Chapter 7 Answers
Question 1
Alice and Bob arrange to meet for lunch on a certain day at noon. However, neither is
known for punctuality. They both arrive independently at uniformly distributed times
between noon and 1 pm on that day. Each is willing to wait up to 15 minutes for the
other to show up. What is the probability they will meet for lunch that day?
Question
$$
2(.75 \cdot .25) + .25^2 = .4375
$$
Answer
There are \(2\) people, they both have a \(75%\) chance of arriving before \(12:45\).
The next person has a \(25%\) chance of arriving \(15\) within minutes after the first person
There is a \(25^{2}\) chance that both of them arrive after \(12:45\).
Explanation
Question 2
Alice, Bob, and Carl arrange to meet for lunch on a certain day. They arrive indepen-
dently at uniformly distributed times between 1 pm and 1:30 pm on that day.
(a) What is the probability that Carl arrives first?
For the rest of this problem, assume that Carl arrives first at 1:10 pm, and condition on
this fact.
(b) What is the probability that Carl will have to wait more than 10 minutes for one of
the others to show up? (So consider Carl’s waiting time until at least one of the others
has arrived.)
(c) What is the probability that Carl will have to wait more than 10 minutes for both
of the others to show up? (So consider Carl’s waiting time until both of the others has
arrived.)
(d) What is the probability that the person who arrives second will have to wait more
than 5 minutes for the third person to show up?
Question
A)
$$
\dfrac{1}{3}
$$
B) I am assuming the question is asking for Carl to wait for exactly one person, not at least one.
$$
\dfrac{1}{2}
$$
C)
$$
\dfrac{1}{4}
$$
D)
$$
\dfrac{3}{4} \cdot \dfrac{1}{3} = \dfrac{1}{4}
$$
Answer
A) pretty self-explanatory
B) break the remaining \(20\) minutes into two \(10\) minute segments. There are \(4\) ways that you could place the \(2^{nd}\) and \(3^{rd}\) people into.
There are \(2\) ways for one person to arrive before \(10\) minutes and the other to arrive \(10\) minutes. There is one way for both people to arrive after \(10\) minutes
and one way for both to show up before \(10\) minutes.
C) Using the same logic as part 2, except this time we want to know the probability that both people are in the second segment.
D) The \(2^{nd}\) person must arrive within the first \(15\) minutes of the remaining \(20\) minutes, otherwise it would be impossible to wait more than \(5\) minutes.
The probability that the second person arrives within the first \(15\) minutes is \(\dfrac{3}{4}\), and the probability that the third person arrives \(5\) minutes after that is \(\dfrac{1}{4}\)
Explanation
Question 3
One of two doctors, Dr. Hibbert and Dr. Nick, is called upon to perform a series of
n surgeries. Let H be the indicator r.v. for Dr. Hibbert performing the surgeries, and
suppose that E(H) = p. Given that Dr. Hibbert is performing the surgeries, each surgery
is successful with probability a, independently. Given that Dr. Nick is performing the
surgeries, each surgery is successful with probability b, independently. Let X be the
number of successful surgeries.
(a) Find the joint PMF of H and X.
(b) Find the marginal PMF of X.
(c) Find the conditional PMF of H given X = k.
Question
Question 4
A fair coin is flipped twice. Let X be the number of Heads in the two tosses, and Y be
the indicator r.v for the tosses landing the same way.
(a) Find the joint PMF of X and Y.
(b) Find the marginal PMFs of X and Y.
(c) Are X and Y independent?
(d) Find the conditional PMFs of Y given X = x and of X given Y = y.
Question
A)
$$
f(x,y) = P(X = x \cap Y = y) = \begin{cases}
\dfrac{1}{4} & x = 0, & y = 1 \\
\dfrac{1}{2} & x = 1, & y = 0 \\
\dfrac{1}{4} & x = 2, & y = 1 \\
\end{cases}
$$
B)
$$
P(X = x) = \sum_{Y} P(X = x \cap Y = y) = \begin{cases}
\dfrac{1}{4} & x = 0 \\
\dfrac{1}{2} & x = 1 \\
\dfrac{1}{4} & x = 2 \\
\end{cases}
$$
$$
P(Y = y) = \sum_{X} P(X = x \cap Y = y) = \begin{cases}
\dfrac{1}{2} & y = 0 \\
\dfrac{1}{2} & y = 1
\end{cases}
$$
C)
No, \(X\) and \(Y\) are not independent. If \(X\) is \(1\), then \(Y\) must be \(0\), even though \(Y\) is \(1\) \(50\)% of the time when we do not know \(Y\).
D)
X dependent on Y
$$
P(X = 0 \vert Y = 1) = .5
$$
$$
P(X = 1 \vert Y = 0) = 1
$$
$$
P(X = 2 \vert Y = 1) = .5
$$
all other values for X dependent on Y are zero
Y dependent on X
$$
P(Y = 1 \vert (X = 0) \cup (X = 2)) = 1
$$
$$
P(Y = 0 \vert X = 1) = 1
$$
all other values for Y dependent on X are zero
Answer
Question 5
A fair die is rolled, and then a coin with probability p of Heads is flipped as many times
as the die roll says, e.g., if the result of the die roll is a 3, then the coin is flipped 3
times. Let X be the result of the die roll and Y be the number of times the coin lands
Heads.
(a) Find the joint PMF of X and Y . Are they independent?
(b) Find the marginal PMFs of X and Y.
(c) Find the conditional PMFs of Y given X = x and of Y given X = x.
Question
A)
$$
f_{X,Y}(x,y) = P(X = x, Y = y) = \dfrac{ {x \choose y} p^y (1-p)^{x-y} }{6}
$$
Where \(1 \le x \le 6\) and \(y \le x\)
They cannot be independent because a higher value of \(x\) allows a higher value for \(y\).
B)
marginal probability for \(X\)
$$
f_X(x) = P(X = x) = \sum_{y \in Y} P(X = x,Y = y) = \dfrac{1}{6}
$$
with support \(X \in [1,6]\)
marginal probability for \(Y\)
$$
f_Y(y) = P(Y = y) = \sum_{x \in X} P(X = x, Y = y) = \sum_{n=x}^{6} {n \choose y} p^y (1-p)^{x-y}
$$
Answer
A)
The value for \(X\) is chosen uniformly at random within the range \([1,6]\).
The value for \(Y\) is chosen randomly from the binomial distribution.
Explanation
Question 6
A committee of size k is chosen from a group of n women and m men. All possible
committees of size k are equally likely. Let X and Y be the numbers of women and men
on the committee, respectively.
(a) Find the joint PMF of X and Y . Be sure to specify the support.
(b) Find the marginal PMF of X in two different ways: by doing a computation using
the joint PMF, and using a story.
(c) Find the conditional PMF of Y given that X = x.
Question
A)
$$
P_{X,Y}(x,y) = P(X = x, Y = y) = \dfrac{ {n \choose X} { m \choose Y}}{ {n+m \choose X+Y} }
$$
With support \(0 \le X \le n\) and \(0 \le Y \le m\) and \(X+Y = k\)
B)
$$
P(X) = \sum_{y \in Y} = \dfrac{ {n \choose X} { m \choose Y}}{ {n+m \choose X+Y} }
$$
This is the same as joint probability because there is exaclty one \(Y\) value for each \(X\) value.
C)
$$
P(Y = (k-x) | X = x) = 1
$$
Answer
A)
We use hypergeometric for this type of problem.
B)
Because \(X+Y=k\), we know exactly what one is once we know the other.
C)
As stated in part B, once we know \(X\), we also know \(Y\). So there is only one \(X\) value for each \(Y\) value.
In other words, we must select \(n\) women and \(m\) men, where \(n + m = k\). So there is exactly one \(n\) value for each \(m\) value and vis-versa.
Explanation
Question 7
A stick of length L (a positive constant) is broken at a uniformly random point X.
Given that X = x, another breakpoint Y is chosen uniformly on the interval [0, x].
(a) Find the joint PDF of X and Y . Be sure to specify the support.
(b) We already know that the marginal distribution of X is Unif(0, L). Check that
marginalizing out Y from the joint PDF agrees that this is the marginal distribution of
X.
(c) We already know that the conditional distribution of Y given X = x is Unif(0, x).
Check that using the definition of conditional PDFs (in terms of joint and marginal
PDFs) agrees that this is the conditional distribution of Y given X = x.
(d) Find the marginal PDF of Y .
(e) Find the conditional PDF of X given Y = y.
Question
A)
$$
f_{X,Y}(x,y) = f_X(x) f_Y(y) = \dfrac{1}{yx}
$$
With support \(X \in [0,L]\) and \(Y \in [0,X]\)
E)
$$
P(X = x \vert Y = y) = \dfrac{1}{L-Y}
$$
Answer
Question 8
a) Five cards are randomly chosen from a standard deck, one at a time with replacement.
Let X, Y, Z be the numbers of chosen queens, kings, and other cards. Find the joint PMF
of X, Y, Z.
(b) Find the joint PMF of X and Y .
Hint: In summing the joint PMF of X, Y, Z over the possible values of Z, note that most
terms are 0 because of the constraint that the number of chosen cards is five.
(c) Now assume instead that the sampling is without replacement (all 5-card hands are
equally likely). Find the joint PMF of X, Y, Z.
Question
A)
$$
f_{X,Y,Z}(x,y,z) = \dfrac{5!}{x!y!z!} \left(\dfrac{1}{13}\right)^{x+y} \left(\dfrac{11}{13}\right)^z
$$
where \(x+y+z = 5\)
B)
$$
f_{X,Y}(x,y) = \sum_{y \in Y} \dfrac{5!}{x!y!z!} \left(\dfrac{1}{13}\right)^{x+y} \left(\dfrac{11}{13}\right)^z
$$
where \(x+y+z = 5\)
C)
$$
f_{X,Y,Z}(x,y,z) P(X = x, Y = y, Z = z) = \dfrac{ {4 \choose x} {4 \choose y} {44 \choose z} }{ {52 \choose 5} }
$$
where \(x+y+z = 5\)
Answer
For all of these, we specify the support as \(x+y+z = 5\). This means we can just use random variables and not worry about bounds.
A)
Use multinomial and just plug in the numbers.
B)
Use multinomial, sum up over \(Z\)
C)
Think of this as the hypergeometric distribution with \(3\), not \(2\) labeled categories.
Explanation