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University of California, Los Angeles **We aren't endorsed by this school

Course

ECON 1

Subject

Mathematics

Date

Oct 17, 2023

Pages

1

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Math
15B
(Bender)
Midterm
Exam
27
October
2000
1.
(36
pts)
A
five
digit
number
is
a
sequence
of
five
digits,
the
first
of
which
is
NOT
zero.
Thus, 12345
and
10101
are
valid
but
01234
and
1234
are
NOT
valid.
(a)
Ans:
(b)
Ans:
(c)
Ans:
How
many
five
digit
numbers
are
there?
One
way
is
to
note
there
are
9
choices
for
the
first
digit
and
10
for
each
of
the
remaining
giving
9
x
10%.
You
could
also
note
that
it
is
simply
all
numbers
less
than
10°
and
at
least
10?
giving
10°
—
10*
=
9
x
10".
How
many
five
digit
numbers
have
all
digits
different?
(as
in
12345
but
not
10101)
There
are
9
choices
for
the
first
digit,
9
for
the
second
(anything
but
the
first),
8
for
the
third
(anything
but
the
first
two),
and
so
on,
giving
9
x
9
x
8
x
7
x
6.
How
many
five
digit
numbers
have
no
digit
appearing
just
once?
(So
11111
and
10010
are
okay,
but
10111
and
12312
are
not.)
Either
one
digit
appears
5
times
OR
two
digits
appear,
one
twice
and
one
three
times.
There
are
9
possibilities
for
the
first
case.
For
the
second
case,
choose
the
two
positions
for
the
pair
of
digits
in
(2')
ways,
choose
the
first
digit
in
the
number
in
9
ways
(anything
but
zero),
and
choose
the
other
digit
in
9
ways.
Thus
we
have
9
+
(3)
x
9
x9.
2.
(16
pts.)
Nine
people,
including
Alice,
are
to
be
divided
into
two
teams
of
four
people
each,
plus
a
referee.
If
all
divisions
are
equally
likely,
what
is
the
probability
that
Alice
is
the
referee?
(No,
it
doesn't
matter
if
the
teams
are
distinguishable
or
not.)
Be
sure
to
explain
how
you
got
your
answer.
Ans:
There
are
nine
choices
for
the
referee,
all
of
which
are
equally
likely.
Thus,
the
probability
that
Alice
is
the
referee
is
1/9.
3.
(24
pts.)
An
integer
k
from
1
to
9
is
picked
uniformly
at
random.
Let
X(k)
=1if
k
is
odd
and
X
(k)
=
0
if
k
is
even.
Let
Y
(k)
be
the
remainder
when
k
is
divided
by
3.
(a)
Ans:
Draw
a
table
like
the
one
here
and
fill
in
the
probabilities.
First
we
construct
a
table
of
X
and
Y
for
each
possible
outcome.
Since
each
outcome
has
probability
1/9,
a
simple
count
lets
us
fill
the
table.
123456789
X\Yl
o
1
2
X[101010101
0
1/9
1/9
2/9
Y
[120120120
1
2/9
2/9
1/9

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