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B–3 Probability and Relative Frequency 683<br />

From the definition of probability, as given by Eq. (B–1), it is seen that all probability<br />

functions have the property<br />

0 … P(A) … 1<br />

(B–2)<br />

where P(A) = 0 if the event A is a null event (never occurs) and P(A) = 1 if the event A is a<br />

sure event (always occurs).<br />

Joint Probability<br />

DEFINITION.<br />

The probability of a joint event, AB, is<br />

P(AB) = lim<br />

n: q an AB<br />

n b<br />

(B–3)<br />

where n AB is the number of times that the event AB occurs in n trials.<br />

In addition, two events, A and B, are said to be mutually exclusive if AB is a null event, which<br />

implies that P(AB) K 0.<br />

Example B–1 EVALUATION OF PROBABILITIES<br />

Let event A denote an auto accident blocking a certain street intersection during a 1-min interval.<br />

Let event B denote that it is raining at the street intersection during a 1-min period. Then, event<br />

E = AB would be a blocked intersection while it is raining, as evaluated in 1-min increments.<br />

Suppose that experimental measurements are tabulated continuously for 1 week, and it is<br />

found that n A = 25, n B = 300, n AB = 20, and there are n = 10,080 1-min intervals in the week of<br />

measurements. (n A = 25 does not mean that there were 25 accidents in a 1-week period, but that<br />

the intersection was blocked for 25 1-min periods because of car accidents; this is similarly<br />

true for n B and n AB .) These results indicate that the probability of having a blocked intersection is<br />

P(A) = 0.0025, the probability of rain is P(B) = 0.03, and the probability of having a blocked<br />

intersection and it is raining is P(AB) = 0.002.<br />

The probability of the union of two events may be evaluated by measuring the compound<br />

event directly, or it may be evaluated from the probabilities of simple events as given<br />

by the following theorem:<br />

THEOREM.<br />

Let E = A + B; then,<br />

P(E) = P(A + B) = P(A) + P(B) - P(AB)<br />

(B–4)<br />

Proof. Let the event A only occur n 1 times out of n trials, the event B only occurs n 2<br />

times out of n trials, and the event AB occur n AB times. Thus,<br />

P(A + B) = lim an 1 + n 2 + n AB<br />

b<br />

n: q n<br />

= lim an 1 + n AB<br />

b + lim<br />

n: q n<br />

an 2 + n AB<br />

b - lim<br />

n: q n<br />

n b<br />

n: q an AB

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