Math 55: Midterm #2, 3 March 2008
Write your name, your student ID number, your section time and number,
and a threeproblem grading grid, on the cover of your blue book. Hand in
your exam book at 11:00 am. Books, notes, calculators, scratch paper and/or
collaboration are not allowed.
Justify your computations: correct answers
with inadequate explanation may receive partial credit.
1: (a)
This is the number of onto functions from an 8element set to a 7
element set, divided by the total number of functions.
It can be directly
evaluated by choosing a squirrel to receive two pecans, choosing two pecans
for that squirrel to get, and permuting the other 6 pecans/squirrels:
p
=
7
parenleftBigg
8
2
parenrightBigg
6!
7
8
.
A common wrong answer is
parenleftBigg
7
1
parenrightBigg
/
parenleftBigg
8 + 7
−
1
8
parenrightBigg
,
which is wrong because the final configurations don’t have equal probabilities.
(b)
Let
f
i
be the indicator of the event
F
i
that the
i
th squirrel gets a pecan,
so we are asked for the expectation of
f
=
f
1
+
· · ·
+
f
7
. By symmetry, this is
7
E
(
f
1
) = 7
p
(
F
1
) = 7(1
−
p
(
¯
F
1
)). Since there are 6
8
ways for the first squirrel
not to get a pecan,
E
(
f
) = 7
E
(
f
1
) = 7(1
−
(6
/
7)
8
)
.
(c)
This is the expected number of successes in
n
= 8 Bernoulli trials with
success probability
p
= 1
/
7, which is
np
= 8
/
7.
(d)
This is the variance of the number of successes in
n
= 8 Bernoulli trials
with success probability
p
= 1
/
7, which is
npq
= 48
/
49.
2: (a)
A nonMartian student gets a perfect score with probability (2
/
3)
4
.
1
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(b)
Let
P
be the event of a perfect score,
M
the event that the student
is Martian,
N
the event that the student is not Martian. Since
N
and
M
partition the sample space into a disjoint union,
p
(
P
) =
p
(
P
∩
N
) +
p
(
P
∩
M
) =
p
(
P

N
)
p
(
N
) +
p
(
P

M
)
p
(
M
)
by definition of conditional probability. Since the conditionals are given, we
have
p
(
P
) = (2
/
3)
4
·
(81
/
82) + 1
·
(1
/
82) = 17
/
82
.
(c)
By (b) and the definition of conditional probability,
p
(
M

P
) =
p
(
M
∩
P
)
p
(
P
)
=
p
(
P

M
)
p
(
M
)
p
(
P
)
=
1
·
1
/
82
17
/
82
=
1
17
.
4:
(a)
The inclusionexclusion formula reads

P
∪
Q
∪
R

=

P

+

Q

+

R
 − 
P
∩
Q
 − 
P
∩
R
 − 
Q
∩
R

+

P
∩
Q
∩
R

.
(b)
Define
P
=
{
1
≤
n
≤
pqr

p

n
}
and similarly for
Q
and
R
. Then

P

=
⌊
pqr/p
⌋
=
qr
, while

Q

=
pr
and

R

=
pq
. Similarly

P
∩
Q

=
⌊
pqr/pq
⌋
=
r
since
p
,
q
and
r
are all distinct
primes, and

P
∩
R

=
q
,

Q
∩
R

=
p
. Finally

P
∩
Q
∩
R

= 1 so we can
apply inclusionexclusion:
{
1
≤
n
≤
pqr

gcd(
n, pqr
) = 1
}
=
pqr
− {
1
≤
n
≤
pqr

p

n
∨
q

n
∨
r

n
}
=
pqr
− 
P
∪
Q
∪
R

=
pqr
−
qr
−
pr
−
pq
+
r
+
q
+
p
−
1
.
(c)
Putting
p
= 2,
q
= 3,
r
= 5 gives 8 numbers 1, 7, 11, 13, 17, 19, 23, 29
between 1 and
pqr
= 30 inclusive and relatively prime to 30. Our formula
gives
pqr
−
qr
−
pr
−
pq
+
r
+
q
+
p
−
1 = 30
−
15
−
10
−
6 + 5 + 3 + 2
−
1 = 8
.
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 Fall '08
 STRAIN
 Math, Conditional Probability, Probability theory, nonnegative integer solutions

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