NAME
MATH 461 - F13, Test 2, Fall 2015
November 13, 2015
Books, notes and extra papers are not allowed on this test
Please show all your work and explain all answers to qualify for full credit
1. (15 points) Consider a roulette wheel consisting of 38 numbers 1 through 36, 0, and
double 0. If Smith always bets that the outcome will be one of the numbers 1 through
12, what is the probability that
(a) Smith will lose his first 4 bets;
(b) His first win will occur on his fifth bet?
(c) Find the expected number of bets until his first win (including the winning bet).
Solution: First note that the probability of winning in one bet is 12/38 = 6/19.
4
= probability of losing 4 times in a row
(a) 13
19
4 6 (b) 13
= probability of losing in the first four bets and then winning
19
19
(c) Let X denote the number of bets until the first win. Then X has geometric
distribution with parameter p = 12/38 = 6/19. Therefore, EX = 19/6.
2. (15 points) Suppose that X is uniformlydistributed
over the interval (−2, 2). Find the
|X|
probability density function of Y = log 1 − 2 .
Solution: First note that the p.d.f. of X is equal to 1/4 for x ∈ (−2, 2) and 0 otherwise.
Next, X ∈ (−2, 2) =⇒ |X| ∈ [0, 2) =⇒ 1 − |X|/2 ∈ (0, 1] =⇒ log(1 − |X|/2) ≤ 0.
Thus, for a ≤ 0,
P(Y ≤ a) = P(log(1 − |X|/2) ≤ a) = P(1 − |X|/2 ≤ ea ) = P(|X| ≥ 2(1 − ea ))
1
= 2P(X ≥ 2(1 − ea )) = 2 (2 − 2(1 − ea )) = ea .
4
It follows that
fY (a) =
ea , a ≤ 0
0, a > 0
3. (10 points) The time (in hours) required to repair a machine is an exponential random
variable with parameter λ = 41 . Find (a) the probability that the repair will be shorter
than 4 hours; (b) the conditional probability that the repair time is at least 8 hours,
given that its duration exceeds 6 hours.
Solution: Let X = the time in hours required to repair a machine. Then X is exponential with parameter λ = 1/4.
(a) P(X < 4) = 1 − e−4λ = 1 − e−1
(b) P(X > 8 | X > 6) = memoryless property = P(X > 2) = e−1/2
4. (15 points) A certain basketball player knows that on average he will make 90 percent
of free throw attempts. Use normal approximation to find the probability that in 100
attempts he will be successful at most 85 times.
Solution: Let X = the number of free throw attempts the player will make. Then X
is binomial with parameters (100, 0.9). We approximate X with the normal variable
with parameters (100(0.9), 100(0.9)(0.1)) = (90, 9).
85.5 − 90
X − 90
√
≤
)
P(X ≤ 85) = (continuity correction) = P(X ≤ 85.5) = P( √
9
9
4.5
≈ P(Z ≤ − ) = P(Z ≤ −1.5) = Φ(−1.5) = 1 − Φ(1.5)
3
≈ 1 − 0.93319 = 0.06681
5. (15 points) The joint density function of X and Y is
x + y 0 < x < 1, 0 < y < 1,
f (x, y) =
0
otherwise.
(a) Find P(X + Y < 1);
(b) Find EY ;
(c) Find Var(Y ).
Solution:
(a)
ZZ
Z
P(X + Y < 1) =
1
Z
(x + y) dy dx =
(x + y) dy dx
x+y<1
1
y2
xy +
2
0
1−x
1−x
1
0
(1 − x)2
=
=
x(1 − x) +
dx
2
0
0
0
3
1
Z 1 2
x
1
x
x
1 1
1
=
− +
dx = − +
=− + =
2
2
6
2 0
6 2
3
0
Z
Z
(b)
1
1
1
xy 2 y 3
=
y(x + y) dy dx =
+
2
3
0
0
0
2
1
Z 1
x 1
x
x
7
=
+
dx =
+
=
2 3
4
3 0 12
0
Z
EY
Z
Z
1
0
(c)
EY
2
Z
1
1
Z
1
Z
2
y (x + y) dy dx =
=
0
Z
0
0
1
0
1
2
1
x
x
5
+
=
6
4 0 12
0
2
5
7
11
−
=
Var(Y ) = EY 2 − (EY )2 =
12
12
144
=
x 1
+
3 4
xy 3 y 4
+
3
4
dx =
6. (15 points) The lifetime of light bulb A has exponential distribution with expectation 15
hours, while the lifetime of light bulb B has exponential distribution with expectation
18 hours. If both light bulbs are switched on at the same time, find the probability
that light bulb A lives longer.
Solution: Let X be the lifetime of A and Y the lifetime of B. Then X is exponential(λ),
with λ = 1/15, Y is exponential(µ) with µ = 1/18. By independence, the joint density
function of X and Y is f (x, y) = λe−λx µe−µy . Hence
Z Z
Z ∞Z ∞
−λx
−µy
P(Y ≤ X) =
λe µe
dx dy =
λe−λx µe−µy dx dy
y≤x
Z
e
=
0
0
∞
−λy
−µy
µe
µ
dy =
=
λ+µ
y
1
18
1
15
+
1
18
=
10
180
12
180
+
10
180
=
5
.
11
7. (15 points) A man and a woman agree to meet at a certain location about 12:30 p.m.
Assume that the man arrives at a time uniformly distributed between 12:15 and 12:45,
and the woman independently arrives at a time uniformly distributed between 12:00
and 1 p.m. Given that the woman arrives after 12:30, find the conditional probability
that she will still have to wait for at least five minutes.
Solution: Let X be the time man arrives and Y the time the woman arrives. We may
assume that X is uniform on (15, 45) and Y uniform on (0, 60). It is assumed that X
and Y are independent. We need to compute the conditional probability
P(X > Y + 5 | Y > 30) =
P(X > Y + 5, Y > 30)
.
P(Y > 30)
1
1
Clearly, P(Y > 30) = 1/2. The joint density of X and Y is f (x, y) = 30
× 60
for
15 < x < 45, 0 < y < 60, and zero otherwise. To compute the numerator it helps to
draw the picture. We have
ZZ
Z 40 Z 45
1
dx dy
P(X > Y + 5, Y > 30) =
=
30<y<60,y+5<x<45
30
y+5 1800
40
Z 40
1
1
y2
=
(40 − y)dy =
40y −
1800 30
1800
2 30
1
1
=
50 =
.
1800
36
(This can be also computed by drawing a picture and finding the area of the appropriate
1
.
region.) Finally, the required conditional probability is 18