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1. PROBABILITIES

Classical or a priori probabilities are defined in terms of the possible outcomes of a trial, recognized in advance as equally probable. In the toss of a coin, the probability of getting a head is 1/2: the number of outcomes that give a head, 1, divided, by the total number of possible outcomes, head or tail, 2. In the toss of a die, the probability of getting one dot is 1/6: the number of outcomes that give one dot, 1, divided by the total number of possible outcomes, one through six dots, 6. In general, the probability of event a is

Equation A.1 (A.1)

where a is the number of equally probable outcomes that satisfy criteria a, and n is the total number of equally probable outcomes.

In the examples just given, the outcomes are mutually exclusive; i.e., only one outcome is possible at a time. If events a and b are mutually exclusive, then

Equation A.2 (A.2)

and

Equation A.3 (A.3)

For the toss of a coin, p(head or tail) = p(head) + p(tail) = 1, and p(not head) = 1 - p(head) = 1/2. For the toss of a die, p(1 dot or 2 dots) = p(1 dot) + p(2 dots) = 1/3, and p(not 1 dot) = 1 - p(1 dot) = 5/6.

In these examples, the outcomes also are statistically independent; i.e., the occurrence of one event does not affect that of another. If events a and b are statistically independent, then

Equation A.4 (A.4)

The probability of obtaining heads in each of two tosses of a coin is (1/2)(1/2) = 1/4. The probability of obtaining a single dot in each of two tosses of a die is (1/6)(1/6) = 1/36.

Events are conditional if the probability of one event depends on the occurrence of another. If the probability that b will occur, given that a has occurred, is p(b/a), then the probability that both will occur is

Equation A.5 (A.5)

For example, the probability of drawing two aces from a deck of cards is (4/52)(3/51) = 1/221. If the first card were put back into the deck and the deck reshuffled, then the probability of drawing the second ace would not be conditioned on the drawing of the first, and the probability would be (4/52)(4/52) = 1/169, in accord with Eq. A.4.

Here are some rules of thumb for dealing with compound events, when the events are both mutually exclusive and statistically independent:

Equation A.6 (A.6)

Equation A.7 (A.7)

Equation A.8 (A.8)

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