1. An urn contains six chips numbered 1 through 6.
Three are drawn out, assuming the
order of the chips is irrelevant. Characterize S and list all outcomes that are in the event
A = “Second smallest chip is a 3”.
We have that S = {(i, j, k) : 1 ≤ i ≤ j ≤ k ≤ 6}. Thus j = 3 and i = 1, 2 and k = 4, 5, 6.
Therefore
A = {(1, 3, 4), (1, 3, 5), (1, 3, 6), (2, 3, 4), (2, 3, 5), (2, 3, 6)}.
2. Let P be the set of right triangles with a 5cm hypotenuse and whose height and length are
a and b, respectively. Characterize the outcomes in P .
Let a right triangle be
with a < b < c. Then P = {(a, b, 5) : a2 + b2 = 52 , a < b}.
3. In the game of craps, the person rolling two fair six sided dice (the shooter) wins outright
if their first toss is a 7 (sum of both numbers) or an 11. If their first toss is a 2, 3, or
12, they loses outright. If their first roll is something else, say, a 9, that number becomes
their ‘point’ and they keep rolling the dice until they either roll another 9, in which case
they win, or a 7, in which case they lose. Characterize the sample outcomes contained in
the event B = “Shooter wins with a point of 9”.
B = {(9, 9), (9, no 7 - no 9, 9), (9, no 7 - no 9, no 7 - no 9, 9), . . .)}
4. Define A = {x : 0 ≤ x ≤ 1}, B = {x : 0 ≤ x ≤ 3} and C = {x : −1 ≤ x ≤ 2}. Determine
the following sets of points:
(a) AC ∩ B ∩ C
(b) AC ∪ (B ∩ C)
(c) A ∩ B ∩ C C
(d) [(A ∪ B) ∩ C C ]C
a. AC = R \ [0, 1] thus AC ∩ B = (1, 3] and AC ∩ B ∩ C = (1, 3] ∩ C = (1, 2].
b. AC ∪ (B ∩ C) = R.
c. A ∩ B ∩ C C = ∅.
d. [(A ∪ B) ∩ C C ]C = R \ (2, 3].
1
5. Let A, B and C be any three events defined on a sample space S. Show the following:
(a) Outcomes in A ∪ (B ∩ C) are the same as the outcomes in (A ∪ B) ∩ (A ∪ C)
(b) Outcomes in A ∩ (B ∪ C) are the same as the outcomes in (A ∩ B) ∪ (A ∩ C)
a. If s ∈ A ∪ (B ∩ C), then s ∈ A or s ∈ B ∩ C; in either cases, s belongs to A ∪ B and
A ∪ C, so s ∈ (A ∪ B) ∩ (A ∪ C).
Take s ∈ (A ∪ B) ∩ (A ∪ C), if s ∈ A, then s ∈ A ∪ (B ∩ C); if s ∈
/ A, it must belong to
B ∩ C. In both cases, s ∈ A ∪ (B ∩ C).
b. If s ∈ A∩(B ∪C), then s belongs to both A and B ∪C, then s ∈ (A∩B) or s ∈ (A∩C),
i.e. s ∈ (A ∩ B) ∪ (A ∩ C).
Take s ∈ (A ∩ B) ∪ (A ∩ C), then s ∈ (A ∩ B) or s ∈ (A ∩ C); in either cases, s belongs
to A and (B ∪ C), i.e. s ∈ A ∩ (B ∪ C).
6. Prove the DeMorgan’s laws.
Theorem (De Morgan’s laws) For any two events A and B:
(a) The complement of their intersection is the union of their complements: (A ∩ B)C =
AC ∪ B C
(b) The complement of their union is the intersection of their complements: (A ∪ B)C =
AC ∩ B C
a. Let M = (A ∩ B)C and N = (AC ∪ B C ). Let s ∈ M , i.e. s ∈ (A ∩ B)C . And
s∈
/ (A ∩ B)
s∈
/ A or s ∈
/B
s ∈ AC or s ∈ B C
s ∈ AC ∪ B C
s∈N
which means that M ⊂ N . Then let s ∈ N , i.e. s ∈ AC ∪ B C . And
s ∈ AC or s ∈ B C
s∈
/ A or s ∈
/B
s∈
/ (A ∩ B)
s ∈ (A ∩ B)C
s∈M
which means that N ⊂ M .
b. Let P = (A ∪ B)C and Q = AC ∩ B C . Let s ∈ P , i.e. s ∈ (A ∪ B)C . And
s∈
/ (A ∪ B)
s ∈ AC and s ∈ B C
s ∈ AC ∩ B C
s∈Q
which means that P ⊂ Q. Then let sinQ, i.e. s ∈ AC ∩ B C . And
s ∈ AC and s ∈ B C
s∈
/ A and s ∈
/B
s∈
/ (A ∪ B)
s ∈ (A ∪ B)C
s∈P
which means that Q ⊂ P .
2
7. Suppose that three fair dice are tossed. Let Ai be the event that a 6 shows on the ith die,
i = 1, 2, 3. Does P (A1 ∪ A2 ∪ A3 ) = 1/2? Explain.
A1 ∪A2 ∪A3 is the event “at least one 6 in 3 throws” or A1 ∪A2 ∪A3 = (no six in 3 throws)C .
Let Bi = (no six in throw i). Then
P (A1 ∪ A2 ∪ A3 ) = 1 − P (B1 ∩ B2 ∩ B3 )
= 1 − P (B1 )P (B2 )P (B3 )
3
5 1
=1− ̸=
6 2
1
P (A1 ∪ A2 ∪ A3 ) = P (A1 ) + P (A2 ) + P (A3 ), where P (Ai ) = 6 only holds if A1 , A2 , A3
are mutually exclusive, which is not the case here.
8. Let A1 , A2 , . . . , An be a partition of S, that is a series of events for which Ai ∩ Aj = ∅ if
i ̸= j and A1 ∪ A2 ∪ . . . ∪ An = ∪ni=1 Ai = S. Let B be any event defined on S. Express B
as a union of intersections.
B = (B ∩ A1 ) ∪ (B ∩ A2 ) ∪ . . . (B ∩ An ) = ∪ni=1 (B ∩ Ai )
9. If P (A|B) < P (A), show that P (B|A) < P (B), assuming both P (A) > 0 and P (B) > 0.
P (A∩B)
If P (A|B) = P (B) < P (A) then P (A ∩ B) < P (A)P (B).
P (A∩B) P (A)P (B)
Thus P (B|A) = P (A) < P (A) = P (B).
10. Two fair dice are rolled. What is the probability that the number on the first die is at least
as large as 4 given that the sum of the two dice is 8?
A1 = {(i, j) : i ≥ 4} and A2 = {(i, j) : i + j = 8} = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} The
outcomes in A2 which belongs also to A1 are {(4, 4), (5, 3), (6, 2)}. Then
P (A1 ∩ A2 ) 3
P (A1 |A2 ) = = .
P (A2 ) 5
11. An urn contains five white chips, four black chips and three red chips. Four chips are
drawn sequentially and without replacement. What is the probability of obtaining the
sequence (white, red, white, black)?
The total number of chips is 5 + 4 + 3 = 12.
Use the formula
P (A1 ∩ A2 ∩ . . . ∩ An ) = P (An |A1 ∩ A2 ∩ . . . ∩ An−1 ) × P (An−1 |A1 ∩ A2 ∩ . . . ∩ An−2 )
× . . . × P (A1 )
With A1 = (white chip is drawn on first selection)
A2 = (red chip is drawn on second selection)
A3 = (white chip is drawn on third selection)
A4 = (black chip is drawn on fourth selection)
Thus
P (A1 ∩ A2 ∩ A3 ∩ A4 ) = P (A1 )P (A2 |A1 )
P (A3 |A1 ∩ A2 )P (A4 |A1 ∩ A2 ∩ A3 ∩ A4 )
5 3 4 4
= = 0.02
12 11 10 9
3
12. Medical records show that 0.01% of the general adult population not belonging to a high-risk
group (for example, intravenous drug users) are HIV-positive. Blood tests for the virus
are 99.9% accurate when given to someone infected and 99.99% accurate when given to
someone not infected. What is the probability that a random adult not in a high-risk group
will test positive for HIV?
Let H denote the event “outcome of test is HIV-positive” and I denote the event “person
has HIV, in general adult population”.
Thus
P (H|I) = 0.999
P (H C |I C ) = 0.9999
P (I) = 0.0001
The question asks to find P (H).
P (H) = P (H ∩ I) + P (H ∩ I C )
= P (H|I)P (I) + P (H|I C )P (I C )
= 0.999 · 0.0001 + 0.0001 · 0.9999
= 0.00019989 = 0.02%
13. Josh takes a twenty-question multiple-choice exam where each question has five possible
answers. Some of the answers he knows, while others he selects the answer just by making
lucky guesses. Suppose that the conditional probability of him knowing the answer to a
randomly selected question given that he got it right is 0.92. How many of the twenty
questions was he prepared for?
Let K denote “knowing the answer” and R denote “getting the right answer”.
The following probabilities are given by the question
P (K|R) = 0.92
P (R|K) = 1
P (R|K C ) = 0.2
x
P (K) =
20
The question asks to find x.
According to the Bayes’ theorem
P (R|K)P (K)
P (K|R) =
P (R|K)P (K) + P (R|K C )P (K C )
x
20
x 1 20−x = 0.92
20 + 5 20
Then
x = 0.92 · (x − 0.2x + 4)
(1 − 0.92 · 0.8)x = 0.92 · 4
x = 14.
4
14. Suppose that P (A) = 1/4 and P (B) = 1/8.
(a) What does P (A ∪ B) equal if
i. A and B are mutually exclusive?
ii. A and B are independent?
(b) What does P (A|B) equal if
i. A and B are mutually exclusive?
ii. A and B are independent?
1 1 3
a-i. P (A ∪ B) = P (A) + P (B) = 4 + 8 = 8
a-ii.
P (A ∪ B) = P (A) + P (B) − P (A ∩ B)
= P (A) + P (B) − P (A)P (B)
1 1 11 8+4−1 12 − 1 11
= + − = = =
4 8 48 32 32 32
P (A∩B) 0
b-i. P (A|B) = P (B) = 1 = 0.
8
P (A∩B) P (A)P (B)
b-ii. P (A|B) = P (B) = P (B) = P (A) = 41 .
15. Suppose that two fair dice (one red and one green) are rolled. Define the events
A1 : a 1 or a 2 shows on the red die
A2 : a 3, 4 or 5 shows on the green die
A3 : the dice total is 4, 11, or 12
Show that these events satisfy P (A ∩ A2 ∩ A3 ) = P (A1 )P (A2 )P (A3 ) but it is not true that,
for all i ̸= j, P (Ai ∩ Aj ) = P (Ai )P (Aj ).
We have that the orange rectangle represents A1 , the green rectangle represents A2 , and
the pink blocks represent A3 .
1 12 18 6 6
We have that P (A1 ∩ A2 ∩ A3 ) = P ({1, 3}) = 36 = P (A1 )P (A2 )P (A3 ) = 36 36 36 = 216 =
1
36 and
6 12 18 6
P (A1 ∩ A2 ) = = P (A1 )P (A2 ) = =
36 36 36 36
2 12 6 2
P (A1 ∩ A3 ) = = P (A1 )P (A3 ) = =
36 36 36 36
2 18 6 3
P (A2 ∩ A3 ) = ̸= P (A2 )P (A3 ) = = .
36 36 36 36
