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# Properties of Outcomes Help

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By McGraw-Hill Professional
Updated on Aug 26, 2011

## Properties of Outcomes—Law of Large Numbers

Here are some formulas that describe properties of outcomes in various types of situations. Don't let the symbology intimidate you.

### Law of Large Numbers

Suppose you toss an ''unweighted'' die many times. You get numbers turning up, apparently at random, from the set {1, 2, 3, 4, 5, 6}. What will the average value be? For example, if you toss the die 100 times, total up the numbers on the faces, and then divide by 100, what will you get? Call this number d (for die). It is reasonable to suppose that d will be fairly close to the mean, µ:

d ≈ µ
d ≈ (1 + 2 + 3 + 4 + 5 + 6)/6
= 21/6
= 3.5

It's possible, in fact likely, that if you toss a die 100 times you'll get a value of d that is slightly more or less than 3.5. This is to be expected because of ''reality imperfection.'' But now imagine tossing the die 1000 times, or 100,000 times, or even 100,000,000 times! The ''reality imperfections'' will be smoothed out by the fact that the number of tosses is so huge. The value of d will converge to 3.5. As the number of tosses increases without limit, the value of d will get closer and closer to 3.5, because the opportunity for repeated coincidences biasing the result will get smaller and smaller.

The foregoing scenario is an example of the law of large numbers. In a general, informal way, it can be stated like this: ''As the number of events in an experiment increases, the average value of the outcome approaches the theoretical mean.'' This is one of the most important laws in all of probability theory.

## Independent Outcomes

Two outcomes H1 and H2 are independent if and only if the occurrence of one does not affect the probability that the other will occur. We write it this way:

p(H1 H2) = p(H1) p(H2)

Figure 3-1 illustrates this situation in the form of a Venn diagram. The intersection is shown by the darkly shaded region.

A good example of independent outcomes is the tossing of a penny and a nickel. The face (''heads'' or ''tails'') that turns up on the penny has no effect on the face (''heads'' or ''tails'') that turns up on the nickel. It does not matter whether the two coins are tossed at the same time or at different times. They never interact with each other.

To illustrate how the above formula works in this situation, let p(P) represent the probability that the penny turns up ''heads'' when a penny and a nickel are both tossed once. Clearly, p(P) = 0.5 (1 in 2). Let p(N) represent the probability that the nickel turns up ''heads'' in the same scenario. It's obvious that p(N) = 0.5 (also 1 in 2). The probability that both coins turn up ''heads'' is, as you should be able to guess, 1 in 4, or 0.25. The above formula states it this way, where the intersection symbol can be translated as ''and'':

p(P N) = p(P)p(N)
= 0.5 × 0.5
= 0.25

## Mutually Exclusive Outcomes

Let H1 and H2be two outcomes that are mutually exclusive; that is, they have no elements in common:

H1 H2 =

In this type of situation, the probability of either outcome occurring is equal to the sum of their individual probabilities. Here's how we write it, with the union symbol translated as ''either/or'':

p(H1 H2) = p(H1) + p(H2)

Figure 3-2 shows this as a Venn diagram.

When two outcomes are mutually exclusive, they cannot both occur. A good example is the tossing of a single coin. It's impossible for ''heads'' and ''tails'' to both turn up on a given toss. But the sum of the two probabilities (0.5 for ''heads'' and 0.5 for ''tails'' if the coin is ''balanced'') is equal to the probability (1) that one or the other outcome will take place.

Another example is the result of a properly run, uncomplicated election for a political office between two candidates. Let's call the candidates Mrs. Anderson and Mr. Boyd. If Mrs. Anderson wins, we get outcome A, and if Mr. Boyd wins, we get outcome B. Let's call the respective probabilities of their winning p(A) and p(B). We might argue about the actual values of p(A) and p(B). We might obtain empirical probability figures by conducting a poll prior to the election, and get the idea that pemp(A) = 0.29 and pemp(B) = 0.71. We can be certain, however, of two facts: the two candidates won't both win, but one of them will. The probability that either Mrs. Anderson or Mr. Boyd will win is equal to the sum of p(A) and p(B), whatever these values happen to be, and we can be sure it is equal to 1 (assuming neither of the candidates quits during the election and is replaced by a third, unknown person, and assuming there are no write-ins or other election irregularities). Mathematically:

p(A B) = p(A) + p(B)
= pemp(A) + pemp(B)
= 0:29 + 0:71
= 1

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