1103 11.7 Basics of Probability Consider another event G for this experiment, “the result is a 2.” G= 526 K = 52, 4, 66 G´K = 52, 4, 66 P1G2 = 1 6 P1K2 = 3 6 = 1 2 P1G´K2 = 3 6 = 1 2 In this case, P1G2 + P1K2 ≠P1G´K2. As Figure 17 suggests, the difference in the two preceding examples comes from the fact that events H and K cannot occur simultaneously. Such events are mutually exclusive events. In fact, H¨K = ∅, which is true for any two mutually exclusive events. Events G and K, however, can occur simultaneously. Both are satisfied if the result of the roll is a 2, the element in their intersection 1G¨K = 5262. 2 3 4 5 6 1 G H K Figure 17 Probability of Compound Events For any events E and F, the following holds true. P1E or F2 =P1E∪F2 =P1E2 +P1F2 −P1E∩F2 EXAMPLE 4 Finding Probabilities of Compound Events One card is drawn from a standard deck of 52 cards. What is the probability of the following compound events? (a) The card is an ace or a spade. (b) The card is a 3 or a king. SOLUTION (a) The events “drawing an ace” and “drawing a spade” are not mutually exclusive. It is possible to draw the ace of spades, an outcome satisfying both events. P1ace or spade2 = P1ace2 + P1spade2 - P1ace and spade2 Probability of compound events = 4 52 + 13 52 - 1 52 Find and substitute known probabilities. = 16 52 Add and subtract fractions. = 4 13 Write in lowest terms. There are 4 aces, 13 spades, and 1 ace of spades. CAUTION When finding the probability of a union, remember to subtract the probability of the intersection from the sum of the probabilities of the individual events.
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