thus, the x-axis can be thought of as a mirror that casts a “shadow path” of the original path by reflecting it on this mirror until it hits the x-axis for the first time. After the first time the original path hits the x-axis at B=(b,0), the shadow path is exactly the same as the original path. In Figure 8.5, the segment A′B is the shadow path of the original segment AB. After B the two segments converge and continue as one path to N. Thus, the number of paths from A to N that touch or cross the x-axis is the same as the number of paths from A′ to N.

Let Na,n(k,v) denote the number of possible paths between the points (a,k) and (n,v). (A guide to understanding this parameter is that the subscript indicates the starting and ending times while the argument indicates the initial and final positions.) Then Na,n(k,v) can be computed as follows. Let a path consist of m steps to the right and l steps to the left. Thus, the total number of steps is m+l= n−a, and the difference between the number of rightward steps and the leftward steps is m−l=v−k. From this we obtain

(8.20)m=1 2{n−a+v−k}

Because Na,n(k,v) can be defined as the number of “successes,” m, in m+l=n−a binomial trials, we have that

(8.21)Na,n (k,v)=(n−am)

where m is as defined in Eq. (8.20).

Consider the event {Yn=x|Y0=0}; that is, the position of the walker after n steps is x, given that he started at the origin. In this case we have that a=0,k=0,v=x. Thus, m=(n −x)/2 and

N0,n(0,x)=(n−am)=(nn+x2)=(nn− x2)

where (n+x)/2 is an integer. Thus, if p is the probability of a step of 1 and q is the probability of a step of −1,

P[Yn=x]=N0,n(0,x)p(n+x)/2q(n−x)/2=(nn+x 2)p(n+x)/2q(n−x)/2

as we derived earlier in the chapter.

According to the reflection principle, the number of paths from A to N that touch or cross the time axis is equal to the number of paths from A′ to N; that is, if we denote the number of paths that touch or cross the time axis in Figure 8.5 by Na,n1(k,v), then

Na,n1(k,v)=Na,n( −k,v)

If we assume that k and v are positive numbers as shown in Figure 8.5, then the number of paths from (a,k) to (n,v) that do not touch or intersect the time as, denoted by Na,n0(k,v), is the complement of the number of paths, Na,n1(k,v), that touch or intersect the time axis. Thus,

Na,n0(k,v)=Na,n (k,v)−Na,n1(k,v)=Na,n(k,v)−Na,n( −k,v)

To apply this principle to the first passage time problem we proceed as follows. Consider a random walk that starts at A=(0,0), crosses or touches a line y>v, and then ends at D=(n,v). This is illustrated in Figure 8.6. From our earlier discussion, the total number of paths between (0,0) and (n,x) is N0,n(0,v).

Consider a reflection on the line Y=y, as shown in Figure 8.6. Assume that the last point at which the path from (0,0) to (n,v) intersects this line is the point B=(k,y). Then the reflection of the path from (0,0) to (n,v) on this line from the point B is shown in dotted line. The terminal point of this line is C=(n,2y−v). According to the reflection principle, the number of paths from A to D that intersect or touch the line Y=y is

N0,n 1(0,v)=N0,n(0,2y−v)

To compute the first passage time, we note that for the walker to be at the point v at time n, he must be either at the point v−1 at time n−1 or at the point v+1 at time v−1. Since his first time of reaching the point v is n, we conclude that he must be at v−1 at time n−1. Thus, the number of paths from A to D that do not touch or cross Y=v before time n−1, N0,n0(0,v) , is

N0,n0(0,v) =N0,n(0,v)−N0,n1(0,v)=N0,n−1(0,v−1)−N0,n−1(0,2v−(v−1))=N0,n−1(0,v−1)−N0,n−1 (0,v+1)=(n−1n+v−22)−(n−1n+v2)=(n−1)!(n+v−22)!(n−v2)!−(n−1)!( n+v2)!(n−v−22)! =(n−1)!(n+v 2−1)!(n−v2)!−( n−1)!(n+v2)!(n−v2−1)!=(n−1)!(n +v2)!(n−v2)!{ n+v2−n−v2}={vn}n!(n+v2)! (n−v2)!={vn}(nn+v2) =vnN0,n(0,v)

where N0,n(0,v) is the total number of paths from A to D. Thus, the probability that the first passage time from A to D occurs in n steps is

p Tv(n)=P[Tv=n]= vn(nn+v2) pn+v2(1−p)n−v2

Because a similar result can be obtained when v<0, we have that

pTv(n)=P[Tv=n]=|v|n(nn+v2)pn+v2(1−p)n−v2

Note that n+v must be an even number. Also, recall that the probability that the walker is at location v after n steps is given by

P[Yn=v]=(nn+v2)pn+v2(1−p)n−v2

Thus, the PMF for the first passage time can be written as follows:

pTv(n)=P[Tv=n]=|v|nP[Yn=v]

Probability and Odds Compared

We know from Chapter 3 that the probability of an event is given by the number of ways the event in question can occur in ratio to the number of ways any event can occur. If r out of n patients have a disease, the chance that one of the n patients chosen randomly has the disease is r/n. The odds that the randomly chosen patient has the disease is the number of ways the event can occur in ratio to the number of ways it does not occur, r/(n−r). For example, the probability of a one-spot on the roll of a die is 1/6; the odds are 1-to-5.

12.3.1 Monte Carlo Simulations

In general, suppose we have the ability to recreate (simulate) the experiment an arbitrary number of times and define a sequence of Bernoulli random variables, Xi, that are defined according to

12.15Xi={1,ifAoccursduringtheithexperiment,0,otherwise.

Hence, Xi is simply an indicator function for the event A. If the experiments are independent, then the probability of the event A, pA, can be estimated according to

This is nothing more than estimating the mean of an IID sequence of random variables. From the development of Chapter 7, we know that this estimator is unbiased and that as n → ∞ the estimate converges (almost everywhere via the strong law of large numbers) to the true probability.

In practice, we do not have the patience to run our simulation an infinite number of times nor do we need to. At some point, the accuracy of our estimate should be “good enough,” but how many trials is enough? Some very concrete answers to this question can be obtained using the theory developed in Chapter 7. If the event A is fairly probable, then it will not take too many trials to get a good estimate of the probability, in which case runtime of the simulation is not really too much of an issue. However, if the event A is rare, then we will need to run many trials to get a good estimate of pA. In the case when n gets large, we want to be sure not to make it any larger than necessary so that our simulation runtimes do not get excessive. Thus, the question of how many trials to run becomes important when simulating rare events.

Assuming n is large, the random variable

12.17Pr(|pˆA−pA|<0.01pA)=0.9 =1−α.

From the results of Chapter 7, Section 7.5, we get

12.18ε0.1=0.01pA =σXnc0.1=pA(1−pA)nc0.1,

where the value of c0.1 is taken from Table 7.1 as c0.1 = 1.64. Solving for n gives us an answer for how long to run the simulation:

12.19n=(100c0.1)2(1−pA)pA=(164)2 (1−pA)pA≈(164)2pA.

Or in general, if we want the estimate,

12.20n=(100cαβ )2(1−pA)pA≈(100cαβ)2pA.

This result is somewhat unfortunate because in order to know how long to run the simulation, we have to know the value of the probability we are trying to estimate in the first place. In practice, we may have a crude idea of what to expect for pA which we could then use to guide us in selecting the number of trials in the simulation. However, we can use this result to give us very specific guidance in how to choose the number of trials to run, even when pA is completely unknown to us. Define the random variable NA to be the number of occurrences of the event A in n trials, that is,

Note that E[NA] = npA. That is, the quantity npA can be interpreted as the expected number of occurrences of the event A in n trials. Multiplying both sides of Equation (12.20) by pA then produces

12.22E[NA]=(100cαβ)2 (1−pA)≈(100cαβ)2.

Hence, one possible procedure to determine how many trials to run is to repeat the experiment for a random number of trials until the event A occurs some fixed number of times as specified by Equation (12.22). Let Mk be the random variable which represents the trial number of the kth occurrence of the event A. Then, one could form an estimate of pA according to

It turns out that this produces a biased estimate; however, a slightly modified form,

12.24pˆA=k−1 Mk−1,

produces an unbiased estimate (see Exercise 7.12).

Example 12.7

Suppose we wish to estimate the probability of an event that we expect to be roughly on the order of p ~ 10−4. Assuming we want 1% accuracy with a 90 % confidence level, the number of trials needed will be

n=1p(100cα β)2=104(100*1.641)2=268,960,000.

Alternatively, we need to repeat the simulation experiment until we observe the event

NA=(100cαβ)2=26,896

times. Assuming we do not have enough time available to repeat our simulation over 1/4 of a billion times, we would have to accept less accuracy. Suppose that due to time limitations we decide that we can only repeat our experiment 1 million times, then we can be sure that with 90% confidence, the estimate will be within the interval (p − ɛ, p + ɛ), if ɛ is chosen according to

ε=p(1−p )ncα≈pncα=1.6410−2103=1.64×10−5=0.164 p.

With 1 million trials we can only be 90% sure that the estimate is within 16.4% of the true value.

The preceding example demonstrates that using the Monte Carlo approach to simulating rare events can be very time consuming in that we may need to repeat our simulation experiments many times to get a high degree of accuracy. If our simulations are complicated, this may put a serious strain on our computational resources. The next subsection presents a novel technique, which when applied intelligently can substantially reduce the number of trials we may need to run.

Conditional Probabilities

If events are not entirely independent, the probability of an event B may depend upon whether or not some other event A is known to have occurred. We may be able to use this idea to simplify various probability computations.

Example 18.1.2 Balls in a Box—Brute Force

Suppose we put 4 white and 2 black balls into a box and then draw out at random two of the 6 balls, one after the other. We want to know the probabilities of the possible color sequences of the balls we draw out: ww (both white), wb (white, then black), bw (black, then white), and bb (both black).

One straightforward approach to this problem would be to count the possibilities in a sample space in which each member is an ordered pair of balls to be removed. Let’s think of the balls as numbered (1–4 white, 5 and 6 black), and place the members of our sample space in a 6× 6 array in which the row denotes the first ball removed and the column denotes the second ball removed. Each element in the array then denotes a possible removal, except for the diagonal elements, which would describe the impossible act of removing the same ball twice. After removing the diagonal elements, each remaining array position corresponds to an equally likely two-ball draw, with different portions of the array corresponding to different color patterns. We have

We now count the two-ball events of each color pattern, finding 12 ww, 8 wb, 8 bw, and 2 bb. There are 30 possibilities in all. Therefore,

(18.5)p(ww)=1230=25,p(wb)=p(bw)=830=415,p(bb) =230=115.▪

Computations of the type illustrated in Example 18.1.2 can become complicated and cumbersome. An alternate approach is to use the notion of conditional probability, defined as the probability of an eventB, given that an event A is known to have occurred. The notation we use1 to denote this probability is p (B∣A). With this notation, the probability of A, then B, is

(18.6)p(AB)=p(A)p(B∣A).

Here P(AB ) indicates that A and B occur in the order A, then B. For example, this formula states that if A occurs with 60% probability, and if, when A is known to have occurred, B then occurs with 50% probability, the event sequence A,B has probability (60%)(50%)=(0.6)(0.5)=0.30, which is 30%.