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ISyE 6739 — Test 1 Solutions — Summer 2014
This is a take-home test. Please work alone, and don’t spend more than a few hours on
it. Send me an email if you have any questions.
1. TRUE or FALSE? Ā ∪ B̄ ∪ C̄ = A ∩ B ∩ C.
Solution: TRUE (DeMorgan).
♦
2. Suppose that A, B, and C are three independent events such that P(A) = 1/4,
P(B) = 1/6, and P(C) = 1/2. Find the probability that exactly one of the events
will occur.
Solution: We want
P(A ∩ B̄ ∩ C̄) + P(Ā ∩ B ∩ C̄) + P(Ā ∩ B̄ ∩ C)
= P(A)P(B̄)P(C̄) + P(Ā)P(B)P(C̄) + P(Ā)P(B̄)P(C) (by independence)
= (1/4)(5/6)(1/2) + (3/4)(1/6)(1/2) + (3/4)(5/6)(1/2) = 23/48. ♦
3. Suppose that A, B, and C are three disjoint events such that P(A) = 1/4,
P(B) = 1/6, and P(C) = 1/2. Find the probability that exactly one of the events
will occur.
Solution: Since the events are disjoint, the desired probability is simply
P(A) + P(B) + P(C) = 11/12. ♦
4. In a certain population, 70% of the people like The Beatles, 60% like The Zombies,
and 50% like The Rolling Stones. In addition, 50% like The Beatles and The
Zombies, 40% like The Beatles and The Rolling Stones, 40% like The Zoms and
The Stones, and 30% like all three music groups. What percentage of people like
at least one of the three groups?
2
Solution:
P(B) + P(Z) + P(S) − P(BZ) − P(BS) − P(ZS) + P(BZS) = 0.80. ♦
5. Consider the set-up in Problem 4. Find the probability that a random person likes
exactly one of the three groups.
Solution: Several ways to do this. Working from the inside-out of the Venn
diagram, we eventually get
P(B ∩ Z̄ ∩ S̄) + P(B̄ ∩ Z ∩ S̄) + P(B̄ ∩ Z̄ ∩ S) = 0.1 + 0 + 0 = 0.1. ♦
6. TRUE or FALSE? If P(A ∩ B) = P(A)P(B) and P(A ∩ C) = P(A)P(C) and
P(B ∩ C) = P(B)P(C), then A, B, and C are independent.
Solution: FALSE. (We also require P(A ∩ B ∩ C) = P(A)P(B)P(C).)
♦
7. Consider a box containing 4 red sox, 2 blue sox, and 5 green sox. Suppose I select
3 sox from the box. What’s the probability that the third sock is green?
Solution: Since we don’t know what happened with the first two sox, the desired
probability is 5/11. ♦
8. My friends have four kids. I know that there is at least one boy and at least one
girl. What’s the probability that are exactly two boys and two girls?
Solution: Similar to (but a little tougher than) an example in class,
P(2 B’s and 2 G’s | ≥ 1 B and ≥ 1 G)
# ways you can have exactly 2 B’s
=
# ways you can have exactly 1, 2, or 3 B’s
4
=
4
1
+
2
4
2
+
4
3
= 3/7. ♦
3
9. A 4-sided die is thrown 5 times. Find the probability that each face appears at
least once.
Solution: Find all the ways to scramble tosses of the form AABCD, where A
represents the face that is tossed twice
and B, C, D are the other faces. There are
5
4 possible choices for A, and then 2 places in the sequence of 5 tosses to put the
2 A’s. Finally, there are 3! ways to arrange B, C, and D in the remaining slots in
the sequence. Thus, the desired probability is
4 52 3!
= 0.234. ♦
45
10. Pick 6 cards from a standard deck. Find the probability of getting exactly two
triples (3’s-of-a-kind).
Solution: Find all the ways to select the cards. Then the desired probability is
13 4 2
[choose two ranks][choose 3 cards from each rank]
2
3 . ♦
=
52
52
6
6
11. The year on planet Bieber is 40 days long, and all birthdays are equally likely.
There are 6 people in the room. What’s the probability that at least two of them
share the same birthday?
Solution: Following the example in class, the desired probability is
1 − P(nobody has the same birthday)
40(39) · · · (35)
P40,6
= 1−
= 0.3253. ♦
= 1−
6
40
406
n
n
n+1
12. TRUE or FALSE? If n ≥ k, then
+
=
.
k
k−1
k
Solution: TRUE. Here’s the proof.
n!
n!
n
n
+
=
+
k
k−1
k! (n − k)! (k − 1)! (n − k + 1)!
4
n! k
n! (n − k + 1)
+
k! (n − k + 1)! k! (n − k + 1)!
n! (n − k + 1 + k)
=
. ♦
k! (n − k + 1)!
=
13. Suppose I toss 6 dice.
4,5,5,4,2,2?
What’s the probability that I’ll see three pairs, e.g.,
Solution:
The # ways to toss 6 dice is 66 .
The # ways to pick the three ranks is 63 = 20.
The # ways to place the first rank is 62 = 15.
Now there are only 4 slots left. The # ways to place the second rank is 42 = 6.
Now there are only 2 slots left. The # ways to place the third rank is 22 = 1.
Thus, the desired probability is 1800
= 0.0386. ♦
66
14. It snows in beautiful Siberacuse 70% of the time. If it is snowing, then Justin will
wear his boots with probability 0.8. If it is not snowing, then Justin (who is pretty
stupid) will still wear his boots with probability 0.4. If Justin is seen wearing his
boots, what’s the probability that it’s snowing?
Solution: By Bayes,
P(S|B) =
14
(0.8)(0.7)
P(B|S)P(S)
=
. ♦
=
(0.8)(0.7) + (0.4)(0.3)
17
P(B|S)P(S) + P(B|S̄)P(S̄)
15. 4 couples are at a masquerade ball and are dressed up in costumes so that the
partners don’t recognize each other. If the men and women are randomly paired
together for dinner, what’s the probability that at least one pair will be a properly
matched couple?
Solution: This is a straightforward application of the envelope problem. The
probability of a match is
1−
(Note that this answer ≈ 1 −
1
e
1
1
1
+ −
= 0.625.
2! 3! 4!
≈ 0.632.)
♦
5
16. TRUE or FALSE? If X is a continuous random variable with probability density
function f (x), it is possible to have f (x) > 1 for some x.
Solution: TRUE. (f (x) isn’t a probability.)
♦
17. Suppose X is a random variable representing the number of cars entering a
parking lot over a 5-minute interval. Research has revealed that X is Poisson with
parameter λ = 2.5. Find P(−0.5 ≤ X ≤ 2).
Solution: P(X = 0 or 1 or 2) =
P2
x=0
e−2.5 2.5x
x!
= 0.5438.
♦
18. Using the same set-up as in Question 17, find P(0 ≤ X ≤ 1 | 0 ≤ X ≤ 2).
Solution:
P((X = 0 or 1) ∩ (X = 0 or 1 or 2))
P(X = 0 or 1 or 2)
P1 e−2.5 2.5x
P(X = 0 or 1)
x!
=
= Px=0
= 0.5283. ♦
2
e−2.5 2.5x
P(X = 0 or 1 or 2)
x=0
x!
19. Consider the continuous random variable X having p.d.f.
cx if 0 ≤ x ≤ 1
f (x) =
.
0
otherwise
(a) Find c.
Solution: Since 1 =
R
R
f (x) dx =
R1
0
cx dx, we immediately have c = 2.
(b) Find P(0 ≤ X ≤ 0.5).
Solution:
R 0.5
0
2x dx = 0.25.
♦
(c) Find E[X].
Solution:
R
R
xf (x) dx =
R1
0
2x2 dx = 2/3.
♦
♦
6
(d) Find Var(X).
R1
Solution: E[X 2 ] = 0 2x3 dx = 1/2, so we have Var(X) = E[X 2 ] − (E[X])2 =
1/2 − (2/3)2 = 1/18. ♦
(e) Find E[2X − 2].
Solution: 2E[X] − 2 = −2/3.
♦
(f) Find Var(2X − 2).
♦
Solution: 4Var(X) = 2/9.
(g) Use Chebychev’s inequality to get an upper bound on P(|X − E[X]| > 0.25).
Solution: We have
P(|X − E[X]| > 0.25) ≤
Var(X)
1/18
=
= 8/9. ♦
2
1/16
20. If X is a discrete random variable that can take on the values −0.2, 1.0, and 2.5,
each with probability 1/3, find E[sin(πX/2)] (where the sine evaluation is done in
radians.
Solution: By LOTUS,
E[sin(πX/2)] =
X
sin(πx/2)f (x)
x
sin(−0.1 π) + sin(0.5 π) + sin(1.25 π)
3
= −0.0054. ♦
=
21. Suppose X ∼ Unif(0, 1), i.e., f (x) = 1, 0 < x < 1. Find the p.d.f. of Y = e3X+2 .
Solution: The c.d.f. of Y is
G(y) = P(Y ≤ y) = P e3X+2 ≤ y = P(3X + 2 ≤ `n(y))
`n(y) − 2
`n(y) − 2
= P X≤
=
, e2 ≤ y ≤ e5 ,
3
3
7
where the last equality follows because X ∼ Unif(0, 1). Thus, Y has p.d.f.
d
1
g(y) =
G(y) =
, e2 ≤ y ≤ e5 . ♦
dy
3y
22. Suppose X ∼ Unif(0, 1), find the p.d.f. of Y = −`n(X) + 1.
Solution: The c.d.f. of Y is
G(y) = P(Y ≤ y) = P(−`n(X) + 1 ≤ y)
= P X ≥ e1−y = 1 − e1−y , y ≥ 1,
after a bit of algebra and the fact that X ∼ Unif(0, 1). Thus, Y has p.d.f.
d
g(y) =
G(y) = e1−y , y ≥ 1,
dy
which is a shifted Exp(1) distribution. ♦
23. TRUE or FALSE? LOTUS states that E[g(X)] = g(E[X]) for any reasonable
function g(·).
Solution: FALSE. (The equality itself is even false!)
♦
Γ(α+β) α−1
x (1 − x)β−1 for 0 < x < 1, where α and β
24. Suppose X has p.d.f. f (x) = Γ(α)Γ(β)
R ∞ y−1 −t
are positive constants and Γ(y) ≡ 0 t e dt. Find E[X].
Solution: Applying some algebraic elbow grease, we have
Z
E[X] =
xf (x) dx
R
Z
Γ(α + β) ∞ α
=
x (1 − x)β−1 dx
Γ(α)Γ(β) 0
Γ(α + β) Γ(α + 1)Γ(β)
=
(by definition of the beta function)
Γ(α)Γ(β) Γ(α + 1 + β)
Γ(α + β) Γ(α + 1)
α
=
=
,
Γ(α + 1 + β) Γ(α)
α+β
by a well-known property of the gamma function.
♦