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| Addition Rule |
If A and B are
events, the probability of obtaining either of them is: P(A or B) = P(A) + P(B) - P(A and B) If the events A and B are mutually exclusive( that is, if both events cannot occur simultaneously), the last term [P(A and B)] will be 0. Thus the addition rule with mutually exclusive events becomes: P(A or B) = P(A) + P(B) Example: Suppose a high school consists of 25% juniors, 15% seniors, and the remaining 60% is students of other grades. The relative frequency of students who are either juniors and seniors is 40%. We can add the relative frequencies of juniors and seniors because no student can be both junior and senior. P(J or S) = 0.25 + 0.15 The above formula can be expanded to consider more than two exclusive events: P(A or B or C or D... or Z) = P(A) + P(B) + P(C) + ... + P(Z) |
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Venn |
Diagrams
that picture the sample space, S, as an
area in the plane and events as areas within S
are called Venn Diagrams.
In the first Venn diagram (left), events A and B are mutually exclusive. In the second Venn diagram (right), events A and B are not mutually exclusive. Example: Suppose that we draw one card from a deck of 52 playing cards. What is the probability that the card will be either a king or a heart? The probability of drawing a king is 4/52; the probability of drawing a heart is 13/52; and the probability of drawing the king of hearts is 1/52. Therefore: P(K or H) = 4/52 + 13/52 - 1/52 = 4/13 |
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Independent |
In the preceding
section, we were concerned with determining the
probability of obtaining one event or another based upon
a single draw. Now we will learn how events A and B both
occur simultaneously. Example: What is the probability that two tails occurs when two coins are tossed? Let A represent the occurrence of a tail on the first coin and B represent the occurrence of a tail on the second coin. In this example, the occurrence of A is not dependent upon the occurrence of B and vice versa. Events A and B are said to be Independent. That is, the outcome of the first toss has no effect on the outcome of the second toss. The probability of the simultaneous occurrences of two independent events is the product of the probabilities of each event: P(A and B) =
P(A) . P(B) Example: Suppose we have two dice. A is the event that 4 shows on the first die, and B is the event that 4 shows on the second die. If both dice are rolled at once, what is the probability that two 4s occur? P(A) = 1/6 |
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| Multiplication
Rule for Independent Events |
P(A and B) =
P(A) . P(B) The above formula can be expanded. If A , B , C , ... , Z are independent events, then: P(A and B and C and ... and Z) = P(A) . P(B) . P(C) ... P(Z) |
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| Dependent Variables |
When events are dependent,
each possible outcome is related to the other. Given two
events A and B, the probability of obtaining both A and B
is the product of the probability of obtaining one of the
events times the conditional probability of obtaining the
other event, given the first event has occurred. P(A and B) = P(A) . P(B|A) This rule says that for both of the two events to occur, the first one must occur ( (P(A) ) and then, given the first event has occurred, the second event occurs ( (P(B|A) ). Example: What is the probability of drawing two aces from a deck of playing cards? Since there are 4 aces in a 52 deck of cards, the probability of drawing one ace is 4/52. Having removed one ace and not replacing it reduces the probabilities of drawing another ace on the second draw. The 51 cards remaining contain 3 aces and therefore the probability of drawing an ace on the second draw is 3/51. Therefore, we can multiply these probabilities and determine the probability of drawing two aces: 4/52 . 3/51 = 1/221 |
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Multiplication
Rule |
The multiplication
rule for dependent events can be extended to several
dependent events: P(A
and B and C) = P(A) . P(B|A) . P(C|A and B) |
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