Stat 400/Math 463 Homework 1 Solutions
Spring 2023 - Yu
Exercise 1
Assume that the probability Nick finds Jess crying at 9pm given that she has not watched Dirty Dancing
that evening is 0.2. The probability that Jess is crying at 9pm given that she has watched Dirty Dancing is
0.9. Assume that there is an overall probability of 0.7 that Jess watches Dirty Dancing on any given evening.
a) (1 pt) If Jess is crying at 9pm, find the probability that she has not watched Dirty Dancing that evening.
Solution:
Denote the following Random Variables:
D = Jess Has Watched Dirty Dancing
C = Jess Is Crying At 9pm
Then, P (D) = 0.7 =⇒ P (Dc ) = 0.3, and
P (C|Dc ) = 0.2
P (C|D) = 0.9
P (C |D ) = 0.8
P (C c |D) = 0.1
c
c
By Bayes Theorem,
P (C|D) ∗ P (D)
P (C)
P (C|D) ∗ P (D)
=
P (C|D) ∗ P (D) + P (C|Dc )P (Dc )
(0.9)(0.7)
=⇒ =
(0.9)(0.7) + (0.2)(0.3)
= 0.913
P (D|C) =
(Law of Total Probability)
∴ P (Dc |C) = 1 − 0.913
= 0.0869
b) (1 pt) If Jess is not crying at 9pm, find the probability that she has already had watched Dirty Dancing
that evening.
Solution:
1
P (C c |D) ∗ P (D)
P (C c |D) ∗ P (D) + P (C c |Dc ) ∗ P (Dc )
(0.1)(0.7)
=
(0.1)(0.7) + (0.8)(0.3)
P (D|C c ) =
= 0.2258
Exercise 2
Draw one card at random from a standard deck of cards. The sample space S is the collection of the 52 cards.
Assume that the probability set function assigns 1/52 to each of the 52 outcomes. Let the following events be
defined:
• A = {Queen or King}
• B = {Spade}
• C = {Black Jack or Queen}.
Find:
a) (0.5 pt) P [A ∪ B]
Solution:
P (A ∪ B) = P (A) + P (B) − P (A ∩ B)
8 + 13 − 2
=
52
19
≈ 0.3654
=
51
b) (0.5 pt) P [A ∩ B]
Solution:
Note that A and B are independent events
P (A ∩ B) = P (A) ∗ P (B)
8
1
=
52
4
=
1
≈ 0.0385
26
c) (0.5 pt) P [A ∪ (B ∩ C)]
Solution:
P (A ∪ (B ∩ C)) = P (A) + P (B ∩ C) − P (A ∩ B ∩ C)
8
2
1
=
+
−
52 52 52
9
=
≈ 0.1731
52
*Note: The intersection of B and C is Jack of Spades & Queen of Spades. P [B ∩ C] =
2
2
52 .
d) (0.5 pt) P [AC ∪ B]
Solution:
P (Ac ∪ B) = P (Ac ) + P (B) − P (Ac ∩ (B)
= P (Ac ) + P (B) − P (Ac )P (B)
44 13
44
1
=
+
−
52 52
52
4
=
(because A and B are independent)
23
≈ 0.1731
26
Exercise 3
Suppose S = {1, 2, 3, 4, 5, ...} and for any element of S,
P [k] = c
5k
k!
Find the value of c that makes this a valid probability distribution. (2 pts)
Solution:
Recall:
X
P (s) = 1
s∈S
=⇒
∞
X
P (k) = 1
k=1
∞
X
c
k=1
∞
X
c
k=1
5k
=1
k!
5k
=1
k!
∞
X
5k
50
c·
( − )=1
k!
0!
k=0
c(e5 − 1) = 1
∴c=
1
e5 − 1
≈ 0.006784
Note: we will start using f(x) instead of P(s) notation from here on out
Exercise 4
Suppose S = {2, 3, 4...} and
P [k] =
c
(5/2)k
a) (1.5 pts) Find a value of c that makes this a valid probability distribution.
3
(Exponential Series)
Solution:
∞
X
c
=1
(5/2)k
k=2
∞
X
2 k
=1
5
k=2
(2/5)2
c
=1
1 − (2/5)
4
c
=1
15
c
∴c=
15
4
b) (1 pt) Find P [Outcome is greater than 3].
Solution:
P (K > 3) ⇐⇒ 1 − P (K ≤ 3)
=⇒ P (K ≤ 3) = P (K = 2) + P (K = 3)
15/4
15/4
+
(5/2)2
(5/2)3
21
=
25
=
∴ P (K > 3) = 1 −
=
21
25
4
25
Or. . .
P (K > 3) =
∞
X
15/4
(5/2)k
k=4
∞
15 X 2 k
∗
4
5
k=4
15 (2/5)4
=
4 1 − 2/5
=
=
4
25
Exercise 5
Suppose P [A] = 0.5, P [B C ] = 0.3, and P [A ∩ B] = 0.2
a) (0.5 pts) Find P [B|A]
4
(Geometric Series)
Solution:
P (A ∩ B)
P (A)
0.2
=
0.5
2
=
= 0.4
5
P (B|A) =
b) (0.5 pts) Find P [B C |AC ]
Solution:
P (Ac ∩ B c )
P (Ac )
P (A ∪ B)c
=
P (Ac )
P (B c |Ac ) =
1 − P (A) + P (B) − P (A ∩ B)
=
P (Ac )
1 − (0.5 + 0.7 − 0.2)
=
0.5
= 0
c) (0.5 pts) Find P [AC |B]
Solution:
P (Ac ∩ B)
P (B)
P (B|Ac )P (Ac )
=
P (B)
P (Ac |B) =
1 − P(Bc |Ac ) P (Ac )
=
P (B)
0.5
=
0.7
5
=
≈ 0.7143
7
5