Basic Concepts of Probability
6.7 The Multiplication Rule
always works; but Formula 6.2 works whenP(AandB) = 0. In these situations the last expression in Formula 6.4 simply falls out of the formula, leaving us with Formula 6.2.
■QuestionSuppose we are about to win a game if we roll either an even number or a number greater than 4. What are our chances of winning?
Solution
Step 1.First determine theP(A). Let us call rolling an even number eventA. Let us further assume that we know that the chance of rolling an even number is .5.
Step 2.Second determine theP(B). Let us call rolling a number greater than 4 eventB. Let us further assume that we know that the chance of rolling a number greater than 4 is .3333.
Step 3.Finally determineP(A and B). This, when translated into our problem, means the chance of rolling a number that is both an even number and a number greater than 4. Let us assume we know that the chance of this occur- ring is .1667. (The number 6 is the only value on a standard die that is both an even numberanda number that is greater than 4. Since all elemental events are equally likely, the chance of rolling a 6 is 1/6 or .1667.)
P A or B =P A +P B −P A and B P A or B = 5 + 3333− 1667 = 6667or2 3
The probability of rolling a die and getting either an even number or a number greater than 4 is .6667. We have a 2/3rd chance of winning on the next roll.■
The addition rule for more than two events when the events are not mutually exclusive is much more complicated. Since this chapter is merely an introduc- tion to probability, this topic will not be covered.
Probabilistic independencebetween events is found when knowledge of the occurrence of one event has no effect on determining the probability of occurrence of a second event. For instance, if we wanted to determine the like- lihood of selecting a card from a deck of cards that is both a spade and a face card, the occurrence of a spade does not change the likelihood that the card will be a face card. If we did not know the card selected is a spade, the probability of a face card (we will include the ace for the sake of argument) is 16/52 or 4/13. If, however, we know the card selected is a spade, the probability that the card is also a face card has not changed–it is still 4/13. In this case, we can say that eventA (spade) and eventB(face card) are independent of each other; the occurrence of one event did not change the likelihood of the other event occurring. (Even though we presented this relationship from the perspective of selecting a face card, the same relationship can be shown from the perspective of selecting a spade. Namely, there is a 1 in 4 chance of selecting a spade knowing nothing else.
If we learn that we have a face card, the ratio stays the same–we still have a 1 in 4 chance of selecting a spade. If eventAis independent of eventB, then eventBis independent of event A – this is a shared property.) The formula for the multiplication rule for independent events follows.
Multiplication rule for two independent events P(A and B) =P(A)P(B) (Formula 6.5)
Formula 6.5 is read as“The probability of eventsAandB co-occurring is equal to the probability of event A occurring multiplied by the probability of event Boccurring.”Be sure to note that this formula is true only if the two events are inde- pendent. We will develop a more general rule for multiplication in the next section.
Let us take a look at an example. If we roll a die and flip a coin, certainly the outcome of each is independent of the outcome of the other. So, what is the probability of getting both a heads on the coin flipanda 5 on the roll of the die? Well, theP(heads) = 1/2 andP(5) = 1/6. So, using our rule,P(heads and 5) =P(heads)P(5) = (1/2)(1/6) = 1/12. If we think about this, we can see that it is correct. The elementary events in this example are the combined events of the two actions–one from the coin flip and one from the roll of the die. There are 12 of them (e.g.head and 1,head and 2…,tail and 1,tail and 2…). Further- more, all of the elementary events are equally likely, so the probability of any one of them (e.g.head and 5) is 1/12, just as we found using the rule.
Let us look at another example. Suppose we know that the probability of Lisa earning an“A”in a collegiate course is .9, and we know that the probability of Jason earning an “A” in the course is .4. Suppose we also know that these students have no interaction at all; in other words, their grades should be independent. So, the probability that both will get an A is (.9)(.4) = .36. As we would expect, it must be less than the probability of either one of them getting an“A.”
■QuestionWhat is the probability of randomly selecting a 4 and an 8 on two successive draws from a deck of cards? Since sampling with replacement is used, one card is randomly drawn from the deck and then put back into the deck, and then a second card is randomly selected.
Solution
Step 1.First determine theP(A). Let us call drawing a 4 eventA. Since all cards are equally likely to be selected, we can use Formula 6.1.
P A = number of favorable events total number of events = 4
52= 0769
The probability of randomly selecting a 4 is .0769 or about 7.7%.
Step 2.DetermineP(B). Let us call drawing an 8 eventB. Note thatP(B) is unaf- fected by step 1. This is because the card drawn for step 1 has been replaced (the“sampling with replacement”procedure); therefore, the second draw is from a complete deck of cards.
P B = number of favorable events total number of events = 4
52= 0769
The probability of randomly selecting a 8 is .0769 or about 7.7%.
Step 3.The multiplication rule is now applied.
P A and B =P A P B
P A and B = 0769 0769 = 0059
The probability of drawing a 4 replacing the card and then drawing an 8 is .0059. Stated differently, only 59 times out of 10 000 would these two events be expected to occur in this sequence.■
Formula 6.5 can be extended to any number of independent events. For example, if we wanted to know the probability of drawing an ace, a spade, and the 6 of clubs on three successive draws, we would merely multiply the three probabilities of occurrence.
Probabilistic dependencebetween events is found when knowledge of the occurrence of one event changes the determination of the probability of occurrence for the second event. For instance, if we wanted to determine the likelihood of selecting a person from a population that is both a biological female and under 6 ft tall, the occurrence of a biological female would change the like- lihood that the person will also be under 6 ft tall. If we stipulate that about 8% of people are 6 ft tall or taller, we would have about a 92% chance of randomly selecting a person who was less than 6 ft tall. But if we stipulate that the person we have selected is a biological female, the likelihood of having a person who is under 6 ft tall has just jumped to about 99%. In this case, the occurrence of one event (female) changed the likelihood of the other event occurring. Yes, this is a
6.7 The Multiplication Rule 177
bit confusing because both events are technically occurring at the same time (at the point of selecting the individual), but the way to calculate the likelihood of this co-occurrence necessarily involves exploring the relationship between the two events. To determine the probability of two dependent events, we need to use the more general multiplication rule. (Why we use the more general mul- tiplication rule for dependent events will be clarified at the end of the next section.)
Multiplication rule for two events P(A and B) =P(A|B)P(B) (Formula 6.6)
The symbolP(A|B) is read as“The probability of eventAoccurring given the occurrence of eventB”. [It does not mean thatP(A) is to be divided byP(B).] The symbolP(A|B) is also referred to as a conditional probability, a concept that will be further explored in the next section of this chapter. A worked problem clari- fies the use of Formula 6.6.
■QuestionWhat is the probability of randomly selecting a person from the campus student population that is both a psychology major and a biological female?
GivenSuppose it is known that 10% of the student population are psychology majors, 80% of the psychology majors are biological females, and 60% of the entire student population are biological females.
Solution
DetermineP(A),P(B), andP(A|B). Let us call being a biological femaleP(A).
Let us call being a psychology majorP(B). This would mean thatP(A|B) would mean the probability of being a biological female given that one is a psychology major. (This is the only way to assign the events for this problem–we do not have enough data to findP(A|B) if eventsAandBare switched.) Since we are given all of the needed values, there is no need for preliminary steps.
If being a psychology major and being a biological female were independent of each other (where the occurrence of one of these events did not change the rate of occurrence for the other), then we could use Formula 6.5 and determine that
P A and B =P A P B P A and B = 6 1 = 06
However, these events are not independent and so we must use Formula 6.6.
In this case
P A and B =P A B P B P A and B = 8 1 = 08
The probability of randomly selecting a student on campus who is both a psychology major and a biological female is 8%. Since there is a dependency between eventsAandB, we shouldnotuse Formula 6.5, and it would notbe true that the probability of randomly selecting a student on campus who is both a psychology major and a biological female is 6%.■
Formula 6.6 allows us to incorporate into the calculations the realization that most psychology majors are biological females. And so, of the popu- lation of students, the likelihood of event A and event B co-occurring in the same person is increased a bit by the fact that events A and Balready seem to co-occur a bit more than one might expect if the two events were randomly represented in the population (or said in other words, if the two events were independent of each other). Of course in other situations, events A and B may be dependent, but in such a way that the occurrence of A decreasesthe likelihood of eventBoccurring. The key term regarding whether events are independent of each other or not is whether the like- lihood of one event changes (increases or decreases) if the other event is known to have occurred.