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1.1. (a) The cases are (1, 2), (2, 1), (3, 2), (2, 3), (4, 3), (3, 4), (5, 4), (4, 5), (6, 5), (5, 6). Consequently, probability is 10/36 = 5/18. (b) We have 9 cases when the largest number is 5, and 11 cases when the largest number is 6. Consequently, probability is 20/36 = 5/9. 1.2. P ({c}) = 1 − P ({a, b}) = 0.3, P ({a}) = 1 − P ({b, c}) = 0.4, hence P ({a}) = 1 − P ({b}) − P ({c}) = 0.3. 1.3. P (A ∪ B) = P (A) + P (B) − P (A ∩ B) ≤ 1, hence P (A ∩ B) ≥ P (A) + P (B) − 1 = 1.1 − 1 = 0.1. P (A ∩ B) ≤ P (A) = 0.4. The maximal value is 0.4 (when A ⊂ B). The minimal value is 0.1 (when A ∪ B = Ω. 1.4. (a) P (A) = 4 · 1/8 = 1/2, since there are 3 cases when they have one girl and one case when they have no girls. P (B) = 6/8 = 3/4, since there are six cases when there are children of both sexes (three cases when only one boy, and three cases when only one girl). The event A ∩ B is having exactly one girl, and hence P (A ∩ B) = 3/8. In this case we have P (A ∩ B) = 3/8 = 1/2 · 3/4 = P (A) · P (B), hence the events are independent. (b) Similar computation shows that P (A) = 5/16, P (B) = 1 − 2/16 = 7/8 (since there are only two cases when they do not have children of both sexes). P (A ∩ B) = 4/16 = 1/4. We see that P (A ∩ B) 6= P (A) · P (B), hence the events are not independent. 1.5. Probability of an even sum is 1/2, since it is even exactly in half of the cases (for every fixed value of one die, there are three values of the = 1/3, second die such that the sum is even). Probability of B is 2+5+4+1 36 since 3 = 1 + 2 = 2 + 1, 6 = 1 + 5 = 2 + 4 = 3 + 3 = 4 + 2 = 5 + 1, 9 = 3 + 6 = 4 + 5 = 5 + 4 = 6 + 3, and 12 = 6 + 6. A ∩ B means that the sum belongs to {6, 12}, and probability of this is 5+1 36 = 1/6. We see that P (A ∩ B) = P (A)P (B), i.e., that the events are independent. 1.6. Possible values of the sum are 3, . . . , 12. Let us count in how many ways each of these numbers can be represented as a sum of three numbers from 1 to 4. 3 1+1+1 1 4 1+1+2 3 5 1 + 1 + 3, 1 + 2 + 2 6 6 1 + 1 + 4, 1 + 2 + 3, 2 + 2 + 2 3 + 6 + 1 = 10 7 1 + 2 + 4, 1 + 3 + 3, 2 + 2 + 3 6 + 3 + 3 = 12 8 1 + 3 + 4, 2 + 2 + 4, 2 + 3 + 3 6 + 3 + 3 = 12 9 1 + 4 + 4, 2 + 3 + 4, 3 + 3 + 3 3 + 6 + 1 = 10 10 2 + 4 + 4, 3 + 3 + 4 6 11 3 + 4 + 4 3 12 4 + 4 + 4 1 It follows that distribution of the sum X is s 3 4 5 6 7 8 P (X = s) 1/64 3/64 6/64 10/64 12/64 12/64 1 9 10 11 12 10/64 6/64 3/64 1/64