Final Exam - NEXT WEEK! The final exam is worth 400 points, twice as much as each of the semester exams. The remaining homework assignments count toward 50 of these points. The 350 points on the exam administered next week cover the most recent material (Sections 19 and 20) and all previous material in the course. The colored review sheet that I distributed contains practice review exercises for the final exam. You may bring this colored review sheet to the final exam and use it during the exam. Write anything on it you want beforehand. There is a link on the webpage schedule to the numerical answers only for these practice review exercises. It is up to you to do the exercises and check that you can get the correct numerical answers. You can email me if you have any questions.

Note: all homework submitted this week and the last quiz will be available for students to pick from my secretary Sheran Swank before the final exam.

Math 106 – Combinatorics – Quiz #6 Review Sheet Name __________________________________________________ 1. (a) (b) A fair coin is flipped 20 times. What is the probability of observing exactly 10 heads? U = E = {all strings of size 20 consisting of Hs and Ts} {all strings of size 20 consisting of 10 Hs and 10 Ts} #U = #E = 220 = 1,048,576 #E P(E) = —– = #U 184,756 ———— = 0.176 or 17.6% 1,048,576 C(20,10) = 184,756 What is the probability of observing at least 15 heads? U = E = {all strings of size 20 consisting of Hs and Ts} {all strings of size 20 consisting of Hs and Ts with at least 15 Hs} #U = #E = 220 = 1,048,576 #E P(E) = —– = #U 21,700 ———— = 0.021 or 2.1% 1,048,576 C(20,15) + … + C(20,20) = 21,700

(c) (d) What is the probability of observing at most 15 heads? U = E = {all strings of size 20 consisting of Hs and Ts} {all strings of size 20 consisting of Hs and Ts with at most 15 Hs} #U = #E = #E P(E) = —– = #U 1,042,380 ———— = 0.994 or 99.4% 1,048,576 220 = 1,048,576 220 – [C(20,16) + … + C(20,20)] = 1,042,380 What is the probability of observing no consecutive heads, that is, no heads are adjacent in the 20 flips? U = E = {all strings of size 20 consisting of Hs and Ts} {all strings of size 20 consisting of Hs and Ts with no adjacent Hs} 17,711 ———— = 0.017 or 1.7% 1,048,576 #U = #E = 220 = 1,048,576 #E P(E) = —– = #U F(20) = 17,711

2 (a) (b) A penny, a nickel, a dime, a quarter, a one-dollar bill, a five-dollar bill, a ten-dollar bill, a twenty-dollar bill, a fifty-dollar bill, and a 100-dollar bill are randomly placed in a row. What is the probability that all the coins are together? U = E = {all permutations of the coins & bills} {all permutations of the coins & bills where all the coins are together} #U = #E = P(10,10) = 10! = 3,628,800 #E P(E) = —– = #U 120,960 ———— = 0.033 or 3.3% 3,628,800 P(7,7)P(4,4) = 7!4! = 120,960 What is the probability that all the coins are together and all the bills are together? U = E = {all permutations of the coins & bills} {all permutations of the coins & bills where all the coins are together and all the bills are together} #U = #E = P(10,10) = 10! = 3,628,800 P(2,2)P(6,6)P(4,4) = 2!6!4! = 34,560 #E P(E) = —– = #U 34,560 ———— = 0.0095 or 0.95% 3,628,800

(c) (d) (e) What is the probability that the bills are not all together? U = E = {all permutations of the coins & bills} {all permutations of the coins & bills where all the bills are not together} #U = #E = P(10,10) = 10! = 3,628,800 P(10,10) – P(5,5)P(6,6) = 10! – 5!6! = 3,542,400 #E P(E) = —– = #U 3,542,400 ———— = 0.976 or 97.6% 3,628,800 What is the probability that none of the coins are adjacent (that is, no two coins are next to each other)? U = E = {all permutations of the coins & bills} {all permutations of the coins & bills where none of the coins are adjacent} #U = #E = P(10,10) = 10! = 3,628,800 C(7,4)P(6,6)P(4,4) = 604,800 #E P(E) = —– = #U 604,800 ———— = 0.167 or 16.7% 3,628,800 What is the probability that none of the bills are adjacent (that is, no two bills are next to each other)?

U = E = {all permutations of the coins & bills} {all permutations of the coins & bills where none of the bills are adjacent} #U = #E = P(10,10) = 10! = 3,628,800 #E P(E) = —– = #U ———— = 0 3,628,800