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Part 3 Module 5 The Multiplication Rule

Learn how to calculate the probability of the conjunction of two or more events using the general multiplication rule. Examples included.

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Part 3 Module 5 The Multiplication Rule

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  1. Part 3 Module 5The Multiplication Rule Recall the following advice from our study of counting in Part 1 Module 6: Generally: 1. “or” means “add” 2. “and” means “multiply.” We have seen, in Part 3 Modules 3 and 4, that the fact #1 above carries over to probability: P(E or F) = P(E) + P(F) for mutually exclusive events; P(E or F) = P(E) + P(F) – P(E and F) in general. We are about to see that fact #2 above also carries over to probability.

  2. “and” means “multiply” In probability, just as in counting, “and” is associated with multiplication. We have the following fact that can be used to calculate the probability of the conjunction of two or more events: For any events E, F P(E and F) = P(E) × P(F, given E)

  3. The Multiplication Rule This fact is called the general multiplication rule: For any events E, F P(E and F) = P(E) × P(F, given E) In words: to find the probability that two events will both occur, multiply the probability that the first event will occur, times the probability that the second event will occur, assuming that the first event did occur.

  4. At the Forest Folks Gathering there are 4 hobbits (H) 8 gnomes (G) They will randomly select one person serve as Wolverine Groomer and a different person to serve as Cat Washer. 1. What is the probability that the Wolverine Groomer is a hobbit (H1) and the Cat Washer is a gnome (G2)? A. .1667 B. .2424 C. .2222 D. None of these 2. What is the probability that both selectees are gnomes (G1 and G2)? 3. What is the probability that at least one selectee is a hobbit? Example

  5. Example http://www.math.fsu.edu/~wooland/prob/prob22.html History indicates that when Homerina the basketball player goes to the foul line to shoot a pair of free throws, she will make the first shot 87% of the time. If she makes the first shot, then she will make the second shot 78% of the time. On the other hand, if she misses the first shot, then she will make the second shot 46% of the time. What is the probability that Homerina will miss the first shot and miss the second shot?A. 0.0598B. 0.0939C. 0.0909D. 0.0702E. 0.4698

  6. Solution History indicates that when Homerina the basketball player goes to the foul line to shoot a pair of free throws, she will make the first shot 87% of the time. If she makes the first shot, then she will make the second shot 78% of the time. On the other hand, if she misses the first shot, then she will make the second shot 46% of the time. What is the probability that Homerina will miss the first shot and miss the second shot? P(miss both shots) = P(miss 1st shot) × P(miss 2nd shot, given missed 1st shot) The data tells use that 87% of the time she makes the first shot; this means that she will miss the first shot 13% of the time. P(miss 1st shot) =.13 That data tell us that if she misses the first shot, then 46% she will make the second shot, so, if she misses the first shot, 54% of the time she will miss the second shot. P(miss 2nd shot, given missed 1st shot) = .54 So, P(miss both shots) = .13 × .54 = .0702

  7. Follow-up question History indicates that when Homerina the basketball player goes to the foul line to shoot a pair of free throws, she will make the first shot 87% of the time. If she makes the first shot, then she will make the second shot 78% of the time. On the other hand, if she misses the first shot, then she will make the second shot 46% of the time. Use the answer to the previous question to answer this question: What is the probability that she will make at least one of the two shots? Answer: Either (Make both shots) or (Make first and miss second) or (Miss first and make second) .87 × .78 + .87 × .22 + .13 × .46 = .9892 Alternate solution Since “At least one” is the opposite of “none,” P(Make at least one shot) = 1 – P(Miss both shots) = 1 – (.13 × .54) = 1 – .0702 = .9892

  8. Another issue At the Forest Folks Gathering there are 4 hobbits (H) 8 gnomes (G) They will select three persons to ride for random urinalysis. What is the probability that all three selectees are gnomes? Answer: 8/12 × 7/11 × 6/10 = 336/1320 = .255 Notice, however, that there is a mistake in the reasoning (even though it still produces the right answer). Since the three selectees are all being treated the same as one another, order doesn’t matter (there is no first, second or third; all that matters is which three persons are selected). In this case, the number of outcomes to the experiment is C(12, 3) = 220. The number of outcomes favorable to the event is C(8,3) = 56 So the correct calculation is 56/220 = .255 The incorrect analysis at the top of the page yields the correct answer anyway, because the error (acting as if order matters) occurs in both the numerator and the denominator, and so cancels itself. In situations like this, we can assume that order matters (even though it doesn’t) and still get the right answer.

  9. Independent Events Two or more events are independent if they don’t affect or influence one another. More formally, E and F are independent events if the occurrence of E does not affect the probability of F, and the occurrence of F does not affect the probability of E. Example: suppose we have a red die and a green die. The experiment involves rolling each die. Let E be the event that when we roll the red die we get an even number, and let F be the event that when we roll the green die we get a “2.” These are independent events: the probability of one event is not affected by whether or not the other event occurs. Suppose we want to find the probability that both events occur…

  10. The Multiplication Rule for Independent Events If E, F are independent events P(E and F) = P(E) × P(F) In words: to find the probability that two independent events will both occur, multiply the probability that the first event will occur, times the probability that the second event will occur. Referring to the previous question: The probability that the red die roll will produce an even number is 3/6 and the probability that the green die roll will produce a “2” is 1/6, so the probability that both events occur is 3/6 × 1/6 = 3/36 = .0833

  11. Pseudo-independent events A survey a preschoolers indicates that 10% agree with the statement “broccoli tastes good.” If two preschoolers are randomly selected, what is the probability that both of them agree with the statement? We must emphasize that there is not enough information to answer this question exactly. In order to do so, we would have to know the exact size of the population (N) and exactly how many agreed with the statement (A), in which case P(Both agree) = A/N × (A–1)/(N–1) which is also the same as C(A,2)/C(N,2) In a case like this, where an experiment involves selections from a large population of unspecified size, we will approximate the answer by pretending that the selections are independent, even though, strictly speaking, they aren’t. Thus, P(both agree) = P(First person agrees) × P(Second person agrees) = .1 × .1 = .01 Again, this answer (.01) is not the exact answer, but it is the best we can do working only with the percent information, not knowing the size of the population. For large populations, the resulting error is not significant.

  12. Another example Suppose a three-digit number is randomly created using digits from this set: {1, 3, 6, 8, 9} What is the probability that the number has no repeated digits?

  13. “At least one” is the complement of “none” New question: Suppose a three-digit number is randomly created using digits from this set: {1, 3, 6, 8, 9} What is the probability that the number has at least one repeated digit? Solution: This is a much more complicated question to deal with directly, but it is easy to answer if we realize that we can use the answer to the previous question. Since “at least one” is the opposite (complement) of “none”, we have P(“at least one repeated digit”) = 1 – P(“no repeated digits”) = 1 – .48 (.8 was the answer to the previous question) = .52

  14. Yet another example A box contains a $1 bill, a $5 bill, a $10 bill, a $20 bill, a $50 bill and a $100 bill. Three bills are chosen without replace and their monetary sum is determined. What is the probability that the monetary sum will be $151.

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