# 1.1. (a) The cases are (1, 2), (2, 1), (3,... (6, 5), (5, 6). Consequently, probability is 10/36 = 5/18. ```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 &middot; 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 &middot; 3/4 =
P (A) &middot; 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) &middot; 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
```