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Section 11.5 Probability with the Fundamental Counting Principle, Permutations, and Combinations 1. 2. 10/2/2011 Example 1 - Probability with Permutations. Objectives Compute probabilities with permutations. Compute probabilities with combinations. 4 students, Corey (C), Matt (M), Shanity (S), and Hope (H), have volunteered to work problems at the board. The order will be decided by drawing names out of a hat. What is the probability that a. They go in alphabetical order? b. C is first and H is last? Since order matters, we count permutations. 1 10/2/2011 Section 11.5 2 Section 11.5 Solution Solution (continued) Recall the theoretical probability of any event, E, is a) Since there is only one way to arrange the students in alphabetical order (n(E) = 1) P (alphabetical order) = 1 24 P(E) = number of outcomes in event E = total number of possible outcomes n( E ) n( S ) b) Use the fundamental counting principle to find the number of possible outcomes with C first and H last: Count total number of possible outcomes in both cases as 4P4 = 4! = 24 = n(S) We still need to calculate n(E) for each event. Position: 1 2 3 4 n(C first, H last) = Number of ways to choose 1 2 1 1 1 . 2 .1 . 1 = 2 10/2/2011 Section 11.5 4 Example 2 – Choosing conference attendees Solution So, P(C first, H last) = n(C first, H last) 2 1 = = total possible outcomes 24 12 A club consists of five men and seven women. Three members are selected at random to attend a conference. Find the probability that the selected group consists of a) 3 men b) 1 man, 2 women 10/2/2011 Section 11.5 5 10/2/2011 Section 11.5 6 Example 2 – Choosing conference attendees Example 2 , continued Solution: a) P(3 men) = number of ways of selecting 3 men Out of the 12, select 3 for total possible number ways to select a conference, n(S). total number of possible combinations Order of selection does not matter, so this is a problem involving combinations. 10/2/2011 Section 11.5 7 Example 2 continued Example 2 continued Number of ways to select 3 men, n(E): 5 C3 = b) Choosing 1 man and 2 women is a 2step process (it doesn’t matter which step is 1st). 5! 5! = = 10 (5 − 3)!3! 2!3! Number of ways to pick any 3 club members, n(S): 12! 12! C3 = = = 220 12 (12 − 3)!3! 9!3! Therefore: P(3 men ) = n( E ) = 10 = 1 n( S ) 220 22 10/2/2011 Out of 5 men, select 3 for n(E) where E is the event that 3 men are chosen for the conference. To count the number of possible outcomes for the event, we will need to know how many ways to complete each step, and use the Fundamental Counting Principle. ≈ .05 Section 11.5 9 Example 2 continued Time permitting b) P(1 man, 2 women) = number of ways of selecting 1 man, 2 women total number of possible combinations Calculate P( 2 men, 1 women) and P(3 women) to verify that Choose: Number of ways 1 man 5C1 =5 2 women 7C2=21 The number of was to choose 1 man, 2 women = 5 . 21 = 105, so P(1 man, 2 women) = 105 ≈ .48 220 sum of probabilities of every possible outcome = 1 Example 3 continued Example 3 - Winning the Lottery In Virginia’s Cash 5 lottery game, players choose 5 numbers from 1 to 34. The prizes on a $1 ticket are: $100,000 if you match all 5 in any order, $100 for 4, $5 for 3. Solution: The probability of each of this events is equal to: number of ways of winning total number of possible outcomes For each case, the total number of possible outcomes is the same. Because the order of the 5 numbers does not matter, the total possible outcomes involves counting combinations, and equals: Calculate the probability of each of these results. n 10/2/2011 Section 11.5 13 C r = 34 C5 = 34! 34! = = 278,256 (34 − 5)!5! 29!5! 10/2/2011 Section 11.5 Example 3 continued Example 3 continued To find the P(match 4) and P(match 3), we need to know the number of ways to match 4 and 3. Solution (cont’d) There is only one way of winning (choose all 5 correct numbers). So if the event, E, is match all 5 numbers, n(E) = 1. P(match all 5) = These outcomes can be viewed as 2-step processes. To match 4, we need to choose 4 of the 5 winning numbers, and 1 of the 29 losing numbers. We will use the fundamental counting principle. 1 ≈ .0000036 278,256 Choose: Number of ways 10/2/2011 14 Section 11.5 4 correct 5C4 =5 29C1 = 29 15 Example 3 continued Example 3 continued There are 5 . 29 = 145 possible outcomes with 4 numbers correct (n(4 correct) = 145), so To match 3: 145 ≈ .00052 P(match all 4) = 278,256 1 wrong Choose: Number of ways 3 correct 5C3 = 10 2 wrong 29C2=406 There are 10 . 406 = 4060 combinations with 3 numbers correct, so P(match all 3) = 4060 ≈ .015 278,256 Compare to Virginia Lottery Web Page Our results Virginia Lottery probabilities 5 out of 5 1 278,256 1 278,256 4 out of 5 145 ≈ .00052 278,256 1 ≈ .00052 1,919 3 out of 5 4060 ≈ .015 278,256 1 ≈ .014 69