Download Sample solutions

MA 3B SOLUTION 1
1. Homework 1
Exercise 1.1. A coin is tossed three times.
(a) List all outcome in the sample space for this experiment.
(b) What is the probability that there are two consecutive tails?
(c) What is the probability that all three tosses yield distinct outcomes?
Proof. (a) {HHH, HHT, HT H, HT T, T HH, T HT, T T H, T T T }. 23 = 8 elements in this sample space.
(b) Elements with two consecutive tails are {HT T, T T H, T T T }. 83 .
(c) There are no elements with three different outcomes. 08 = 0.
Exercise 1.2. There are five balls numbered 1 to 5 in a urn. Two balls are drawn at random without
replacement. B1 is number on first ball. B2 is number on second ball.
(a) Make a table of all the outcomes in the sample space for this experiement. How many points does it
contain?
(b) What is the numerical probability that the largest number drawn will be 3?
(c) What is the numerical probability that the largest number drawn will be 4 or more?
(d) What is the numerical probability that the sum of the numbers drawn will be even.
(e) What is the numerical probability that the sum of the numbers drawn will be a multiple of three.
Proof. (a) Let (B1 , B2 ) be a tuple where the first coordinate is the number of the first ball, B1 , and the
second cordinate is the number of the second ball B2 . Put into a table the following set:
{(1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 4), (3, 5)
(4, 1), (4, 2), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4)}
5!
(5−2)!
There are
= 5 × 4 = 20 possiblities.
(b) The number of outcomes where B1 = 3 and B2 < 3 is 1 × 2. Similarly, the number of outcomes where
4
B1 < 3 and B2 = 3 is 2 × 1. There are 4 total outcome where the largest number is 3. 20
= 51 .
(c) There are 3 × 2 many outcomes where B1 and B2 are less than 4. There are 20 − 6 = 14 outcomes
14
7
where the largest number is 4 or greater. 20
= 10
.
(d) An even sum can only be obtained if both B1 and B2 are even or both B1 and B2 are odd. There are
8
2 × 1 even-even. There are 3 × 2 odd-odd. 20
= 25 .
(e) Since B1 , B2 ≤ 5, the obtainable multiple of 3 are 3, 6, and 9. 3 is obtainable from (1, 2) and (2, 1). 6
8
= 25 .
is obtainable from (1, 5), (2, 4), (4, 2), (5, 1). 9 is obtainable from (4, 5) and (5, 4). 20
Exercise 1.3. A day-care center has six children from three different families. One family has one child in
the center, another has two, and the third family has three. All three families have all their children in the
center. What is the probability that a randomly selected child is the first-born of its family when chosen as
followed?
(1) There is an urn for each family, which contains all the children in that family. A family is selected at
random, and then a child is selected randomly from the family urn.
(2) All the children are put into one urn, and a child is slected at random.
Proof. Let A be the family with one child, A1 . Let B be the family with two children listed in order of birth
as B1 and B2 . Let C be the family with three children listed in order as C1 , C2 , and C3 . The sample space
is Ω = {A1 , B1 , B2 , C1 , C2 , C3 }.
(a) One wants to determine the probability function µ on the algebra P(Ω), the collection of all subsets
of Ω. Since it is equally likely to select any of the three families, µ({A1 }) = µ({B1 , B2 }) = µ({C1 , C2 , C3 }).
By additivity, µ({A1 }) = µ({B1 , B2 }) = µ({C1 , C2 , C3 }) = 31 . Within a family, a child from that family
1
is chosen with equal chance, so µ({B1 }) = µ({B2 }). By additivity, µ({B1 }) = µ({B2 }) =
µ({C1 }) = µ({C2 }) = µ({C3 }) implies µ({C1 }) = µ({C2 }) = µ({C3 }) = 91 . So
1
6.
Similiarly,
1 1 1
11
+ + =
3 6 9
18
(b) If all children are place in single urn, then probability of any singleton from Ω is 16 . So
µ({A1 , B1 , C1 }) = µ({A1 }) + µ({B1 }) + µ{{C1 }) =
µ({A1 , B1 , C1 }) = µ({A1 }) + µ({B1 }) + µ({C1 }) =
1 1 1
1
+ + =
6 6 6
2
2