Section 4.7 Counting – Day 1.

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Presentation transcript:

Section 4.7 Counting – Day 1

Learning Targets In many probability problems, the big obstacle is finding the total number of outcomes, and this section presents several methods for finding such numbers without directly listing and counting the possibilities.

Why Do We Need Counting Methods? When finding a basic probability, what are the two things we need to know? Number of ways event A can occur P(A) = Number of different simple events Sometimes however, it is not practical to construct a list of the outcomes. So finding the total often requires the methods of this section.

Identity Theft A criminal (Mr. Tu) is found using your social security number and claims that all of the digits were randomly generated. What is the probability of getting your social security number when randomly generating digits? Is the criminal’s (Mr. Tu) claim that your number was randomly generated likely to be true?

What We Need … The Fundamental Counting Rule For a sequence of two (or more) events in which the first can occur m ways and the second can occur n ways, the events together can occur m∙n ways. Apply it! What is the probability of Mr. Tu randomly generating your social security number? 9 10 10 10 10 10 10 10 10 10 P(get your SSN) = 1 9∗10 9

Consider the following question given on a history test: Teapot History Facts Consider the following question given on a history test: Arrange the following event in chronological order. Boston Tea Party The invention of the teapot The Teapot Dome Scandal Mr. Tu buying his first teapot The Civil War Assuming you make a random guess, what is your probability of getting the correct answer? 5 4 3 2 1 120 1/120

Additional Information The solution to the previous problem represents a solution that can be generalized by using the following notation and factorial rule. Notation The factorial symbol (!) denotes the product of decreasing positive whole numbers. For example, 4! = 4∙3∙2∙1 = 24. By special definition, 0! =1. Factorial Rule A collection of n different items can be arranged in order n! different ways. (This factorial rule reflects that the first item may be selected n different ways, the second item may be selected n-1 ways, and so on.

Summer Vacation During your upcoming summer vacation, you are planning to visit these six national landmarks: Mr. Tu’s birthplace, childhood home, high school, college, mansion, and Rocky’s palace. You would like to plan the most efficient route so you can spend as much time at these landmarks as possible. How many different route are there? 6 5 4 3 2 1 6! = 720 What if you only have time to visit 4 of these locations? How many possible routes are there now?

When Items Are All Different When Some Items Are Identical To Others IMPORTANT: ORDER MATTERS Permutations Rule When Items Are All Different When Some Items Are Identical To Others There are n different items. We select r of the n items (without replacement). We consider rearrangements of the same items to be different. There are n items, and some are the same. We select all of the n items (without replacement). We consider rearrangements of the same items to be different.

Example Gender Selection Horse Racing 14 couples tried to have baby girls by using the Microsort gender selection method. How many ways can 11 girls and 3 boys be arranged in sequence? The 132nd running of the Kentucky Derby had a field of 20 horses. If a better randomly selects two of those horses to come in 1st and 2nd, what is the probability of winning? 20P2 = P(20, 2) We have n = 14 babies, with n1 = 11 alike (girls) and n2= 3 other alike (boys). The number of permutations is computed as follows: **Calculator: Math  PRB  #4 for !

Two selections are the same because the order does not matter. What Now? How many ways can 3 class representatives be chosen from a group of 12 students? Two selections are the same because the order does not matter. VS.

When Order Doesn’t Matter Combinations When Order Doesn’t Matter There are n different items available. We select r of the n items (without replacement). We consider rearrangements of the same items to be the same. (The combination ABC is the same as CBA).

**Calculator: Math  PRB  #3 Example: The Virginia Win for Life lottery game requires that you select the correct 6 numbers between 1 and 42. Find the number of possible combinations 42C6 = 5,245,786 **Calculator: Math  PRB  #3

Solve it! How many ways can 3 class representatives be chosen from a group of 12 students? 12C3 = C(12, 3) = 220 VS.

Permutations versus Combinations When different orderings of the same items are to be counted separately, we have a permutation problem, but when different orderings are not to be counted separately, we have a combination problem.

Example: A clinical test on humans of a new drug is normally done in three phases. Phase I is conducted with a relatively small number of healthy volunteers. Let’s assume that we want to treat 8 healthy humans with a new drug, and we have 10 suitable volunteers available. a) If the subjects are selected and treated in sequence, so that the trial is discontinued if anyone displays adverse effects, how many different sequential arrangements are possible if 8 people are selected from the 10 that are available? Because order does count, we want the number of permutations of r = 8 people selected from the n = 10 available people. We get: b) If 8 subjects are selected from the 10 that are available, and the 8 selected subjects are all treated at the same time, how many different treatment groups are possible? Because order does not count, we want the number of combinations of r = 8 people selected from the n = 10 available people. We get:

Example: The Florida Lotto game is typical of state lotteries Example: The Florida Lotto game is typical of state lotteries. You must select six different numbers between 1 and 53. You win the jackpot if the same six numbers are drawn in any order. Find the probability of winning the jackpot. Because the order of the selected numbers does not matter, you win if you get the correct combination of six numbers. Because there is only one winning combination, the probability of winning the jackpot is 1 divided by the total number of combinations. With n = 53 numbers available and r = 6 numbers selected, the number of combinations is: With 1 winning combination and 22,957,480 different possible combinations, the probability of winning the jackpot is 1/22,957,480.

More Practice -- Lucky Letters You must select letters from the following: AAABBC Lottery 1: Order the above letters into any arrangement your group desires. To win, your order must exactly match the winning order. (3!2!)/6! You must select letters from the following: ABCD Lottery 2: Select two letters from the group above. To win, the house must pick the same two letters, order does not matter. 1/4C2

Lucky Letters You must select letters from the following: AAABBC Lottery 1: Order the above letters into any arrangement your group desires. To win, your order must exactly match the winning order and color. 2!/6! You must select letters from the following: ABCD Lottery 2: Select two letters from the group above. To win, the house must pick the same two letters, order and color does not matter. 1/ 4C2

PowerPlay Number Swap Lottery 1: You must select 2 different numbers between 1 and 5. You win the jackpot if the same 2 numbers are drawn in any order. 1/5C2 Lottery 2: You must select 3 different numbers between 1 and 9. You win the jackpot if the same 3 numbers are drawn in the same order as you selected. 1/9P3

PowerPlay Number Swap Lottery 1: You must select 4 different numbers between 1 and 5. You win the jackpot if the same 4 numbers are drawn in the same order as you selected. 1/5P4 Lottery 2: You must select 3 different numbers between 1 and 6. You win the jackpot if the same 3 numbers are drawn in any order. 1/6C3

The 132nd running of the Kentucky Derby had a field of 20 horses. A Day At The Races The 132nd running of the Kentucky Derby had a field of 20 horses. Bet 1: In horse racing, a trifecta is a bet that the first three finishers in a race are selected and they are selected in the correct order. You place this bet. 1/20P3 = 1/P(20, 3) = 1/6840 Bet 2: In horse racing, a quinela is a bet that the first two finishers are selected, and they can be selected in any order. You place this bet. 1/20C2 = 1/C(20, 2) = 1/190

Real Life PowerPlay Lottery 1: You must select 6 different numbers between 1 and 53. You win the jackpot if the same six numbers are drawn in any order. 1/53C6 = 1/C(53, 6) = 1/22957480 Lottery 2: You must select 6 different numbers between 1 and 53. You win the jackpot if the same six numbers are drawn in the same order as you selected. 1/53P6 = 1/P(53, 6)  1.6529 x 10-10

Recap In this section we have discussed: The fundamental counting rule. The factorial rule. The permutations rule (when items are all different). The permutations rule (when some items are identical to others). The combinations rule.

Homework Pg. 190-191 #16,18, 19, 25, 26