Math 3090 An Introduction to Probability
Final Exam — Solutions
1. A continuous random variable X has cdf
for x ≤ 0, for 0
F(x) = < x < 1, for x ≥
ax2 1.
b
(a) Determine the constants a and b.
(b) Find the pdf of X. Be sure to give a formula for fX(x) that is valid for all x.
(c) Calculate the expected value of X.
(d) Calculate the standard deviation of X.
Answer:
(a) We must have a = limx→−∞ = 0 and b = limx→+∞ = 1, since F is a cdf. (b) For all
x = 06 or 1, F is differentiable at x, so
f(x) = F 0(x) = ( 2x if 0 < x < 1, 0
otherwise.
(One could also use any f that agrees with this definition for all x = 06 or 1.)
.
0556,
and
2. Suppose the number of children born in Ruritania each day is a binomial random
variable with mean 1000 and variance 100. Assume that the number of children born on
any particular day is independent of the numbers of children born on all other days. What
is the probability that on at least one day this year, fewer than 975 children will be born in
Ruritania?
,Answer: We approximate the number of children born each day by a normal random
variable. Letting X denote the number of children born on some specified day, and Z denote
a standard normal, we have
9946. Since each such random variable X (one for
each day) is assumed independent of the others, the probability that 975 or more children
will be born on every day of this year is .9946365 ≈ .1386, and the probability that, on at
least one day this year, fewer than 975 children will be born is close to 1 − .1386 ≈ 86%.
3. Suppose that the time until the next telemarketer calls my home is distributed as an
exponential random variable. If the chance of my getting such a call during the next hour is
.5, what is the chance that I’ll get such a call during the next two hours?
Answer: First solution: Letting λ denote the rate of this exponential random variable X, we
have .5 = FX(1) = 1−e−λ, so λ = ln2 and FX(2) = 1−e−2λ = 1−(e−λ)2 = 1−(.5)2 = .75. Second
solution: We have P(X ≤ 2) = P(X ≤ 1) + P(1 < X ≤ 2). The first term is .5, and the second can
be written as P(X > 1 and X ≤ 2) = P(X > 1)P(X ≤ 2 | X > 1). The first of these factors equals 1
− P(X ≤ 1) = 1 − .5 = .5, and the second (by virtue of the memorylessness of the exponential
random variable) equals P(X ≤ 1) = .5. So P(X ≤ 2) = .5 + (.5)(.5) = .75.
4. Suppose X is uniform on the interval from 1 to 2. Compute the pdf and expected
value of the random variable Y = 1/X. Answer: We have
fX(x) = ( 1 if 1 < x < 2,
0 otherwise.
Putting g(t) = 1/t we have Y = g(X); since g is monotone on the range of X with inverse
function , Theorem 7.1 tells us that
if 1,
Yy
0 otherwise.
(Check: = 1.) We have
(Check:
5. I toss 3 fair coins, and then re-toss all the ones that come up tails. Let X denote the
number of coins that come up heads on the first toss, and let Y denote the number of re-
tossed coins that come up heads on the second toss. (Hence 0 ≤ X ≤ 3 and
0 ≤ Y ≤ 3 − X.)
, (a) Determine the joint pmf of X and Y , and use it to calculate E(X + Y ).
(b) Derive a formula for E(Y |X) and use it to compute E(X + Y ) as E(E(X + Y |X)).
Answer:
(a) P(X = j,Y = k) equals
whenever 0 ≤ j ≤ 3 and 0 ≤ k ≤ 3 − j (and
equals zero otherwise), so the joint pmf f = fX,Y has the following values:
Hence .
(Alternatively: X + Y is the total number of coins that come up heads on the first toss or,
failing that, heads on the re-toss. Each of the three coins has a chance of contributing 1
to this total, so by linearity of expectation, the expected value of the total is + + = .
(b) For each fixed x (0 ≤ x ≤ 3), when we condition on the event X = x, Y is just a binomial
random variable with
, and therefore with expected value ). Hence
.
6. Let the continuous random variables X,Y have joint distribution
( 1/x if 0 < y < x < 1,
fX,Y (x,y) =
0 otherwise.
(a) Compute E(X) and E(Y ).
(b) Compute the conditional pdf of Y given X = x, for all 0 < x < 1.
(c) Compute E(Y |X = x) for all 0 < x < 1, and use this to check your answers to part
(a).
(d) Compute Cov(X,Y ).
Answer:
, .
(b) The marginal pdf for , which equals
for 0 < x < 1 (and equals zero otherwise). That is, X is uniform on the interval from 0
to 1. Hence for each 0 < x < 1, the conditional pdf for Y given X = x is fY |X(y|x) = fX,Y
(x,y)/fX(x), which is for 0 < y < x and 0 otherwise.
. (We can also derive this answer
from the fact that the conditional distribution of Y given X = x was shown in (b) to be
uniform on the interval (0,x), and from the fact that the expected value of a random
variable that is uniform on an interval is just the midpoint of the interval.)
To check the formula for E(Y |X), we re-calculate
), which agrees with .
, so Cov(X,Y ) = E(XY )−E(X)E(Y ) =
7. I repeatedly roll a fair die. If it comes up 6, I instantly win (and stop playing); if it
comes up k, for any k between 1 and 5, I wait k minutes and then roll again. What is the
expected elapsed time from when I start rolling until I win? (Note: If I win on my first roll,
the elapsed time is zero.)
Answer: Let T denote the (random) duration of the game, and let X be the result of the first
roll. Then
5)+
)+15), so 6E(T) = 5E(T) + 15 and E(T) = 15 (minutes).
8. Suppose that the number of students who enroll in Math 431 each fall is known (or
believed) to be a random variable with expected value 90. It does not appear to be normal,
so we cannot use the Central Limit Theorem.
(a) If we insist on being 90% certain that there will be no more than 35 students in
eachsection, should UW continue to offer just three sections of Math 431 each fall, or
would our level of aversion to the risk of overcrowding dictate that we create a fourth
section?
(b) Repeat part (a) under the additional assumption that the variance in the
enrollmentlevel is known to be 20 (with no other additional assumptions).
Answer:
(a) Since we do not know the variance, the best we can do is use Markov’s inequality:
85; this is much bigger than .10, so to be on the safe side we
should create a fourth section.
Final Exam — Solutions
1. A continuous random variable X has cdf
for x ≤ 0, for 0
F(x) = < x < 1, for x ≥
ax2 1.
b
(a) Determine the constants a and b.
(b) Find the pdf of X. Be sure to give a formula for fX(x) that is valid for all x.
(c) Calculate the expected value of X.
(d) Calculate the standard deviation of X.
Answer:
(a) We must have a = limx→−∞ = 0 and b = limx→+∞ = 1, since F is a cdf. (b) For all
x = 06 or 1, F is differentiable at x, so
f(x) = F 0(x) = ( 2x if 0 < x < 1, 0
otherwise.
(One could also use any f that agrees with this definition for all x = 06 or 1.)
.
0556,
and
2. Suppose the number of children born in Ruritania each day is a binomial random
variable with mean 1000 and variance 100. Assume that the number of children born on
any particular day is independent of the numbers of children born on all other days. What
is the probability that on at least one day this year, fewer than 975 children will be born in
Ruritania?
,Answer: We approximate the number of children born each day by a normal random
variable. Letting X denote the number of children born on some specified day, and Z denote
a standard normal, we have
9946. Since each such random variable X (one for
each day) is assumed independent of the others, the probability that 975 or more children
will be born on every day of this year is .9946365 ≈ .1386, and the probability that, on at
least one day this year, fewer than 975 children will be born is close to 1 − .1386 ≈ 86%.
3. Suppose that the time until the next telemarketer calls my home is distributed as an
exponential random variable. If the chance of my getting such a call during the next hour is
.5, what is the chance that I’ll get such a call during the next two hours?
Answer: First solution: Letting λ denote the rate of this exponential random variable X, we
have .5 = FX(1) = 1−e−λ, so λ = ln2 and FX(2) = 1−e−2λ = 1−(e−λ)2 = 1−(.5)2 = .75. Second
solution: We have P(X ≤ 2) = P(X ≤ 1) + P(1 < X ≤ 2). The first term is .5, and the second can
be written as P(X > 1 and X ≤ 2) = P(X > 1)P(X ≤ 2 | X > 1). The first of these factors equals 1
− P(X ≤ 1) = 1 − .5 = .5, and the second (by virtue of the memorylessness of the exponential
random variable) equals P(X ≤ 1) = .5. So P(X ≤ 2) = .5 + (.5)(.5) = .75.
4. Suppose X is uniform on the interval from 1 to 2. Compute the pdf and expected
value of the random variable Y = 1/X. Answer: We have
fX(x) = ( 1 if 1 < x < 2,
0 otherwise.
Putting g(t) = 1/t we have Y = g(X); since g is monotone on the range of X with inverse
function , Theorem 7.1 tells us that
if 1,
Yy
0 otherwise.
(Check: = 1.) We have
(Check:
5. I toss 3 fair coins, and then re-toss all the ones that come up tails. Let X denote the
number of coins that come up heads on the first toss, and let Y denote the number of re-
tossed coins that come up heads on the second toss. (Hence 0 ≤ X ≤ 3 and
0 ≤ Y ≤ 3 − X.)
, (a) Determine the joint pmf of X and Y , and use it to calculate E(X + Y ).
(b) Derive a formula for E(Y |X) and use it to compute E(X + Y ) as E(E(X + Y |X)).
Answer:
(a) P(X = j,Y = k) equals
whenever 0 ≤ j ≤ 3 and 0 ≤ k ≤ 3 − j (and
equals zero otherwise), so the joint pmf f = fX,Y has the following values:
Hence .
(Alternatively: X + Y is the total number of coins that come up heads on the first toss or,
failing that, heads on the re-toss. Each of the three coins has a chance of contributing 1
to this total, so by linearity of expectation, the expected value of the total is + + = .
(b) For each fixed x (0 ≤ x ≤ 3), when we condition on the event X = x, Y is just a binomial
random variable with
, and therefore with expected value ). Hence
.
6. Let the continuous random variables X,Y have joint distribution
( 1/x if 0 < y < x < 1,
fX,Y (x,y) =
0 otherwise.
(a) Compute E(X) and E(Y ).
(b) Compute the conditional pdf of Y given X = x, for all 0 < x < 1.
(c) Compute E(Y |X = x) for all 0 < x < 1, and use this to check your answers to part
(a).
(d) Compute Cov(X,Y ).
Answer:
, .
(b) The marginal pdf for , which equals
for 0 < x < 1 (and equals zero otherwise). That is, X is uniform on the interval from 0
to 1. Hence for each 0 < x < 1, the conditional pdf for Y given X = x is fY |X(y|x) = fX,Y
(x,y)/fX(x), which is for 0 < y < x and 0 otherwise.
. (We can also derive this answer
from the fact that the conditional distribution of Y given X = x was shown in (b) to be
uniform on the interval (0,x), and from the fact that the expected value of a random
variable that is uniform on an interval is just the midpoint of the interval.)
To check the formula for E(Y |X), we re-calculate
), which agrees with .
, so Cov(X,Y ) = E(XY )−E(X)E(Y ) =
7. I repeatedly roll a fair die. If it comes up 6, I instantly win (and stop playing); if it
comes up k, for any k between 1 and 5, I wait k minutes and then roll again. What is the
expected elapsed time from when I start rolling until I win? (Note: If I win on my first roll,
the elapsed time is zero.)
Answer: Let T denote the (random) duration of the game, and let X be the result of the first
roll. Then
5)+
)+15), so 6E(T) = 5E(T) + 15 and E(T) = 15 (minutes).
8. Suppose that the number of students who enroll in Math 431 each fall is known (or
believed) to be a random variable with expected value 90. It does not appear to be normal,
so we cannot use the Central Limit Theorem.
(a) If we insist on being 90% certain that there will be no more than 35 students in
eachsection, should UW continue to offer just three sections of Math 431 each fall, or
would our level of aversion to the risk of overcrowding dictate that we create a fourth
section?
(b) Repeat part (a) under the additional assumption that the variance in the
enrollmentlevel is known to be 20 (with no other additional assumptions).
Answer:
(a) Since we do not know the variance, the best we can do is use Markov’s inequality:
85; this is much bigger than .10, so to be on the safe side we
should create a fourth section.
