Stat 330 (Spring 2015): Homework 1
Due: January 23, 2015
Show all of your work, and please staple your assignment if you use more than one sheet. Write your name,
the course number and the section on every sheet. Problems marked with * will be graded and one additional
randomly chosen problem will be graded.
1. A coin is tossed three times, and the sequence of heads and tails is recorded.
(a) Determine the sample space, Ω.
(b) List the elements that make up the following events: i. A = at least two heads, ii. B = the first
two tosses result in heads, iii. C = the last toss results in a tail
(c) List the elements of the following events: i. A, ii. A ∩ C. iii. A ∪ C
Answer:
(a) Let H and T stand for the events of head and tail, respectively. Ergo, the sample space is
Ω = {HHH, HHT, HT H, HT T, T HH, T HT, T T H, T T T }
(b)
i. A = at least two heads = {HHH, HHT, T HH, HT H}
ii. B = the first two tosses result in heads = {HHT, HHH}
iii. C = the last toss results in a tail = {HHT, HT T, T HT, T T T }
(c)
i. Ā = Ω \ A = {HT T, T HT, T T H, T T T } = at most one head
ii. A ∩ C = {HHT }
iii. A ∪ C = {HHH, HHT, T HH, HT H, HT T, T HT, T T T } = at least two heads or the last toss
results in a tail
2. * Let A, B ⊂ Ω. Draw Venn diagrams that shows the sets A ∪ B, A ∩ B, and A ∩ B̄. (Many drawings
are correct, depending on your choices of A and B.) Then, use Kolmogorov’s axioms to show that the
following statements are true.
Answer:
The Venn diagrams is shown below
Figure 1: Venn diagram of problem 2
(a) For any two events A, B ⊂ Ω,
P (A ∩ B̄) = P (A) − P (A ∩ B)
Answer:
For each event A, B ⊂ Ω, it is easy to see that
A = (A ∩ B̄) ∪ (A ∩ B)
1
Stat 330 (Spring 2015): Homework 1
Due: January 23, 2015
and
(A ∩ B̄) ∩ (A ∩ B) = ∅
so that by axiom (iii)
P (A) = P (A ∩ B̄) + P (A ∩ B) ⇒ P (A ∩ B̄) = P (A) − P (A ∩ B).
(b) For two events A, B ⊂ Ω with B ⊂ A,
P (A ∩ B̄) = P (A) − P (B)
Answer:
Notice A ∩ B = B for any sets A, B ⊂ Ω with B ⊂ A, so that following part (a),
P (A ∩ B̄) = P (A) − P (A ∩ B) = P (A) − P (B).
(c) For events A, B, C ⊂ Ω verify the following extension of the addition rule:
P (A ∪ B ∪ C) = P (A) + P (B) + P (C) − P (A ∩ B) − P (A ∩ C) − P (B ∩ C) + P (A ∩ B ∩ C)
Answer:
For any set D, E ⊂ Ω, in class we derive that
P (D ∪ E) = P (D) + P (E) − P (D ∩ E).
Let D = A and E = B ∪ C, we have
P (A ∪ B ∪ C) = P (D ∪ E)
= P (D) + P (E) − P (D ∩ E)
= P (A) + P (B ∪ C) − P (A ∩ (B ∪ C))
= P (A) + P (B) + P (C) − P (B ∩ C) − P ((A ∩ B) ∪ (A ∩ C))
= P (A) + P (B) + P (C) − P (B ∩ C) − [P (A ∩ B) + P (A ∩ C) − P (A ∩ B ∩ A ∩ C)]
= P (A) + P (B) + P (C) − P (B ∩ C) − P (A ∩ B) − P (A ∩ C) + P (A ∩ B ∩ C)
3. How many 6-character license plate combinations are possible if there must be three letters followed by
three numbers?
Answer: As order matters (e.g., ‘AFH 239’ is different from ‘FHA 392’) we use permutations. There are
26 possibilities for the first three positions, and 10 possibilities for the last three positions. The number
of permutations is 263 × 103 = 17.58 million different license plates.
4. In how many ways can 12 people be divided into three groups of 4 for an evening of bridge?
Answer: The tables are indistinguishable. So let’s think about the number of combinations to seat
the other 11 people with player #1. There are 11
3 = 165 ways to seat 3 more people at whichever table
has player #1.
Now take any player that is not at player #1’s table. There are 73 = 35 ways to seat 3 more people at
this table. Whoever remains is seated at the other table. The multiplication rule gives the total number
of possible seating arrangements: 165 × 35 = 5775.
8
12!
(Note, if the tables were distinguishable, then there would be 12
4
4 = 4! 4! 4! = 34,650 arrangements.)
In how many ways can this be done if the 12 consist of six pairs of partners that must play with each
other?
Answer: Now there are six partnerships to be seated at three tables. There are binom51 = 5 ways to
seat another partnerships with partnership #1. Of the remaining four partnerships, there are only three
ways to have unique arrangements. Thus there are 15 unique table arrangements for the six partnerships:
(12)(34)(56)
(13)(24)(56)
(14)(23)(56)
(15)(23)(46)
(16)(23)(45)
(12)(35)(46)
(13)(25)(46)
(14)(25)(36)
(15)(24)(36)
(16)(24)(35)
(12)(36)(45)
(13)(26)(45)
(14)(26)(35)
(15)(26)(34)
(16)(25)(34)
2
Stat 330 (Spring 2015): Homework 1
Due: January 23, 2015
5. A drawer of socks contains seven black socks, eight blue socks, and nine green socks. Two socks are
chosen in the dark.
(a) What is the probability that they match?
Answer: The probability that you draw 2 black socks, 2 blue socks, or 2 green socks is
7 8 9
7 8 9
7 8 9
7
8
9
85
2 0 0
0 2 0
0 0 2
2 + 2 + 2
+
+
=
=
= .308.
24
24
24
24
276
2
2
2
2
Whenever you have groups of indistinguishable objects and you want to know the probability of
selecting a certain number of each group to make k total selections, use this formula:
np
n1 n2
k1 k2 · · · kp
n
k
where you select k1 out of n1 objects for the first group, etc. for the p groups, for a total of k
selections out of n objects.
For example, the probability that a poker hand has 3 spades, 2 hearts, and no diamonds or clubs is
13 13 13 13
3
2
0
52
5
0
.
(b) What is the probability that a black pair is chosen?
7
Answer: The probability of drawing a black sock first is 24
. Conditional on this, the probability
6
of drawing a black sock next is 23 as there is one less black sock in the drawer. Therefore, the
24
7
7 6
7
probability of drawing a black pair is 24
23 = 92 = .0761. This is also equal to 2 / 2 using the
notation of (a).
6. * (Baron’s book): 2.1
Answer:
7. (Baron’s book): 2.2
Answer:
3
Stat 330 (Spring 2015): Homework 1
Due: January 23, 2015
8. (Baron’s book): 2.4
Answer:
4