Jan 31 Homework Solutions
Math 151, Winter 2012
Chapter 3 Problems (pages 102-110)
Problem 61
Genes relating to albinism are denoted by A and a. Only those people who receive the a
gene from both parents will be albino. Persons having the gene pair A, a are normal in
appearance and, because they can pass on the trait to their offspring, are called carriers.
Suppose that a normal couple has two children, exactly one of whom is an albino. Suppose
that the nonalbino child mates with a person who is known to be a carrier for albinism.
(a) What is the probability that their first offspring is an albino?
For the first offspring to be an albino, both parents would have to be carriers. We
have
P (offspring is aa) = P (offspring is aa|both parents are Aa)P (both parents are Aa).
If both parents are Aa, then the offspring is albino with probability 1/4. Hence
1
P (offspring is aa|both parents are Aa) = .
4
Now let’s compute P (both parents are Aa). One parent of the offspring is known
to be a carrier. We know that the other parent is normal. We also know that the
other parent must be the offspring of two carriers, because the other parent has two
normal parents and an albino sibling.
Hence
=
=
=
=
=
=
=
P (both parents are Aa)
P (other parent is Aa)
since one parent is known to be a carrier
P (other parent is Aa|other parent is normal)
since other parent is known to be normal
P (other parent is Aa and normal)
by definition
P (other parent is normal)
P (other parent is Aa)
since any Aa person is normal
P (other parent is normal)
P (other parent is Aa)
P (other parent is AA) + P (other parent is Aa)
1/2
1/4 + 1/2
2
.
3
1
Thus
P (offspring is aa) = P (offspring is aa|both parents are Aa)P (both parents are Aa)
1 2
= ·
4 3
1
= .
6
(b) What is the conditional probability that their second offspring is an albino given that
their first born is not?
We know
P (2nd is albino|1st is normal) =
P (1st is normal and 2nd is albino)
P (1st is normal)
We calculate the denominator
P (1st is normal) = 1 − P (1st is albino) = 1 −
1
5
=
6
6
using part (a).
If the first offspring is normal and the second offspring is albino, then their parents
must be carriers, so
P (1st offspring is normal and 2nd is albino)
= P (1st offspring is normal and 2nd is albino|both parents are Aa)P (both parents are Aa).
If the parents are both carriers, the first offspring is normal with probability 3/4
and the second is albino with probability 1/4. By Part (a), both parents are carriers
with probability 2/3. Thus
P (1st offspring is normal and 2nd is albino)
= P (1st offspring is normal and 2nd is albino|both parents are Aa)P (both parents are Aa)
3 1 2
·
·
=
4 4
3
1
=
.
8
In conclusion, we have
P (2nd is albino|1st is normal) =
P (1st is normal and 2nd is albino)
1/8
3
=
= .
P (1st is normal)
5/6
20
Problem 64
A true-false question is to be posed to a husband-and-wife team on a quiz show. Both the
husband and the wife will independently give the correct answer with probability p. Which
of the following is a better strategy for the couple?
(a) Choose one of them and let that person answer the question.
2
(b) Have them both consider the question, and then either given the common answer if
they agree or, if they disagree, flip a coin to determine which answer to give.
Under strategy (a), we have
P (correct answer) = P (correct answer|wife is chosen)P (wife is chosen)
+P (correct answer|husband is chosen)P (husband is chosen)
1
1
= p · + p · = p.
2
2
Under strategy (b), we have
P (correct answer) = P (correct answer|both are correct)P (both are correct)
+P (correct answer|only wife is correct)P (only wife is correct)
+P (correct answer|only husband is correct)P (only husband is correct)
+P (correct answer|both are wrong)P (both are wrong)
1
1
= 1 · p2 + · p(1 − p) + · p(1 − p) + 0 · (1 − p)2
2
2
= p,
using the fact that if only one of the wife or husband is correct, then there is a 1/2
probability the coin will land so that the correct answer is given.
Therefore, the two strategies work equally well.
Problem 69
A certain organism possesses a pair of each of 5 different genes (which we will designate by
the first 5 letters of the alphabet). Each gene appears in 2 forms (which we will designate
by lowercase and capital letters). The capital letter will be assumed to be the dominant
gene, in the sense that if an organism possesses the gene pair xX, then it will outwardly
have the appearance of the X gene. For instance, if X stands for brown eyes and x for blue
eyes, then an individual having either gene pair XX or xX will have brown eyes, whereas
one having gene pair xx will have blue eyes. The characteristic appearance of an organism
is called its phenotype, whereas its genetic constitution is called its genotype. (Thus, 2
organisms with respective genotypes aA, bB, cc, dD, ee and AA, BB, cc, DD, ee would
have different genotypes but the same phenotype.) In a mating between 2 organisms, each
one contributes, at random, one of its gene pairs of each type. The 5 contributions of
an organism (one of each of the 5 types) are assumed to be independent and are also
independent of the contributions of the organism’s mate. In a mating between organisms
having genotypes aA, bB, cC, dD, eE and aa, bB, cc, Dd, ee what is the probability that
the progeny will (i) phenotypically and (ii) genotypically resemble
(a) the first parent?
For the A gene, the possible genotypes are aA and aa, both occurring with probability 1/2. Thus for the A gene, the progeny will have the same phenotype as the
first parent with probability 1/2. For the B gene, the possible genotypes are bb,
bB, and BB with respective probabilities 1/4, 1/2, 1/4. Thus for the B gene, the
progeny will have the same phenotype as the first parent with probability 3/4. Do
3
this for all 5 genes. The probability that the progeny and first parent will have the
same phenotype is
P (matches 1st parent’s phenotype)
= P (A correct)P (B correct)P (C correct)P (D correct)P (E correct)
1 3 1 3 1
=
· · · ·
2 4 2 4 2
9
.
=
128
For the A gene, the progeny will have the same genotype as the first parent with
probability 1/2. For the B gene, the progeny will have the same genotype as the
first parent with probability 1/2. Do this for all 5 genes. The probability that the
progeny and first parent will have the same genotype is
P (matches 1st parent’s genotype)
= P (A correct)P (B correct)P (C correct)P (D correct)P (E correct)
1 1 1 1 1
· · · ·
=
2 2 2 2 2
1
=
.
32
(b) the second parent?
P (matches 2nd parent’s phenotype)
= P (A correct)P (B correct)P (C correct)P (D correct)P (E correct)
1 3 1 3 1
=
· · · ·
2 4 2 4 2
9
=
.
128
P (matches 2nd parent’s genotype)
= P (A correct)P (B correct)P (C correct)P (D correct)P (E correct)
1 1 1 1 1
· · · ·
=
2 2 2 2 2
1
=
.
32
(c) either parent?
4
P (matches either parent’s phenotype)
= P (matches 1st parent’s phenotype) + P (matches 2nd parent’s phenotype)
since the two events are disjoint
9
9
=
+
128 128
9
=
.
64
P (matches either parent’s genotype)
= P (matches 1st parent’s genotype) + P (matches 2nd parent’s genotype)
since the two events are disjoint
1
1
+
=
32 32
1
=
.
16
(d) neither parent?
P (matches neither parent’s phenotype) = 1 − P (matches either parent’s phenotype)
9
=1−
64
55
= .
64
P (matches neither parent’s genotype) = 1 − P (matches either parent’s genotype)
1
=1−
16
15
= .
16
Problem 76
Suppose that E and F are mutually exclusive events of an experiment. Show that if independent trials of this experiment are performed, then E will occur before F with probability
P (E)/[P (E) + P (F )].
Define the event G as the event that neither E nor F occurs. That is, G = E c F c .
P (E occurs before F ) = P (E occurs before F |trial 1 is E)P (trial 1 is E)
+P (E occurs before F |trial 1 is F )P (trial 1 is F )
+P (E occurs before F |trial 1 is G)P (trial 1 is G).
5
If E occurs on trial 1, then E occurs before F , and if F occurs on trial 1, then E does
not occur before F . If G occurs on trial 1, then we start all over in determining whether
E or F occurs first. Thus
P (E occurs before F ) = 1 · P (E) + 0 · P (F ) + P (E occurs before F )P (G).
Solving for P (E occurs before F) yields
P (E occurs before F ) =
P (E)
P (E)
=
.
1 − P (G)
P (E) + P (F )
Problem 78
A and B play a series of games. Each game is independently won by A with probability p
and by B with probability 1 − p. They stop when the total number of wins of one of the
players is two greater than that of the other player. The player with the greater number
of total wins is declared the winner of the series.
(a) Find the probability that a total of 4 games are played.
If A wins both of the first two games, the games end before 4 games are played.
Similarly if B wins the first two games. The probability that A and B are tied after
two games is 2p(1 − p). If A and B are tied after two games, then the only way the
game will stop after 4 games is if either A wins the next two games, which occurs
with probability p2 , or B wins the next two games, which occurs with probability
(1 − p)2 . In other words, if A and B are tied after two games, the probability that
there will be exactly 4 games played is p2 + (1 − p)2 . So
P (exactly 4 games played)
= P (exactly 4 games played|tied after two games)P (tied after two games)
=
p2 + (1 − p)2 · 2p(1 − p)
=
− 4p4 + 8p3 − 6p2 + 2p.
(b) Find the probability that A is the winner of the series.
P (A wins) =
P (A wins|A wins after two games)P (A wins after two games)
+P (A wins|B wins after two games)P (B wins after two games)
+P (A wins|tie after two games)P (tie after two games)
= 1 · p2 + 0 · (1 − p)2 + P (A wins) · 2p(1 − p).
Solving for P (A wins), we get
P (A wins) =
6
p2
.
1 − 2p + 2p2
Problem 80
In a certain contest, the players are of equal skill and the probability is 1/2 that a specified
one of the two contestants will be the victor. In a group of 2n players, the players are
paired off against each other at random. The 2n−1 winners are again paired off randomly,
and so on, until a single winner remains. Consider two specified contestants, A and B,
and define the events Ai , i ≤ n, by A plays in exactly i contests, and E by A and B never
play each other.
(a) Find P (Ai ), i = 1, ...n.
In order for A to play in exactly i contests, player A will have to win the first i − 1
contests and lose the i-th contest. Hence
1 i−1 1
1
· = i.
P (Ai ) =
2
2
2
(b) Find P (E).
First observe if A and B play, then one must beat the other, so
P (E) = 1 − P (E c )
= 1 − P (A ever beats B) − P (B ever beats A)
= 1 − 2P (B ever beats A) by symmetry.
We have
P (B ever beats A) =
n
X
P (B ever beats A|Ai )P (Ai ).
i=1
If A loses the i-th contest, he loses to any one of the 2n − 1 other players with equal
probability, so P (B ever beats A|Ai ) = 2n1−1 . Therefore
P (B ever beats A) =
=
n
X
i=1
n
X
i=1
P (B ever beats A|Ai )P (Ai )
1
1
·
2n − 1 2i
by part (a)
n
1 X 1
= n
2 − 1 i=1 2i
1
1 1 − (1/2)n
·
·
2n − 1 2
1/2
1
= n.
2
=
since it’s a geometric series
Hence
P (E) = 1 − 2P (B ever beats A)
1
=1−2 n
2
1
= 1 − n−1 .
2
7
(c) Let Pn = P (E). Show that
Pn−1 2n − 2
Pn =
+
4
4
2n − 1
3
and use this formula to check the answer you obtained in part (b).
Note: there is a mistake in the recursive formula given in the book. The
formula they give is for E c in place of E. We instead prove the correct
recursive formula for E.
We define Pi to be the event that A and B do not play each other in a tournament
with 2i players. Hence Pn = P (E).
Let F be the event that A and B play in the first contest and observe that
Pn = P (E) = P (E|F )P (F ) + P (E|F c )P (F c ).
Since A plays each of the 2n − 1 other players in the first contest with equal probn
. Note P (E|F ) = 0. To calculate P (E|F c ),
ability, P (F ) = 2n1−1 and P (F c ) = 22n −2
−1
note that if A and B do not play in the first contest, then there are two mutually
exclusive ways that they will not play in the tournament. The first way is that one
of A and B loses in the first round, which happens with probability 43 . The second
way is that both both A and B win in the first round and A and B don’t meet
in the subsequent tournament with 2n−1 players. This happens with probability
1
· Pn−1 = Pn−1
. Hence
4
4
3 Pn−1
P (E|F c ) = +
4
4
and
Pn = P (E)
= P (E|F )P (F ) + P (E|F c )P (F c )
3 P 2n − 2
1
n−1
=0· n
+
+
2 −1
4
4
2n − 1
3 P 2n − 2
n−1
=
+
.
4
4
2n − 1
1
, using induction
Now we can check part (b), in which we found that P (E) = 1− 2n−1
1
on n. The base case is n = 1. Clearly if there are only 2 = 2 players, then A and
B must play in the first round and so our formula
P1 = 1 −
1
2n−1
=1−
1
21−1
=1−1=0
is correct.
1
For the inductive step, suppose that formula Pn−1 = 1 − 2n−2
holds for n − 1. We
1
must show that the formula Pn = 1 − 2n−1 holds for n. Using our formula above,
8
we have
Pn−1 2n − 2
· n
4
4
2 −1
3 1 − 1 2n − 2
2n−2
+
· n
=
4
4
2 −1
3 1
n
1
2 −2
=
+ − n · n
4 4 2
2 −1
2n − 1 2n − 2
· n
=
2n
2 −1
2n − 2
=
2n
1
= 1 − n−1 .
2
Pn =
3
+
Hence by induction on n, we have proven the formula P (E) = Pn = 1 −
part (b) for all positive values of n.
1
2n−1
from
(d) Explain why 2n − 1 games are played. Number these games, and let Bi denote the
event that A and B play each other in game i, i = 1, ..., 2n − 1.
In each game, exactly one player loses. Since there is a single winner remaining
after the tournament, all but one of the 2n players lose a game. Thus the number
of games is 2n − 1.
(e) What is P (Bi )?
Two of the 2n players play in game i and each pair of players play in game i with
equal probability, so
P (Bi ) =
1
2n
=
2
2!(2n − 2)!
2
1
= n n
= n−1 n
.
n
2 !
2 (2 − 1)
2 (2 − 1)
(f) Use part (e) to find P (E).
n
Note E = (∪2i=1−1 Bi )c . Hence
n
P (E) = 1 − P ∪2i=1−1 Bi
=1−
=1−
n −1
2X
P (Bi ) since the events Bi are mutually exclusive
i=1
n −1
2X
i=1
1
2n−1 (2n
= 1 − (2n − 1) ·
=1−
1
2n−1
− 1)
1
2n−1 (2n − 1)
.
Note this is the same answer we found in part (b).
9
Problem 81
An investor owns shares in a stock whose present value is 25. She has decided that she
must sell her stock if it either goes down to 10 or up to 40. If each change of price is either
up 1 point with probability .55 or down 1 point with probability .45, and the successive
changes are independent, what is the probability that the investor retires a winner?
Let Pn be the probability that the investor retires a winner given that she owns stock
whose present value is n, 10 ≤ n ≤ 40. Note P10 = 0 and P40 = 1. By Bayes’ formula,
Pn =
P (winner|price goes up to n + 1)P (price goes up)
+P (winner|price goes down to n − 1)P (price goes down)
= Pn+1 · .55 + Pn−1 · .45.
Let’s try to solve for the difference Pn+1 − Pn . From the equation
Pn = Pn+1 · .55 + Pn−1 · .45
we get that
.45(Pn − Pn−1 ) = .55(Pn+1 − Pn ).
Rearranging, we get
Pn+1 − Pn = (.45/.55)(Pn − Pn−1 ).
Hence
Pn+1 − Pn = (.45/.55)(Pn − Pn−1 ) = (.45/.55)n−10 (P11 − P10 ) = (.45/.55)n−10 P11
since P10 = 0.
To compute P11 , note that
1 = P40 − P10 =
39
X
(Pn+1 − Pn ) =
39
X
n−10
(.45/.55)
n=10
n=10
P11
1 − (.45/.55)30
P11 ,
=
1 − .45/.55
where the last equality comes from the sum of a geometric series. So
P11 =
1 − .45/.55
.
1 − (.45/.55)30
Thus P25 is given by
P25 = P25 − P10
=
=
24
X
(Pn+1 − Pn )
n=10
24
X
(.45/.55)n−10 P11
n=10
1 − (.45/.55)15
P11
1 − .45/.55
1 − (.45/.55)15
1 − .45/.55
=
·
1 − .45/.55
1 − (.45/.55)30
1 − (.45/.55)15
=
1 − (.45/.55)30
≈ 0.9530.
=
10
Chapter 3 Theoretical Exercises (pages 110-113)
Problem 21 The Ballot Problem.
In an election, candidate A receives n votes and candidate B receives m votes, where
n > m. Assuming that all of the (n + m)!/n!m! orderings of the votes are equally likely,
let Pn,m denote the probability that A is always ahead in the counting of the votes.
(a) Compute P2,1 , P3,1 , P3,2 , P4,1 , P4,2 , P4,3 .
For n = 2, m = 1, the possible orderings are AAB, ABA, BAA, so P2,1 = 1/3. For
n = 3, m = 1, the possible orderings are AAAB, AABA, ABAA, BAAA, so P3,1 =
1/2. For n = 3, m = 2, of the 10 possible orderings, A stays ahead for AABAB
and AAABB, so P3,2 = 1/5. For n = 4, m = 1, the possible orderings are AAAAB,
AAABA, AABAA, ABAAA, and BAAAA, so P4,1 = 3/5. For n = 4, m = 2,
of the 15 possible orderings, A stays ahead for AABABA, AABAAB, AAABBA,
AAABAB, AAAABB, so P4,2 = 1/3. For n = 4, m = 3, of the 35 possible
orderings, A stays ahead for AABABAB, AABAABB, AAABBAB, AAABABB,
and AAAABBB, so P4,3 = 1/7.
(b) Find Pn,1 , Pn,2 .
If m = 1, there are n + 1 possible orderings. A must win the first two votes to say
n−1
. If m = 2,
ahead and B can win one of any of the last n − 1 votes. Thus Pn,1 = n+1
n+2
there are 2 possible orderings.
first
votes
must be AAAA, AAAB, or
Then−2
fourn−2
n−2
AABA, so there are n−2
+
+
=
+
2(n
− 2) orderings where A
2
1
1
2
stays ahead. Thus
n−2
+ 2(n − 2)
2
Pn,2 =
n+2
2
(n − 2)(n − 3)/2 + 2(n − 2)
=
(n + 2)(n + 1)/2
(n − 2)(n + 1)/2
=
(n + 2)(n + 1)/2
n−2
=
n+2
(c) On the basis of your results in parts (a) and (b), conjecture the value of Pn,m .
We conjecture that Pn,m =
n−m
n+m
for all values of n and m with n > m.
(d) Derive a recursion for Pn,m in terms of Pn−1,m and Pn,m−1 by conditioning on who
receives the last vote.
By Bayes’ formula
P (A stays ahead) = P (A stays ahead|last vote is A)P (last vote is A)
+P (A stays ahead|last vote is B)P (last vote is B).
11
Regardless of who gets the last vote, A stays ahead until and including the last
vote. The probability that the last vote is for A is n/(n + m) and the probability
that the last vote is for B is m/(n + m), so
Pn,m = Pn−1,m ·
m
n
+ Pn,m−1 ·
.
n+m
n+m
(e) Use part (d) to verify your conjecture in part (c) by an induction proof on n + m.
The base case n + m = 3 is verified in part (a). For the inductive step, suppose
Pn,m = n−m
if n + m ≤ N for some integer N . This is our inductive assumption.
n+m
Now, consider n + m = N + 1. We have
m
n
+ Pn,m−1 ·
n+m
n+m
n−1−m
n
n−m+1
m
·
+
·
n−1+m n+m n+m−1 n+m
by inductive assumption
(n − 1 − m)n + (n − m + 1)m
(n + m − 1)(n + m)
2
n − n − m2 + m
(n + m − 1)(n + m)
(n + m − 1)(n − m)
(n + m − 1)(n + m)
n−m
.
n+m
Pn,m = Pn−1,m ·
=
=
=
=
=
Hence by induction on n + m, we have proven that Pn,m =
of n and m with n > m.
n−m
n+m
holds for all values
Chapter 4 Problems (pages 172-179)
Problem 1
Two balls are chosen randomly from an urn containing 8 white, 4 black, and 2 orange
balls. Suppose that we win $2 for each black ball selected and we lose $1 for each white
ball selected. Let X denote our winnings. What are the possible values of X, and what
are the probabilities associated with each value?
There are at most six possible values for X depending what balls are chosen. If two
white balls are chosen, X = −2. If a white and orange ball is chosen, X = −1. If two
orange balls are chosen, X = 0. If a white and black ball is chosen, X = 1. If a black
and orange ball is chosen, X = 2. If two black balls are chosen, X = 4. Thus X takes
six values −2, −1, 0, 1, 2, 4.
Let’s calculate the probabilities associated with each value.
8
P {X = −2} =
P {X = −1} =
2
14
2
8·2
14 ≈ .1758
2
12
≈ .3077
2
2
14
2
P {X = 0} =
P {X = 1} =
≈ .0110
8·4
14 ≈ .3516
2
P {X = 2} =
4·2
14 ≈ .0879
2
4
2
14
2
P {X = 4} =
≈ .0659
Problem 4
Five men and five women are ranked according to their scores on an examination. Assume
that no two scores are alike and all 10! possible rankings are equally likely. Let X denote
the highest ranking acheived by a woman (For instance, X = 1 if the top-ranked person
is female.) Find P {X = i}, i = 1, 2, 3, ..., 8, 9, 10.
First note there is at least one woman of rank at least 6, so P {X = i} = 0 for
i = 7, 8, 9, 10.
Let i ≤ 6 and suppose the highest rank of a woman is i. Then
there are i − 1 men
5
of higher rank than her, who can be chosen and ordered in i−1 (i − 1)! different ways.
There are 5 choices of women for the i-th rank. There are 10 − i people of lower rank,
who can be ordered in (10 − i)! ways. Thus
5
(i − 1)! · 5 · (10 − i)!
P {X = i} = i−1
10!
for i = 1, 2, ..., 6.
Problem 5
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed n times. What are the possible values of X?
If a coin is tossed n times, the number of tails is some integer i between 0 and n and
the number of heads is n − i. So X takes the values (n − i) − i = n − 2i for i = 0, 1, ..., n.
13