STAT 516 Answers Homework 2 January 23, 2008 Solutions by Mark Daniel Ward PROBLEMS. = {(a 1, a 2,...) : a i < 6 for all i}

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1 STAT 56 Answers Homework 2 January 23, 2008 Solutions by Mark Daniel Ward PROBLEMS 2. We note that E n consists of rolls that end in 6, namely, experiments of the form (a, a 2,...,a n, 6 for n and a i < 6 for all i. The sample space consists of all the E n s, as well as the rolls that never contain a 6. Therefore, the sample space is S = {(a, a 2,...,a n, 6 : n and a i < 6 for all i} {(a, a 2,... : a i < 6 for all i} and ( n= E n c = {(a, a 2,... : a i < 6 for all i} 3. We write A i to denote the event that a randomly-chosen person reads newspaper i. We are given the following: P(A =.0 P(A A 2 =.08 P(A A 2 A 3 =.0 P(A 2 =.30 P(A A 3 =.02 P(A 3 =.05 P(A 2 A 3 =.0 We easily calculate (drawing a Venn diagram is helpful that P(A A 2 A c 3 = P(A A 2 P(A A 2 A 3 =.08.0 =.07 P(A A 3 A c 2 = P(A A 3 P(A A 2 A 3 =.02.0 =.0 P(A 2 A 3 A c = P(A 2 A 3 P(A A 2 A 3 =.0.0 =.03 so the probability of only reading A is P(A A c 2 Ac 3 = P(A P(A A 2 A c 3 P(A A 3 A c 2 P(A A 2 A 3 = =.0 The probability of only reading A 2 is P(A 2 A c Ac 3 = P(A 2 P(A A 2 A c 3 P(A 2 A 3 A c P(A A 2 A 3 = =.9 The probability of only reading A 3 is P(A 3 A c A c 2 = P(A 3 P(A A 3 A c 2 P(A 2 A 3 A c P(A A 2 A 3 = = 0 3a. So the number of people who read only one newspaper is (00,000(P(A A c 2 Ac 3 + P(A 2 A c Ac 3 + P(A 3 A c Ac 2 = 20,000

2 3b. The number of people who read at least two newspapers is (00,000(P(A A 2 A c 3 +P(A A 3 A c 2 +P(A 2 A 3 A c +P(A A 2 A 3 = 2,000 3c. The number of people who read at least one morning paper plus an evening paper is (00,000(P(A A 2 A c 3 + P(A 2 A 3 A c + P(A A 2 A 3 =,000 3d. The number of people who read at least one paper is (00,000(P(A A 2 A 3 = (00,000(P(A + P(A 2 + P(A 3 P(A A 2 P(A A 3 P(A 2 A 3 + P(A A 2 A 3 = (00,000( = 32,000 So the number of people who do not read any newspapers is 00,000 32,000 = 68,000. 3e. The number of people who read only one morning paper and one evening paper is (00,000(P(A A 2 A c 3 + P(A 2 A 3 A c = 0, Note: I use the value of a card to denote whether the card is ace, 2, 3,..., 0, jack, queen, or king. So 3 values are available altogether. There are suits: clubs, diamonds, hearts, spades. 5a. There are ( ( ways to pick a suit and then 3 5 ways to choose 5 cards of that suit. So the probability of a flush is ( ( 3 5 = ( 2 ( b. There are ( 3 3 ways to choose one value for the pair and then three values for the non-paired cards. Then there are ( ( 2 ways to select the suits for the pair, and 3 ways to select the suits for the non-paired cards. So the probability of one pair is ( (3 2 3( 2( 3 = ( c. There are ( ( 3 2 ways to choose two values for the pairs and then one value for the non-paired card. Then there are ( 2 ( 2 ways to select the suits for the pairs, and ways to select the suit for the non-paired card. So the probability of two pairs is 2( (3 ( 2 2 ( = ( d. There are ( ( ways to choose one value for the three-of-a-kind and then two values for the non-matched cards. Then there are ( 3 ways to select the suits for the three-of-akind, and ( 2 ways to select the suits for the non-matched cards. So the probability of a three-of-a-kind is ( (3 2 2( 3( 2 = ( 2 ( e. There are ( 3 ways to choose one value for the four-of-a-kind and then one value for the non-matched card. Then there are ( ways to select the suits for the four-of-a-kind, and ways to select the suit for the non-matched card. So the probability of a four-of-a-kind ( is (3 ( 2 ( ( ( 52 5 =

3 22. The only way that the ordering after one round is the same as the initial ordering is that there is a nonnegative string of h coin flips (say, i of them at the start of the round, followed by a nonnegative string of t coin flips (thus, n i of them; in other words, once a t appears, the remaining of the coin flips must be also t. Let E i denote the event that i flips of h occur, followed by n i flips of t. Then P(E i = ( i ( n i 2 2 =. The E 2 n i s are disjoint. So the probability that the ordering after one round is the same as the initial ordering is P ( n i=0 E i = n i=0 P(E i = n i=0 = n+. 2 n 2 n 27. We write E i to denote the event that the first ball appears on the ith overall draw. (So A s draws correspond to the E i s for odd-numbered i values, and B s draws correspond to the E i s for even-numbered i values. So the probability that A draws the first red ball is ( 3 P(E + P(E 3 + P(E 5 + P(E 7 = 0 ( ( ( ( ( ( ( ( = ( ( ( ( 7 6 ( 3 6 ( 2 5 ( 3 28a. Sampling without replacement: The probability that 3 red balls are drawn is (5 3 ( 9 3 = The probability that 3 blue balls are drawn is 3 (6 ( 9 3 = 20. The probability that 3 green balls 969 are drawn is 3 (8 ( 9 3 = 56. So the probability that 3 balls of the same color are drawn is = Sampling with replacement: The probability that 3 red balls are drawn is ( = The probability that 3 red blue are drawn is ( 6 3 = 26. The probability that 3 red green are drawn is ( 8 3 = 52. So the probability that 3 balls of the same color are drawn is = b. Sampling without replacement: The probability that 3 balls of different colors are drawn is ( 5 ( 6 ( 8 = ( 9 3 3

4 Sampling with replacement: There are 3! orders in which 3 balls of different colors can be drawn. Each of these ways has probability ( ( 5 6 ( So the probability that 3 balls of different colors are drawn is ( ( ( (3! = Each of the b+g people is equally-likely to be in the ith position, and g of these people are girls, so the desired probability is g b+g. 35. The probability that no psychologists are chosen is 3 (30 ( 5 3 = 05. So the probability that 620 at least one psychologist is chosen is 05 = The probability that both socks are red is 2 (3 ( n 2 = 6 probability is, so we solve 6 = to obtain n =. 2 n(n 2 n(n. We are given that this 0. We assume that each person with a broken television is equally likely to call each of the repair people, and all of the repair calls are made independently. We label the four repair people as r, r 2, r 3, r, respectively. All people call a given repair person r i with probability (/, and there are four repair people r i possible. So the probability that just one repair person is called is ( ( = /6. There are two methods by which two repair people can be called. One possibility is that one repair person r i gets three calls, and the other repair person r j gets one call. There are ( ways to choose the identity of the more-popular repair person ri, and then ( 3 ways to choose the identity of the less-popular repair person r j. Once r i and r j are selected, then there are ( 3 ways to choose the three people to call the more-popular repair person r ( i, and the other person must call repair person r j. So the probability of this method is ( 3 ( 3 (/ = 3/6. The other method by which two repair people can be called is that two repair people can gets two calls each. There are ( 2 ways to select the identities of two repair people r i and r j. Without loss of generality, we can write i < j since each of these people get two calls. There are ( 2 ways that repair person ri can get two calls, and then ( the other two people must call repair person r j. So the probability of this method is ( 2 2 (/ = 9/6. So the total probability that two repair people are called is = There is only one method by which three repair people can be called. One repair person r i gets two calls, and two other repair people r j and r k get one call each. There are ( ways to select the identity of repair person r i, and then there are ( 3 2 ways to select the identities of repair people r j and r k. Once the identities of r i, r j, r k are chosen, then there are ( 2 ways to choose who will call the repair person r i ; without loss of generality, we assume that j < k since r j and r k each get one call. There are ( 2 ways remaining that rj can get one call,

5 and then the last remaining person must call repair person r k. So the total probability that three repair people are called is ( 3 2 ( 2( 2( (/ = 9/6. In order for all four repair people to be called, there are ( people who could call r, and the ( 3 people who could call r2, and then ( 2 people who could call r3, and then the last remaining person must call r. So the probability that all four repair people are called is ( 3 2 ( ( (/ = 3/32. In summary, /6 if i = 2/6 if i = 2 P(exactly i repair people are called = 9/6 if i = 3 3/32 if i = 2. The probability that a double 6 appears on one roll is /36. So the probability that a double 6 does not appear on one roll is 35/36. So the probability that a double 6 never appears is (35/36 n. So the probability that a double 6 appears at least once is ( 35 To have ( 35 ln(/2 ln(35/36 36 n. Since ln(/2 ln(35/36 n, we need ( n, i.e., ln ln (( , it follows that we need n 25. n, i.e., ln 2 36 n. 35 n ln, i.e., 36 3a. Method : There are N! ways that all of the people can be arranged altogether. Now we determine the number of ways that A and B can be next to each other in the line. There are N ways to choose the location for the pair A, B. There are 2 ways that the pair A, B can be placed in these two slots, and there are (N 2! ways that the other N 2 people can be placed in the other N 2 slots. So the probability that A, B are next to each other is (N (2(N 2! = 2 N! N Method 2: There are N 2 places that A can stand without being on the end of the line. For each such place, A has probability /N of landing there, and then B has two available places next to A, and thus lands next to A with probability 2/(N. On the other hand, there are 2 places that A can stand on the end of the line. For each such place, A has probability /N of landing there, and then B has one available place next to A, and thus lands next to A with probability /(N. So the probability that A, B are next to each other is (N 2 2 N N + (2 N N = 2 N 3b. If N = 2, then A and B are always next to each other, so the probability is. Otherwise, for N > 2, we observe that, regardless of where A sits, there are N other seats available for B; two of these are next to A. So the probability is 2 N. 5a. Method : The probability that she is successful on the kth try for the first time is ( ( ( ( ( n n 2 n 3 n (k = n n n 2 n (k 2 n (k n 5

6 We require n (k above, i.e., n k. Method 2: She has n keys altogether. Consider the following scenario: After finding the correct keys, she goes through the door, but continues to try the rest of the wrong keys in the door too, until all n keys have been tried. So she makes n tries altogether. The keys are chosen at random, so the correct key occurs with equal probability i.e., with probability /n on any of the n tries. In particular, the probability that the correct key appears on the kth try is /n (if k n, of course. 5b. If she does not discard previously-used keys, then the probability that the correct key is selected for the first time on the kth try is exactly ( n n k ( = n (n k n k 53. Let E i denote the probability that the ith couple sits together. So the desired probability is P(E E 2 E 3 E We compute P(E E 2 E 3 E by using inclusion-exclusion. We have P(E E 2 E 3 E = i= P(E i i<j P(E i E j + i<j<k P(E i E j E k P(E E 2 E 3 E Since the probabilities P(E i are all the same for all i s, and the probabilities P(E i E j are the same for all pairs i, j, and the probabilities P(E i E j E k are the same for all triples i, j, k, then it follows that P(E E 2 E 3 E = P(E ( P(E E ( P(E E 2 E 3 P(E E 2 E 3 E 3 There are 8! ways that all of the people can be arranged altogether. First we determine the number of ways that the st couple are seated together. We can view the st couple as sitting together on a sofa. So there are 6 seats and sofa, which can be arranged in ( 7 = 7 ways. There are 2 ways that the couple can be placed on the sofa, and there are 6! ways that the other 6 people can be placed in the 6 chairs. So the probability that the st couple are seated together is P(E = (7(2(6! 8! = / Next we determine the number of ways that the st and 2nd couples are each seated together. We can view the st and 2nd couples as each sitting on a sofa. So there are seats and 2 sofas, which can be arranged in ( 6 2 = 5 ways. There are 2! ways that the couples can choose their sofas. There are 2 ways that each couple can sit in their sofa. There are! ways that the other people can be placed in the seats. So the probability that the st couple and 2nd couples are seated together is P(E E 2 = (5(2!(2(2(! 8! = / 6

7 Next we determine the number of ways that the st, 2nd, and 3rd couples are each seated together. We can view the st, 2nd, and 3rd couples as each sitting on a sofa. So there are 2 seats and 3 sofas, which can be arranged in ( 5 3 = 0 ways. There are 3! ways that the couples can choose their sofas. There are 2 ways that each couple can sit in their sofa. There are 2! ways that the other 2 people can be placed in the 2 seats. So the probability that the st, 2nd, and 3rd couples are seated together is P(E E 2 E 3 = (0(3!(2(2(2(2! 8! = /2 Next we determine the number of ways that all the couples are each seated together. We can view each couple as sitting on a sofa. There are! ways that the couples can choose their sofas. There are 2 ways that each couple can sit in their sofa. So the probability that all the couples are each seated together is Thus P(E E 2 E 3 E = (!(2(2(2(2 8! = /05 P(E E 2 E 3 E = (/ 6(/ + (/2 /05 = 23/35 so the desired probability is P(E E 2 E 3 E = 2/35 THEORETICAL EXERCISES 5. Since the E i s might have some overlaps, and we insist that the F i s do not have any overlap (i.e., are disjoint, then we just define each F i to include whatever is in E i but was not found in any of the previous E i s, and, therefore, not in any of the previous F i s either! So we define F i = E i \ i j= E i, or in other words, F i = E i E c Ec 2 Ec i.. We know that P(E F = P(E + P(F P(E F, so it follows that P(E F = P(E + P(F P(E F. Since P(E F, then P(E F, so we conclude that P(E F P(E + P(F. In particular, if P(E =.9 and P(F =.8 then P(E F = There are ( M+N r outcomes altogether, all of which are equally likely. The number of ways that it contains exactly k white balls is ( ( M N k r k. So the desired probability is ( M ( N k r k ( M+N r 7

8 6. We use induction to prove that P(E E n P(E + + P(E n (n. For n =, we have P(E P(E, which is true. The n = 2 case was proved already in problem above. Now we use induction. Suppose that we have proved Bonferroni s inequality for the n case, i.e., we already proved P(E E n P(E + + P(E n (n 2. By problem, using E = E E n and F = E n, we have P(E F P(E + P(F. We know by induction that P(E = P(E E n P(E + + P(E n (n 2, so substituting this into P(E in the equation P(E F P(E + P(F yields or more simply P(E F P(E + + P(E n (n 2 + P(F which proves Bonferroni s inequality. P(E E n P(E + + P(E n (n 9. The probability that, in the first k draws, there are exactly r reds is ( n ( m ( n ( m r (k (r r k r = ( n+m k ( n+m k Given that the first k draws have exactly r reds, then there are n (r = n r+ reds remaining and m (k r = m k + r blue balls remaining. The probability that the n r+ next (i.e., kth ball is red is. So the total desired probability is = n r++m k+r ( n ( m r k r ( n+m k n r+ n+m k+ n r + n + m k We consider a sample space with a countably infinite number of points, say x, x 2,.... We note that = P(S = P({x } {x 2 } = i= P(x i. If all points are equally likely, then P(x i is the same for all i, so we write p = P(x i for all i, and thus i= P(x i = i= p = p i= = p =, but we know from above that i= P(x i =, so we have a contradiction. So all points are not equally likely. All points could, on the other hand, have a positive probability of occurring. For an example of this, consider a series of coin flips. Let x i denote the outcome that the st head occurs on the ith flip. Of course, a head might never appear, which we denote by outcome y. So P(x i =, and P(y = 0. Then 2 i P(S = P({y} {x } {x 2 } = P(y+P(x +P(x 2 +P(x 3 + = = 8