Hello, please answer Exercise 1.37 parts a, b, and c. It is highlighted in green. Omit part d. I have first added Example 1.20 as context for the exercise. Enjoy :)

1.4. Consequences of the rules of probability 15 P(red and yellow) = 30 . 10 10 59' This leads to P(exactly one red or exactly one yellow) = 20 20 10 110 59 177 59 177 We used unordered samples, but we can get the answer also by using ordered samples. Example 1.24 below solves this same problem with inclusion-exclusion. Example 1.20. Peter and Mary take turns rolling a fair die. If Peter rolls 1 or 2 he wins and the game stops. If Mary rolls 3, 4, 5, or 6, she wins and the game stops. They keep rolling in turn until one of them wins. Suppose Peter rolls first. (a) What is the probability that Peter wins and rolls at most 4 times? To say that Peter wins and rolls at most 4 times is the same as saying that either he wins on his first roll, or he wins on his second roll, or he wins on his third roll, or he wins on his fourth roll. These alternatives are mutually exclusive. This is a fairly obvious way to decompose the event. So define events A = {Peter wins and rolls at most 4 times) and Ak = {Peter wins on his kth roll). Then A = U. -1 Ak and since the events Ax are mutually exclusive, P(A) = >=1 PLAN). To find the probabilities P(Ax) we need to think about the game and the fact that Peter rolls first. Peter wins on his kth roll if first both Peter and Mary fail k - 1 times and then Peter succeeds. Each roll has 6 possible outcomes. Peter's roll fails in 4 different ways and Mary's roll fails in 2 different ways. Peter's kth roll succeeds in 2 different ways. Thus the ratio of the number of favorable alternatives over the total number of alternatives gives PLAN) = (4 . 2)*-1 . 2 (6 .6)-1.6 (36 12 = ( 3 ) *- ; The probability asked is now obtained from a finite geometric sum: P(A) = _ PLAN = E () -3 =7(1-()4). Above we changed the summation index to j = k - 1 to make the sum look exactly like the one in equation (D.2) in Appendix D and then applied formula (D.2). (b) What is the probability that Mary wins? If Mary wins, then either she wins on her first roll, or she wins on her second roll, or she wins on her third roll, etc., and these alternatives are mutually16 Experiments with random outcomes exclusive. There is no a priori bound on how long the game can last. Hence we have to consider all the infinitely many possibilities. Define the events B = {Mary wins) and By = {Mary wins on her kth roll). Then B = Uo, Be is a union of pairwise disjoint events and the additivety of probability implies P(B) = >k-1 P(Bk). In order for Mary to win on her kth roll, first Peter and Mary both fail k - 1 times, then Peter fails once more, and then Mary succeeds. Thus the ratio of the number of favorable alternatives over the total number of ways k rolls for both people can turn out gives P(B) = (4 . 2)*-1 . 4. 4 k-1 16 1-1 4 (6 .6)k 36 36 The answer comes from a geometric series: P{Mary wins) = P(B) = _ P(BN = > () Note that we calculated the winning probabilities without defining the sample space. This will be typical going forward. Once we have understood the general principles of building probability models, it is usually not necessary to define explicitly the sample space in order to do calculations. Events and complements Events A and A are disjoint and together make up $2, no matter what the event A happens to be. Consequently P(A) + P(AS) = 1. (1.13) This identity tells us that to find P(A), we can compute either P(A) or P(AC). Sometimes one is much easier to compute than the other. Example 1.21. Roll a fair die 4 times. What is the probability that some number appears more than once? A moment's thought reveals that this event contains a complicated collection of possible arrangements. Any one of 1 through 6 can appear two, three or four times. But also two different numbers can appear twice, and we would have to be careful not to overcount. However, switching to the complement provides an easy way out. If we set A = {some number appears more than once} then Ac = {all rolls are different). By counting the possibilities P(A() = 6.543 = " = 18 and consequently P(A) = 13.Exercise 1.37. These questions pertain to Example 1.20 where Peter and Mary take turns rolling a fair die. To answer the questions, be precise about the definitions of your events and their probabilities. (a) As in Example 1.20, suppose Peter takes the first roll. What is the probability that Mary wins and her last roll is a six? (b) Suppose Mary takes the first roll. What is the probability that Mary wins? (c) What is the probability that the game lasts an even number of rolls? Consider separately the case where Peter takes the first roll and the case 36 Experiments with random outcomes where Mary takes the first roll. Based on your intuition, which case should be likelier to end in an even number of rolls? Does the calculation confirm your intuition? (d) What is the probability that the game does not go beyond k rolls? Consider separately the case where Peter takes the first roll and the case where Mary takes the first roll. Based on your intuition, in which case should the game be over faster? Does the calculation confirm your intuition? Hint. Consider separately games that last 2j rolls (even number of rolls) and games that last 2j - 1 rolls (odd number of rolls)