Math 180A (P. Fitzsimmons) Final Exam Solutions Each problem is

Math 180A (P. Fitzsimmons) Final Exam Solutions Each problem is

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.
1
(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.
6