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M362K Homework Assignment 8 - solutions Due October 21, 2011 Problems: 3.1.2, 3.1.4, 3.1.6, 3.1.8, 3.1.14, 4.5.2 from Pitman’s “Probability”. Solution to 3.1.2: a) Two draws are independent in this case, so P[X = k, Y = l] = P[X = k]P[Y = l]. The table looks like this (last row represents the values of X and the first column those of Y ; the numbers in the first row correspond to the (marginal) distribution of X; those in the last column to the (marginal) distribution of Y ): 4 3 2 1 1 4 1 4 1 4 1 4 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 16 1 2 3 4 1 4 1 4 1 4 1 4 Y X The entries in the table that correspond to the event where X ≤ Y are those on upper-left half of the table (including the diagonal). Therefore P[X ≤ Y ] = 85 . b) When there is no replacement, all pairs with different components are equally likely, so the table is given by 1 4 1 4 1 4 1 4 1 12 1 12 1 12 0 0 2 1 12 1 12 1 12 0 1 0 1 12 1 12 1 12 1 12 1 12 1 12 1 2 3 4 4 3 1 4 1 4 1 4 1 4 Y X The entries corresponding to X ≤ Y are the same ones as above, but now the probabilities sum to P[X ≤ Y ] = 12 . Solution to 3.1.4: a) The table for (X1 , X2 ) is given by 6 5 4 3 2 1 1 6 1 6 1 6 1 6 1 6 1 6 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 36 1 2 3 4 5 6 1 6 1 6 1 6 1 6 1 6 1 6 X2 X1 b) The table for (Y1 , Y2 ) is given by 1 36 3 36 5 36 7 36 9 36 11 36 6 0 0 0 0 0 5 0 0 0 0 4 0 0 0 3 0 0 2 0 1 1 36 1 36 2 36 1 36 2 36 2 36 1 36 2 36 2 36 2 36 1 36 2 36 2 36 2 36 2 36 1 36 2 36 2 36 2 36 2 36 2 36 1 2 3 4 5 6 1 36 3 36 5 36 7 36 9 36 11 36 Y2 Y1 Solution to 3.1.6: a) Clearly, supports of both X and Y are {0, 1, 2}. For a pair (k, l) with k, l ∈ {0, 1, 2}, we need to count the number of possible outcomes of three coin tosses (such as HHT or THT) with exactly k Hs among the first two tosses and l among the second two, and multiply by 18 . The table looks like this 1 4 1 2 1 4 2 0 1 1 8 1 8 1 8 1 4 1 8 1 8 1 8 0 0 1 2 0 Y X 2 1 4 1 2 1 4 b) They are not independent since, for example, P[X = 0, Y = 0] = 1 8 6= 1 16 = P[X = 0] × P[Y = 0]. c) The support, i.e., the range, of X + Y is {0, 1, 2, 3, 4} its distribution is given by k 0 1 2 3 4 P[X + Y = k] 1 8 2 8 2 8 2 8 1 8 You get the individual probability P[X + Y = k] by listing all the possible pairs that give you k as the sum. For example P[X + Y = 2] = P[X = 0, Y = 2] + P[X = 1, Y = 1] + P[X = 2, Y = 0] = 0 + 2 8 + 0 = 28 . Solution to 3.1.14: a) Team A wins in g = 4 games with probability p4 (they have to win 4 games and loose none). For g = 5, the team A have to win 4 out of the first 5 games, but not the first 4. The probability here is 4p4 q. For the general g = 4, . . . , 7, the team A needs to win any 3 games out of the first g − 1 g−1 (and lose the rest) and then win the last game. This can be done in 3 ways, and each one has the probability p4 q g−4 , so the required probability is g − 1 4 g−4 p q . 3 P P b) P[ team A wins ] = 7g=4 P[ A needs exactly g games to win ] = 7g=4 is no nicer expression.) g−1 3 p4 q g−4 . (Note: There c) For p = 2/3, the probability is 1808/2187 ∼ = 0.83 d) The point is that the teams could continue playing even after one of them wins 4 times. The losing team could win at most 3 games so the probability of being the first who wins 4 games is the same as the probability of winning at least 4 four games out of 7. (Note: the phrase at least is imporant. The probability of winning exactly 4 games out of 7 is not the same as the probability of being the first to win.) e) Let G be the number of games until the first win. Then, P[G = g] = P[ A wins in exactly g games ] + P[ B wins in exactly g games ] g − 1 4 g−4 g − 1 4 g−4 = p q + q p . 3 3 This expression will not depend on p only if p = 12 . 3 Solution to 4.5.2 (a): 1.0 0.8 0.6 0.4 0.2 -1 1 2 The value of FX is equal to 0 for x < 0 and equal to 1 for x ≥ 3. 4 3 4