MATH 5010–001 SUMMER 2003
SOLUTIONS TO ASSIGNMENT 3
Problems from pp. 104–115
60. If the couple uses strategy (a), then
P (win) = P ( win | husband
+ P ( win | wife is
1
= p×
+ p×
2
is selected) P ( husband is selected)
selected) P (wife is selected)
1
= p.
2
(1)
On the other hand, if they use strategy (b), then
P (win) = P ( win | agree) P (agree) + P ( win | disagree) P (disagree)
1
= P ( win | agree) P (agree) + × P (disagree).
2
Now let H and W respectively denote the events that the husband and the wife get the
correct answer. Then, P (agree) = P (HW) + P (Hc W c ) = p2 + (1 − p)2 = 1 + 2p2 − 2p.
Consequently, we also know that
P (disagree) = 1 − P (agree) = 2p(1 − p).
(2)
It remains to compute P (win | agree). But
P (win and agree)
P (HW)
p2
P (win | agree) =
=
=
.
P (agree)
P (agree)
P (agree)
(3)
Plug (2) and (3) into (1) to obtain
1
P (win) = p2 + 2p(1 − p) = p.
2
So both strategies are equally good.
62. (a) Let Ci denote the event that the ith circuit is closed. Then note that
P (current flows) = P (C1 C2 C5 ∪ C3 C4 C5 )
= P (C1 C2 C5 ) + P (C3 C4 C5 ) − P (C1 C2 C3 C4 C5 )
= p1 p2 p5 + p3 p4 p5 − p1 p2 p3 p4 p5 .
62. (b) Following the hint, we have,
P (current flows) = P (current flows | C3 )P (C3 ) + P (current flows | C3c )P (C3c )
= p3 P (current flows | C3 ) + (1 − p3 )P (current flows | C3c ).
1
(4)
Thanks to the independence of the Cj ’s,
P (current flows | C3 ) = P ((C1 ∪ C2 ) ∩ (C4 ∪ C5 ))
= P (C1 ∪ C2 )P (C4 ∪ C5 )
= (p1 + p2 − p1 p2 ) (p4 + p5 − p4 p5 ) .
(5)
(You can try to find, possibly simpler, representations, but this works well enough.)
Also,
P (current flows | C3c ) = P (C1 C4 ∪ C2 C5 )
= P (C1 C4 ) + P (C2 C5 ) − P (C1 C2 C4 C5 )
= p1 p4 + p2 p5 − p1 p2 p4 p5 .
(6)
Combine (4), (5), and (6) to obtain:
P (current flows) = p3 (p1 + p2 − p1 p2 ) (p4 + p5 − p4 p5 )
+ (1 − p3 ) (p1 p4 + p2 p5 − p1 p2 p4 p5 ) .
75. This is an example of how a bad hint for a bad solution can make a problem more
difficult than need be. This problem will not be graded, but the answer to the main
question is P (E) = 21−n as you will see next. Let EAB denote the event that A and
B meet, and B is the victor. Note that: (i) P (EAB ) + P (EBA ) = P (E); and (ii)
P (EAB ) = P (EBA ). Therefore, it suffices to show that P (EAB ) = 2−n .
75. (a) Note that Ai is the event that contestant A wins the first (i − 1) contests, and
then loses contest number i. Therefore, P (Ai ) = (1/2)i = 2−i .
75. (b) We will need the following combinatorial fact:
Lemma. There are 1 × 3 × 5 × · · · × (m − 1) possible pairings of m distinct points, as
long as m is an even integer.
To understand this lemma, try it first for m = 2: There are two people, labeled 1
and 2, and there is only one pairing: (1, 2). Here, m = 2, and 1 × · · · × (m − 1) = 1
pairing. Next, try it for m = 4. We have people 1, 2, 3, 4, and the possible pairings are
{(1, 2), (3, 4)}, {(1, 3), (2, 4)}, and {(1, 4), (2, 3)}. That is, there are 3 possible pairings,
and 1×· · ·×(m−1) = 1×3 = 3. Similarly, when m = 6, there are 1×3×5 = 15 pairings,
and they are {(1, 2), (3, 4), (5, 6)}, {(1, 2), (3, 5), (4, 6)}, {(1, 2), (3, 6), (4, 5)}, . . .. There
are 3 pairings that involve (1, 2), three that involve (1, 3), three involving (1, 4), three
involving (1, 5), and three involving (1, 6); that adds up to 15 pairings.
Proof. It may help to think of the points as people. Now line these m people up in
a row, and pair them off one by one. To find a pair for the left-most person, you
have m − 1 choices. Once this person is paired-off with someone else, the remaining
left-most person has m − 3 choices for a pair and so on. So the number of possible
pairings is (m − 1)(m − 3) · · · 1. Multiply in reverse order to finish.
2
Note that, given equally-likely pairings of m people, the probability p(m) that any
given two (say people A and B) are paired off is the number of ways to pair-off all
others (that is 1 × 3 × · · · × m − 3) divided by the total number of pairings (that is
1 × 3 × · · · × m − 1). In other words,
p(m) :=
1
.
m−1
(7)
(The symbol “:=” is telling us that this equation is the definition of p(m).) Now to
the problem at hand.
Right off, use Bayes’ rule, together with the computation of (a), in order to simplify
P (EAB ) =
n
X
2−i P (EAB | Ai ).
(8)
i=1
Now, P (EAB | A1 ) is the probability that A and B meet for the first round. Therefore,
P (EAB | A1 ) = p(2n ) =
2n
1
.
−1
(9)
Similarly, P (EAB | A2 ) is the probability that: (i) A and B do not meet during the
first round (prob. = 1 − p(2n )); (ii) B wins his/her first round (prob. = 1/2); and
(iii) A and B meet during the second round (prob. = p(2n−1 )). Thus, by (7),
P (EAB | A2 ) =
=
=
=
=
1 − p(2n )
p(2n−1 )
2
1
1
1
1− n
· n−1
2
2 −1
2
−1
n
1 2 −2
1
· n−1
n
2 2 −1
2
−1
n−1
1 2(2
− 1)
1
· n−1
n
2
2 −1
2
−1
1
.
n
2 −1
(10)
Comparing (9) and (10) should start giving you ideas about what P (EAB | Ai ) is in
general.
I will compute P (EAB | A3 ) by hand before solving the general problem. This would
be enough to understand what is going on. Given A3 , EAB can happen if and only if:
(i) A and B do not meet in round one (prob. = 1 − p(2n ));
(ii) B wins his/her first round (prob. = 1/2);
(iii) A and B do not meet in round two (prob. = 1 − p(2n−1 ));
(iv) B wins his/her second round (prob. = 1/2); and
(v) A and B meet in the third round (prob. = p(2n−2 )).
3
Thus,
P (EAB | A3 ) =
1 − p(2n ) 1 − p(2n−1 )
·
· p(2n−2 ).
2
2
Thanks to (7), this simplifies as:
1
1 − 2n−1
1 − 2n1−1
−1
1
P (EAB | A3 ) =
n−2
2
2
2
−1
n
n−1
−2
1 2 −2
1 2
1
=
·
· n−2
n
n−1
2 2 −1
2 2
−1
2
−1
n−1
n−2
2
−1 2
−1
1
= n
· n−1
· n−2
2 −1 2
−1 2
−1
1
= n
.
2 −1
Now you should check, by induction, that
round i − 1
round 1
}|
{
z
round i
z }| {
z }| {
1 − p(2n−i−2 )
1 − p(2n )
P (EAB | Ai ) =
×···×
× p(2n−i−2 )
2
2
1
.
= n
2 −1
Together with (8), this yields the following:
n
P (EAB ) =
But if r 6= 1, then
term to obtain:
Pn
i=0
n
X
r i = (1 − r n+1 )/(1 − r). Therefore, we can subtract the first
ri =
i=1
1 X −i
2 .
2n − 1 i=1
r − r n+1
r
1 − r n+1
−1 =
=
(1 − r n ).
1−r
1−r
1−r
Pn
2−i = 1 − (1/2)n , and we get
n
1
1
1
1
2 −1
P (EAB ) = n
=
.
1− n = n
2 −1
2
2 −1
2n
2n
This means that
i=1
Theoretical Problems from pp. 115–119
6. First, observe the following set-theoretic fact:
c
(E1 ∪ · · · ∪ En ) = E1c E2c · · · Enc .
4
Therefore, by the independence of the Ek ’s,
c
P [(E1 ∪ · · · ∪ En ) ] =
n
Y
n
Y
P Ejc =
(1 − P (Ei )) .
j=1
j=1
Subtract both sides from 1 to finish.
9.
Let D` (i) denote the event that person i is born on day `, where ` = 1, . . . , 365.
Then,
365
X
P D` (j) D` (i) P (D` (i))
P (Ai,j ) =
=
`=1
365
X
P (D` (j)) P (D` (i)) =
`=1
365
X
`=1
1
1
=
.
2
365
365
1
as well.
This is true for any pair of people, so P (Ar,s ) = 365
Similarly, if i, j, r, and s are four distinct indices, then
365
X
P (Ai,j ∩ Ar,s ) =
P (Dk (i) ∩ Dk (j) ∩ D` (r) ∩ D` (s))
k,`=1
365
X
=
P (D` (i)) P (D` (j)) P (D` (r)) P (D` (s))
k,`=1
365
X
=
k,`=1
1
1
=
.
3654
3652
1
Thus, P (Ai,j ∩ Ar,s ) = P (Ai,j )P (Ar,s ) = 365
2 , and we have established the asserted
pairwise independence. On the other hand, the intersection, over all distinct pairs (i, j),
of Ai,j is the event that everyone has the same birthday as everyone else. The probability
of this event is 365n−1 . (The first person has some birthday, and everyone else has that
get 365−N ,
birthday.) Now note that if we multiply P (Ai,j ) over all
possible pairs we
n
n−1
where N is the number of all possible pairs; i.e., N = 2 . Evidently, 365
= 365N if
and only if N = n − 1; i.e., if and only if n2 = (n − 1). But n2 = n(n−1)
. Therefore, we
2
n
have independence of all pairs if and only if 2 = 1. But this means that there are only
two people in the room. That is, we have independence if and only if there are two people
in the room only.
28. First of all, we need to prove the identity,
Z
0
1
y n (1 − y)m dy =
5
n!m!
.
(n + m + 1)!
(11)
R1
Indeed, let C(n, m) denote the integral, and notice that C(n, 0) = 0 y n dy = n1 , which
is equal to the left-hand side of (11) for m = 0. Now we suppose C(n, m − 1) has the
right form for all n, and prove that C(n, m) does too. To do this, note that
Z
C(n, m) =
0
Z
=
0
1
1
y n (1 − y)m−1 (1 − y) dy
n
y (1 − y)
m−1
Z
dy −
1
0
y n (1 − y)m−1 y dy
= C(n, m − 1) − C(n + 1, m − 1).
By the induction hypothesis, C(n, m−1) = n!(m − 1)!/(n + m)!, whereas C(n+1, m−
1) = (n + 1)!(m − 1)!/(n + m + 1)!. Therefore,
n!(m − 1)! (n + 1)!(m − 1)!
n!(m − 1)!
n+1
C(n, m) =
−
=
1−
(n + m)!
(n + m + 1)!
(n + m)!
n+m+1
n!(m − 1)!
m
n!m!
=
×
=
.
(n + m)!
n+m+1
(n + m + 1)!
This verifies (11). Now to the Laplace rule of succession (Example 5d, page 102).
Let Ci denote the event that the ith coin has been chosen, and Hi the event that the
ith coin tosses heads. Note that P (Hi ) = i/k, and P (Ci ) = (1 + k)−1 (i = 0, . . . , k).
We are asked to compute the conditional probability that the (n + 1)st flip turns
heads, given the first flips resulted in r heads and n − r tails. Let H` and T` denote
the events that the `th toss was heads and tails, respectively. (Do not confuse Hi with
Hi .) Then,
P (H1 · · · Hr Tr+1 · · · Tn Hn+1 ) =
k
X
P H1 · · · Hr Tr+1 · · · Tn Hn+1 Cj P (Cj )
j=0
k
1 X
=
k + 1 j=0
r+1 n−r
j
j
.
1−
k
k
The same expression desribes P (H1 · · · Hr−1 Tr Hr+1 Tr+2 · · · Tn Hn+1 ), etc. Therefore,
the probability that the (n + 1)st toss is heads, and there are exactly r heads prior to
that, is
Z 1
n−r
k r+1 n
1 X j
n
j
→
y r+1 (1 − y)n−r dy,
1−
r k + 1 j=0 k
r
k
0
(12)
as k → ∞. The probability that there are exactly r heads by the nth toss, on the
6
other hand, is:
k
n
1 X n
P H1 · · · Hr Tr+1 · · · Tn Cj
P (H1 · · · Hr Tr+1 · · · Tn ) =
r k + 1 j=0
r
n−r
k r n
1 X j
j
=
1−
r k + 1 j=0 k
k
Z 1
n
→
y r (1 − y)n−r dy,
r
0
(13)
as k → ∞. Combine (12) and (13) to obtain the following limit, as k → ∞, for the
probability in question:
R1
0
y r+1 (1 − y)n−r dy
Pr ≈ R 1
0
=
y r (1 − y)n−r dy
C(r + 1, n − r)
(r + 1)!(n − r)!/(n + 2)!
r+1
=
=
.
C(r, n − r)
r!(n − r)!/(n + 1)!
n+2
7