Exam (elaborations) TEST BANK FOR A First Course in Probability 8th Edition By Sheldon Ross (Solution Manual)

Chapter 1: Problems Problem 1 (counting license plates) Part (a): In each of the first two places we can put any of the 26 letters giving 262 possible letter combinations for the first two characters. Since the five other characters in the license plate must be numbers, we have 105 possible five digit letters their specification giving a total of 262 · 105 = , total license plates. Part (b): If we can’t repeat a letter or a number in the specification of a license plate then the number of license plates becomes 26 · 25 · 10 · 9 · 8 · 7 · 6 = , total license plates. Problem 2 (counting die rolls) We have six possible outcomes for each of the die rolls giving 64 = 1296 possible total outcomes for all four rolls. Problem 3 (assigning workers to jobs) Since each job is different and each worker is unique we have 20! different pairings. Problem 4 (creating a band) If each boy can play each instrument we can have 4! = 24 ordering. If Jay and Jack can play only two instruments then we will assign the instruments they play first with 2! possible orderings. The other two boys can be assigned the remaining instruments in 2! ways and thus we have 2! · 2! = 4 , possible unique band assignments. Problem 5 (counting telephone area codes) In the first specification of this problem we can have 9 − 2 + 1 = 8 possible choices for the first digit in an area code. For the second digit there are two possible choices. For the third digit there are 9 possible choices. So in total we have 8 · 2 · 9 = 144 , possible area codes. In the second specification of this problem, if we must start our area codes with the digit “four” we will only have 2 · 9 = 18 area codes. Problem 6 (counting kittens) The traveler would meet 74 = 2401 kittens. Problem 7 (arranging boys and girls) Part (a): Since we assume that each person is unique, the total number of ordering is given by 6! = 720. Part (b): We have 3! orderings of each group of the three boys and girls. Since we can put these groups of boys and girls in 2! different ways (either the boys first or the girls first) we have (2!) · (3!) · (3!) = 2 · 6 · 6 = 72 , possible orderings. Part (c): If the boys must sit together we have 3! = 6 ways to arrange the block of boys. This block of boys can be placed either at the ends or in between any of the individual 3! orderings of the girls. This gives four locations where our block of boys can be placed we have 4 · (3!) · (3!) = 144 , possible orderings. Part (d): The only way that no two people of the same sex can sit together is to have the two groups interleaved. Now there are 3! ways to arrange each group of girls and boys, and to interleave we have two different choices for interleaving. For example with three boys and girls we could have g1b1g2b2g3b3 vs. b1g1b2g2b3g3 , thus we have 2 · 3! · 3! = 2 · 62 = 72 , possible arrangements. Problem 8 (counting arrangements of letters) Part (a): Since “Fluke” has five unique letters we have 5! = 120 possible arrangements. Part (b): Since “Propose” has seven letters of which four (the “o”’s and the “p”’s) repeat we have 7! 2! · 2! = 1260 , arrangements. Part (c): Now “Mississippi” has eleven characters with the “i” repeated four times, the “s” repeated four times and the “p” repeated two times, so we have 11! 4! · 4! · 2! = 34650 , possible rearranges. Part (d): “Arrange” has seven characters with a double “a” and a double “r” so it has 7! 2! · 2! = 1260 , different arrangements. Problem 9 (counting colored blocks) Assuming each block is unique we have 12! arrangements, but since the six black and the four red blocks are not distinguishable we have 12! 6! · 4! = 27720 , possible arrangements. Problem 10 (seating people in a row) Part (a): We have 8! = 40320 possible seating arrangements. Part (b): We have 6! ways to place the people (not including A and B). We have 2! ways to order A and B. Once the pair of A and B is determined, they can be placed in between any ordering of the other six. For example, any of the “x”’s in the expression below could be replaced with the A B pair x P1 x P2 x P3 x P4 x P5x P6 x .