Basic Probability: Problem Set One
Summer 2017
1.3.1 We have A ∩ B ⊂ B ⇒ P (A ∩ B) ≤ P (B) = 13 .
We also have from the inclusion-exclusion principle that
P (A ∩ B) = P (A) + P (B) − P (A ∪ B)
13
=
− P (A ∪ B)
12
13
1
≥
−1=
12
12
since P (A ∪ B) ≤ 1.
For examples of attaining each bound, let Ω = {1, 2, . . . , 12} let F = P (Ω), and let
1
P (ω) = 12
for each ω ∈ Ω. Further, let A = {ω|ω ≤ 9}.
When B = {ω|ω ≥ 9}, P (A ∩ B) =
each case, P (A) = 34 and P (B) = 13 .
1
12
and for B = {ω|ω ≤ 4}, P (A ∩ B) = 13 . In
For comparable bounds on P (A ∪ B), we have A ⊂ A ∪ B so
3
4
= P (A) ≤ P (A ∪ B).
Also,
13
− P (A ∩ B)
12
13
1
≤
−
=1
12 12
P (A ∪ B) =
since P (A ∩ B) ≥
1
.
12
1.4.3 Let 0, 1, and 2 be the events that the coin has 0, 1, and 2 heads, respectively. Let HL
(resp. HU ) be the events that the lower (upper) side is heads. Note that HU ∩ HL = 2
(the event 2) and P (HL ) = P (HU )
P (HL ) = P (HL |0) P (0) + P (HL |1) P (1) + P (HL |2) P (2)
1 1 2
2
= 0· + · +1·
5 2 5
5
3
=
5
P (HL ∩ HU )
P (HU )
P (2)
=
P (HL )
2/5
=
3/5
2
=
3
P (HL |HU ) =
1
Let HU HL be the event that the first toss is heads up and the second is heads down.
P (HU HL |HU ) =
=
=
=
=
P (HU HL ∩ HU )
P (HU )
P (HU HL )
3/5
5
(P (HU HL |0) P (0) + P (HU HL |1) P (1) + P (HU HL |2) P (2))
3
5
1 1 2
2
0· + · +1·
3
5 4 5
5
5
6
P (HU HL |HU HU ) =
=
=
=
=
P (HU HL ∩ HU HU )
P (HU HU )
P (2)
P (HU HU )
P (2)
P (HU HL )
2/5
1/2
4
5
We do the final question in two parts. First, we condition on what the first coin was.
First, suppose the first coin was 1 (it cannot have been 0).
P (HU HU HU |1 ∩ HU HU ) = P (HU HU HU |10 ∩ HU HU ) P (10 ∩ HU HU )
+P (HU HU HU |11 ∩ HU HU ) P (11 ∩ HU HU )
+P (HU HU HU |12 ∩ HU HU ) P (12 ∩ HU HU )
1 1 1
1
= 0· + · +1·
4 2 4
2
5
=
8
Next, suppose it was 2.
P (HU HU HU |1 ∩ HU HU ) = P (HU HU HU |20 ∩ HU HU ) P (20 ∩ HU HU )
+P (HU HU HU |21 ∩ HU HU ) P (21 ∩ HU HU )
+P (HU HU HU |22 ∩ HU HU ) P (22 ∩ HU HU )
1 1 1
1
= 0· + · +1·
4 2 2
4
1
=
2
2
We have P (2|HU HU ) = P (HU HL |HU HU ) = 54 , and since P (0|HU HU ) = 0, we must
have P (1|HU HU ) = 51
Finally, combining the results:
P (HU HU HU |1 ∩ HU HU ) = P (HU HU HU |1 ∩ HU HU ) · P (1|HU HU )
+P (HU HU HU |2 ∩ HU HU ) · P (2|HU HU )
5 1 1 4
=
· + ·
8 5 2 5
21
=
40
1.5.1
P AC P (B) =
=
=
=
(1 − P (A))P (B)
P (B) − P (A) P (B)
P (B) − P (A ∩ B)
P AC ∩ B
By the exact same logic, replacing B with AC and A with B in the equations above,
AC and B C are independent.
1.5.2 For pairwise independence, consider Aij and Ak` , possibly with i = k but definitely
with j 6= `. If the ith roll comes up x and the k th roll is y (possibly with x = y), then
Aij is the event that the j th roll is x and Ak` is the event that the `th roll is y. Since
the rolls are independent, each of these events has conditional probability 61 and the
1
, no matter what x and y are. Thus, the probability
conditional probability of both is 36
1
1
of each event is 6 and the probability of both is 36
.
For independence of all the events, note that having all the pairs be equal is equivalent
to having all the dice be equal, and that there are six ways to get this outcome, each
T
1
1
having probability 6n , so we have P
i<j Aij = 6n−1 . On the other hand, there are
Q
n(n−1)
1
pairs i < j so i<j P (Aij ) = 6n(n−1)/2
. For n > 2, these are not equal to each
2
other.
1.5.9 Let S be the event that the sum is 7. There are 36 equally likely outcomes of the roll,
and six of them ({(n, 7 − n)|n = 1, 2, . . . , 6}) result in a sum of 7. Therefore, P (S) = 61 .
Choose n for 1 ≤ n ≤ 6, and let T be the event first die comes up n. So P (T ) = 61 .
Now S ∩ T is the same as saying the pair come up (n, 7 − n) which has a probability
1
1
of 36
. So we have P (S ∩ T ) = P (S) P (T ) or 16 · 61 = 36
, so the events S and T are
independent.
1.8.1 (a) Define the events: A1 = 6 turns up on first die and A2 = 6 turns up on second
die, A = 6 turns up exactly once.
3
P (A) = P ((A1 ∪ A2 ) − (A1 ∩ A2 ))
= P (A1 ∪ A2 ) − P (A1 ∩ A2 )
= P (A1 ) + P (A2 ) − 2P (A1 ∩ A2 )
1 1
1
=
+ −2·
6 6
36
5
=
18
(b) A1 = first number is odd, A2 = second number is odd. The rolls are independent,
so A1 and A2 are.
P (A1 ∩ A2 ) = P (A1 ) · P (A2 )
1 1
·
=
2 2
1
=
4
(c) There are three ways for this to happen: A = {(1, 3), (2, 2), (3, 1)}. Each way is
1
.
equally as likely as every other of the 36 joint possibilities, so P (A) = 12
(d) Of the numbers 1 through 6, two each are congruent to 0, 1, and 2 modulo 3. If
the first roll is x modulo 3, then regardless of x there is a one-third chance that
the second is congruent to −x modulo 3 (i.e. that their sum is 0 modulo 3). Thus
the probability that the sum is divisible by 3 is 13
1.8.4 Let P(S) denote the power set of a set S. When F = P(Ω) it is sufficient to define
P (ω) for each ω ∈ Ω.
(a) Let p be the probability of heads on a given flip. Let h(ω) for ω ∈ Ω be the
number of total heads (e.g. h(HHT ) = 2).
Ω = {HHH, HHT, HT H, HT T, T HH, T HT, T T H, T T T }
F = P(Ω)
P (ω) = ph(ω) (1 − p)3−h(ω)
(b) Let d(ω) be the indicator that the first and second balls have different colors (1
if different, 0 o.w.). The color of the first ball is one half probability either way,
and for the second ball to have the same color as the first has probability 13 .
Ω = {U U, U V, V U, V V }
F = P(Ω)
1 + d(ω)
P (ω) =
6
4
(c) The probability of heads coming up first on toss n (ω = n) is the probability of
n − 1 tails followed by a heads, in other words (1 − p)n−1 p. The probability of
heads never coming up (ω = ∞, countable intersection of the nested events “tails
comes up on each of the first n tosses”) is the limit of the probabilities of those
events, limn→∞ (1 − p)n , which is 0 if p > 0 and 1 otherwise.
Ω = N∪∞
F = P (Ω)
and P as defined above.
1.8.15 Let Di be the event that the ith die comes up 1.
(a) If S = 4 then N <= 4 so our sum in the following stops at N = 4.
P (S = 4|N = 2) P (N = 2)
P (N = 2|S = 4) = P4
i=1 P (S = 4|N = i) P (N = i)
1
·1
12 4
= 1 1
1
3
1
· + 12
· 14 + 216
· 81 + 614 · 16
6 2
' 0.197
(b) Again, we must have N <= 4. Let E be the event that N is even.
P (S = 4|E) =
=
=
=
=
P (S = 4 ∩ E)
P (E)
P (S = 4|N = 2) P (N = 2) + P (S = 4|N = 4) P (N = 4)
P∞
i=1 P (N = 2i)
1
1
1
1
· + 64 · 16
12P 4
∞
(1/4)i
i=1
1 1
1 1
3
· +
·
12 4 64 16
0.063
(c)
P (S = 4 ∩ D1 = 1|N = 2) P (N = 2)
P (N = 2|S = 4 ∩ D1 = 1) = P4
i=1 P (S = 4 ∩ D1 = 1|N = i) P (N = i)
1
·1
36 4
=
1
2
1
0 · 12 + 36
· 14 + 216
· 18 + 614 · 16
' 0.852
5
(d) Define:
f (r) = P
N
\
!
Di ≤ r
i=1
=
=
∞ X
r n
n=1
∞ X
n=1
6
2−n
r n
12
r/12
1 − r/12
r
=
12 − r
=
Note f (6) = 1. Then the probability that the highest roll is exactly r is f (r) −
12
f (r − 1), or (12−r)(13−r)
1.8.22 Model this problem as randomly ordering the cherries from 1 to 20, letting the first 15
be those that the pig eats, and the sixteenth be the one we pick up of the remaining 5.
(a) This is the same thing as picking one of the 20 cherries at random - the fact that
five of them happen to be in the pig’s stomach is not relevant. So the answer is
5/20, or .25.
(b) By the above, there are 20!/4 total orderings of the cherries having a stone in
place 16. There are 15! · 5! total orderings having all stones in places 16 through
20. Thus the probability the pig ate no stones given cherry 16 is a stone is 204 ,
(5)
or .000258. The probability he did eat a stone given the cherry we pick is a stone,
then is 1 − .000258 = .999742.
1.8.30 Inductive proof that the probability no students have the same birthday is q(m) =
365!
. q(1) = 1, so the base case works. For the inductive step, assume the
(365−m)!365m
formula works for m − 1. Then there are 365 equally likely choices for the mth student
who enters the room, of which 365 − (m − 1) do not intersect with the set of already
represented birthdays. Thus the new probability of no common birthdays is q(m − 1) ·
365−(m−1)
365!
365!
= ((365−(m−1))!/(365−(m−1))365
m−1 365 = (365−m)!365m = q(m).
365
Thus the formula works for all m, and it follows that the probability of the complement
of this event, that at least 2 students have the same birthday, is 1−q(m). In particular,
for m = 23, q(m) = .493 and 1 − q(m) = .507 > 12 .
1.8.33 There are 52
= 2, 598, 960 total hands of poker, so to compute the probability of
5
a given hand type, we simply divide the number of ways to get that hand type by
2,598,960.
First consider the hands other than flushes and straights. We first choose the duplicated
values (i.e. 2,3,. . . ,A), then the unduplicated values, then the suits of each type.
12 4 43
Pair: 13
= 1098240 ways ⇒ p ' .423
1
3
2 1
6
13
2
42 4
11
1
= 123552 ways ⇒ p ' .0475
4 4 2
Three of a Kind: 1 12
= 54912 ways ⇒ p ' .021
2
3 1
12 4 4
Full House: 13
= 3744 ways ⇒ p ' .0014
1
1
3 2
13 12 4 4
Four of a Kind: 1 1 4 1 = 624 ways ⇒ p ' .00024
Two Pair:
2
1
13
For a straight flush, we must first choose the suit (4 ways) then choose the card values
(A, 2, 3, . . . , 10) that begins the straight. This gives 40 possible hands, so p ' .000015.
For a regular flush, we choose the suit (4 ways) and then the values ( 13
ways), giving
5
5148, then subtract off the 40 straight flushes, giving 5108 ways, so p ' 0.0020.
For a regular straight, we choose the starting value (10 ways), then the suits of each
card (45 ways), then again subtract the 40 straight flushes, giving 10200 ways, so
p ' 0.0039.
2.1.2 Let G be the distribution function of Y .
If a = 0 then Y = b so G(y) is 0 for y < b, 1 for all other values of y.
If a > 0 then:
G(y) = P (Y ≤ y)
= P (aX + b ≤ y)
y−b
= P X≤
a
y−b
= F
a
If a < 0 then:
G(y) = P (Y ≤ y)
= P (aX + b ≤ y)
y−b
= P X≥
a
y−b
= 1−P X <
a
y−b
= 1 − lim− P X ≤
+
→0
a
y−b
= 1 − lim− F
+
→0
a
1
2.7.7 The number of passengers not showing up is T Bin(10, 10
) for Teeny Weeny and
1
B Bin(20, 10 ) for Blockbuster. Teeny Weeny is overbooked if T = 0 and Block
9 10
9 20
buster is overbooked if B ≤ 1. P (T = 0) = 10
' .349 and P (B ≤ 1) = 10
+
9 19 1
20 10
' .392, so Teeny Weeny is overbooked slightly less often.
10
2.7.8 The number of heads out of six tosses of a fair coin is Bin(6, 12 ), so the probability of
6
6
5 or more heads is 21 + 6 21 ' .109.
7
3.1.3 This is identical to the situation where we throw each of n coins twice and define
“success” as coming up heads both times. The probability of success for each coin is p2
2
and there are n such coins so the sought random variable is distributed as
and
Bin(n, p 2) n−k
n 2k
the mass function is the associated binomial mass function, f (k) = k p (1 − p ) .
8