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20. There are five bands trying to be the closing band in a concert. They are all in the concert but each band wants to go last. The producer is going to draw at random the order of the bands. Then bands are A, B, C, D, and E. What is the probability that band A gets drawn last and band B gets drawn second? a. What are the total possible outcomes of drawing the bands? Show your work here: b. List the arrangements in the chart below where band A is chosen last and band B is chosen second. 1" 2nd 3rd 4th 5h (last) c. Find the probability that band A is chosen to perform last and band B is chosen second.

20. There are five bands trying to be the closing band in a concert. They are all in the concert but each band wants to go last. The producer is going to draw at random the order of the bands. Then bands are A, B, C, D, and E. What is the probability that band A gets drawn last and band B gets drawn second? a. What are the total possible outcomes of drawing the bands? Show your work here: b. List the arrangements in the chart below where band A is chosen last and band B is chosen second. 1" 2nd 3rd 4th 5h (last) c. Find the probability that band A is chosen to perform last and band B is chosen second.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Exercise 20 Explanation** In this problem, we are determining the probability of specific arrangements for the order of performance of five bands: A, B, C, D, and E. The main goal is to calculate the likelihood that band A is the last to perform and band B is second. **Question Breakdown** **a. Total Possible Outcomes:** - You need to calculate the total number of permutations of all five bands. - Since there are five bands, the number of possible outcomes is \(5!\) (factorial of 5). - \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). **b. Arrangements for Band A Last and Band B Second:** - You will list the possible sequences that achieve the condition where Band A is last (5th) and Band B is second (2nd). - This involves determining the permutations of the remaining three bands (C, D, E) for the 1st, 3rd, and 4th positions. - There are \(3!\) permutations for these three bands. - \(3! = 3 \times 2 \times 1 = 6\). | 1st | 2nd | 3rd | 4th | 5th (last) | |-----|-----|-----|-----|------------| | C | B | D | E | A | | C | B | E | D | A | | D | B | C | E | A | | D | B | E | C | A | | E | B | C | D | A | | E | B | D | C | A | **c. Probability Calculation:** - The probability of band A being last and band B being second is calculated as the number of favorable outcomes divided by the total outcomes. - The number of favorable outcomes is 6 (from part b). - Total outcomes for all permutations is 120 (from part a). - Probability = \(\frac{6}{120} = \frac{1}{20}\). For further clarity or additional assistance, please consult your teacher or refer to a probability textbook or resource.
Transcribed Image Text:**Exercise 20 Explanation** In this problem, we are determining the probability of specific arrangements for the order of performance of five bands: A, B, C, D, and E. The main goal is to calculate the likelihood that band A is the last to perform and band B is second. **Question Breakdown** **a. Total Possible Outcomes:** - You need to calculate the total number of permutations of all five bands. - Since there are five bands, the number of possible outcomes is \(5!\) (factorial of 5). - \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). **b. Arrangements for Band A Last and Band B Second:** - You will list the possible sequences that achieve the condition where Band A is last (5th) and Band B is second (2nd). - This involves determining the permutations of the remaining three bands (C, D, E) for the 1st, 3rd, and 4th positions. - There are \(3!\) permutations for these three bands. - \(3! = 3 \times 2 \times 1 = 6\). | 1st | 2nd | 3rd | 4th | 5th (last) | |-----|-----|-----|-----|------------| | C | B | D | E | A | | C | B | E | D | A | | D | B | C | E | A | | D | B | E | C | A | | E | B | C | D | A | | E | B | D | C | A | **c. Probability Calculation:** - The probability of band A being last and band B being second is calculated as the number of favorable outcomes divided by the total outcomes. - The number of favorable outcomes is 6 (from part b). - Total outcomes for all permutations is 120 (from part a). - Probability = \(\frac{6}{120} = \frac{1}{20}\). For further clarity or additional assistance, please consult your teacher or refer to a probability textbook or resource.
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