1. No category

Math 180A
(P. Fitzsimmons)
Final Exam
Solutions
Each problem is worth 10 points.
1. Suppose that A and B are events with P[B c ] = 7/12, P[B \ A] = 1/6, and P[A \ B] = 1/3.
(a) Find P[A].
(b) Find P[A ∪ B].
(Recall that B \ A denotes the event B ∩ Ac , etc.)
Solution. First observe that P[B] = 1 − P[B c ] = 1 − 7/12 = 5/12. Therefore, because 5/12 =
P[B] = P[AB] + P[B \ A] = P[AB] + 1/6, we have P[AB] = 5/12 − 1/6 = 3/12.
(a) P[A] = P[AB] + P[A \ B] = 3/12 + 1/3 = 7/12.
(b) P[A ∪ B] = P[A] + P[B \ A] = 7/12 + 1/6 = 9/12 = 3/4. [Alternatively, P[A ∪ B] =
P[B] + P[A \ B] = 5/12 + 1/3 = 9/12 = 3/4.]
2. A bowl contains 5 red marbles and 3 green marbles. Marbles are drawn at random from the
bowl, without replacement. Let Rk be the event that the marble chosen on draw k is red, and let
Gk be the event that the marble chosen on draw k is green.
(a) Find the probability that the first red marble to be drawn is the third marble drawn.
(b) Find P[R3 |G1 ].
Solution. (a) This is the same as the probability of G1 G2 R3 , which is
P[G1 G2 R3 ] = P[G1 ]P[G2 |G1 ]P[R3 |G1 G2 ] =
5
3 2 5
· · =
= 0.08928.
8 7 6
56
(b) First,
P[G1 R3 ] = P[G1 G2 R3 ] + P[G1 R2 R3 ]
5
+ P[G1 ]P[R2 |G1 ]P[R3 |G1 R2 ]
=
56
5
3 5 4
=
+ · ·
56 8 7 6
15
.
=
56
Consequently
P[R3 |G1 ] =
15 . 3
5
= = 0.71428.
56 8
7
3. The joint probabilities for a pair of discrete random variables are given by the formula
(
x(x + y)
, x = 1, 2, 3, y = 1, 2, 3;
P(X = x, Y = y) =
78
0,
otherwise.
(a) Find the marginal probabilities P(X = x) and P(Y = y) for x, y = 1, 2, 3.
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(b) Find E[X].
(c) Find E[X|Y = 2].
[You may find it helpful to display the joint probability function in the form of a table.]
Solution. (a) The marginal probabilities for X are
P[X = 1] =
3
X
P[X = 1, Y = j] =
2+3+4
9
=
78
78
P[X = 2, Y = j] =
24
6 + 8 + 10
=
78
78
P[X = 3, Y = j] =
12 + 15 + 18
45
=
.
78
78
j=1
P[X = 2] =
3
X
j=1
P[X = 3] =
3
X
j=1
The marginal probabilities for X are
P[Y = 1] =
3
X
P[X = j, Y = 1] =
2 + 6 + 12
20
=
78
78
P[X = j, Y = 2] =
26
3 + 8 + 15
=
78
78
P[X = j, Y = 3] =
4 + 10 + 18
32
=
.
78
78
j=1
P[Y = 2] =
3
X
j=1
P[Y = 3] =
3
X
j=1
(b) Using (a)
E[X] = 1 · P[X = 1] + 2 · P[X = 2] + 3 · P[X = 3]
9
24
45
=1·
+2·
+3·
78
78
78
9 + 48 + 135
192
32
=
=
=
78
78
13
= 2.46.
(c) Evidently
3
26
8
P[X = 2|Y = 2] =
26
15
P[X = 3|Y = 2] =
26
P[X = 1|Y = 2] =
Therefore
E[X|Y = 2] = 1 ·
8
15
64
32
3
+2·
+3·
=
=
= 2.46.
26
26
26
26
13
4. A fair six-sided die is tossed repeatedly. Let N be the number of different faces that appear in
the first eight rolls. Calculate E(N ). [Hint: Indicator random variables.]
2
Solution. Let Fk be the event that face k shows up at least once in the eight tosses. Since the
complement of Fk is the event that face k is never tossed,
P[Fk ] = 1 − (5/6)8 ,
k = 1, 2, 3, 4, 5, 6.
As hinted, N = IF1 + IF2 + · · · IF6 , and so
E[N ] = E[IF1 ] + · · · E[IF6 ] = 6P[F1 ] = 1(1 − (5/6)8 ) = 4.6046.
5. One thousand cards are drawn (with replacement) from a standard deck of 52 cards. Let X be
the total number of Kings drawn.
(a) Find E[X] and Var[X].
(b) Use the normal approximation to compute P[65 ≤ X ≤ 90].
Solution. (a) The random variable X has the binomial distribution with parameters n = 1000 and
p = 1/13. Therefore µ = E[X] = 1000/13 = 76.92 and Var[X] = 1000(1/13)(12/13) = 71.01. The
√
associated standard deviation is σ = 71.01 = 8.43.
(b) Using the normal approximation to the binomial (including the continuity correction):
64.5 − 76.92
90.5 − 76.92
−Φ
P[65 ≤ X ≤ 90] = Φ
8.43
8.43
= Φ(1.61) − Φ(−1.47) = Φ(1.61) − 1 + Φ(1.47)
= .9463 − 1 + .9292
= .8755.
6. Customers arrive at Bob’s Bargain Bookstore in accordance with a Poisson process of intensity
10 customers per hour. The store opens at 8 am, but Bob doesn’t arrive until 10 am.
(a) What is the average number of customers that arrive before Bob gets to his store (that
is, between 8 am and 10 am)?
(b) What is the probability that no customer arrives before Bob?
(c) Let T be the amount of time (in hours) that Bob waits until he sees a customer walk in
the door. Find P[T > x] for x > 0.
Solution. Let N (s, t] denote the number of customers arriving in the time interval (s, t] (with time
being measured in hours after opening). We know that N (s, t] has the Poisson distribution with
expectation 10 · (t − s).
(a) This is E[N (0, 2]] = 10 · (2 − 0) = 20.
(b) P[N (0, 2] = 0] = exp(−20) = 2 × 109 .
((c) P[T > x] = P[N (2, 2 + x] = 0] = e−10x , for x > 0. (In other words, T has the exponential
distribution with parameter 10.)
3
7. A random variable X has the continuous-type density function
3
− x2 ), 0 < x < 1;
0,
otherwise.
2 (1
f (x) =
(a) Find P[1/3 < X ≤ 2/3].
(b) Find E[X].
(c) Find Var[X].
Solution. (a)
Z
2/3
3
(1 − x2 ) dx
1/3 2
2/3
3
= (x − x3 /3)
2
1/3
3
2 8/27
1 1/27
=
−
−
−
2
3
3
3
3
10
= 0.3704.
=
27
P[1/3 < X ≤ 2/3] =
(b)
Z
1
3
3
E[X] =
x · (1 − x2 ) dx =
2
2
0
3
3 1 1
−
= .
=
2 2 4
8
1
Z
(x − x3 ) dx
0
(c) First,
Z
1
3
3
x2 · (1 − x2 ) dx =
2
2
0
3 1 1
1
=
−
= .
2 3 5
5
E[X 2 ] =
Consequently, Var[X] = (1/5) − (3/8)2 =
19
320
Z
1
(x2 − x4 ) dx
0
= 0.059375.
8. Suppose that the random variable X has the exponential distribution with parameter 2. Let
Y = X 3.
(a) Find the density function fY for Y .
(b) Find the cumulative distribution function FY for Y .
Solution. (a) We are told that fX (x) = 2e−2x for x > 0. Setting y = x3 , we have x = y 1/3 and
dx/dy = 3y12/3 . Therefore
fY (y) = fX (y 1/3 ) ·
1/3
1
2
= 2/3 e−2y ,
2/3
3y
3y
4
y > 0.
(b) Integrating, we find
FY (y) = P[Y ≤ y]
Z y
=
fY (t) dt
0
Z y
2 −2t1/3
=
e
dt.
2/3
0 3t
Making the change of variables u = t1/3 (so that t = u3 and dt = 3u2 du) this integral becomes
Z
y 1/3
2e−2u du = (1 − e−2y
1/3
),
0
for y > 0. Alternatively
FY (y) = P[Y ≤ y] = P[X 3 ≤ y] = P[X ≤ y 1/3 ] = FX (y 1/3 ) = (1 − e−2y
1/3
),
for y > 0.
9. A random point (X, Y ) is chosen from unit square with continuous-type joint density function
c(3x2 + 2y), 0 ≤ x ≤ 1, 0 ≤ y ≤ 1;
f (x, y) =
0,
otherwise.
(a) Find c.
(b) Find the marginal density function fY for Y .
(c) Find E(Y ).
Solution. (a) The double integral of the density must be equal to 1; therefore
1
Z
Z
1=c
1
1
Z
2
(3x + 2y) dx dy =
0
0
(1 + 2y) dy = 1 + 1 = 2,
0
which means that c = 1/2.
(b) Integrate:
1
Z
fY (y) =
f (x, y) dx =
0
1
Z
1
2
(3x2 + 2y) dx
0
1
1
= (1 + 2y) = + y,
2
2
for 0 < y < 1.
(c)
Z
1
E[Y ] =
Z
yfY (y) dy =
0
Z
=
(y/2 + y 2 ) dy
0
=
y(1/2 + y) dy
)
1
1 1
7
+ =
.
4 3
12
5
1
10. Suppose X and Y are independent random variables such that X has uniform (0, 1) distribution, and Y has exponential distribution with mean 1. Calculate:
(a) E(X + Y );
(b) E(XY );
(c) E (X − Y )2 . endproclaim
R1
Solution. Observe that E[X] = 1/2 and E[Y ] = 1. Also, E[X 2 ] = 0 x2 dx = 1/3 and E[Y 2 ] =
Var[Y ] + (E[Y ])2 = 1 + 12 = 2.
(a) E[X + Y ] = E[X] + E[Y ] = 1/2 + 1 = 3/2.
(b) E[XY ] = E[X] · E[Y ] = (1/2) · 1 = 1/2.
((c) E (X − Y )2 = E[X 2 − 2XY + Y 2 ] = E[X 2 ] − 2E[XY ] + E[Y 2 ] = 1/3 − 2(1/2) + 2 = 4/3.
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