IEOR 4701
Assignment 1
Proposed Solutions
Summer 2011
1. Ross, 4.28
A sample of 3 items is selected at random from a box containing 20 items of which 4 are
defective. Find the expected number of defective items in the sample.
Solution.
Let
X
denote the number of defective items in sample.
If we assume that the sampling was done with replacement, then
X
∼
Bin(3
,
4
/
20) and the
expected number of defective items in the sample is 3
·
1
/
5 = 3
/
5
.
Otherwise, then
X
∼
HyperGeo(
N
= 20
, n
= 3
, m
= 4)
,
and
E
{
X
}
=
3
·
4
20
= 3
/
5
.
2. Ross, 4.41
A man claims to have extrasensory perception. As a test, a fair coin is flipped 10 times, and
the man is asked to predict the outcome in advance. He gets 7 out of 10 correct. What is the
probability that he would have done at least this well if he had no ESP?
Solution.
Let
X
= number of right guesses. Then
X
∼
Bin(10
,
1
/
2)
.
P
{
have done as well
}
=
P
{
X
≥
7
}
= 1

P
{
X
≤
6
}
= 1

0
.
8281 = 0
.
1719
.
3. Ross, Theoretical 4.28.
Let
X
be a negative binomial random variable with parameters
r
and
p
, and let
Y
be a
binomial random variable with parameters
n
and
p
. Show that
P
{
X > n
}
=
P
{
Y < r
}
by means of probabilistic interpretation of these random variables. Do
not
attempt to give
any analytical proof of the preceding.
Solution.
We give a description of each of these events:
{
X > n
}
=
{
at least
n
+ 1 trials needed to obtain
r
successes
}
=
{
not more than
n
trials needed to have
r
successes
}
C
=
{
in a sequence of
n
trials there are at least
r
successes
}
C
{
Y < r
}
=
{
less than
r
successes happen in
n
trials
}
=
{
at least
r
successes happen in
n
trials
}
C
4. At 8:59 AM, A, B, C are waiting (in that order) outside a bank that will open at 9 AM. The
bank has two tellers; each of the requirements that A, B, and C are bringing to the bank are
i.i.d. exponentially distributed random variables with mean 1. What is the probability that
A is the last one to leave the bank?
Solution.
Let
T
A
, T
B
,
and
T
C
the times that
A, B,
and
C
take to be served, respectively. We
are interested in computing
P
{
T
A
> T
B
+
T
C
}
.
Intuitively, using the memoryless property of
the exponential distribution and the fact that the three random variables are i.i.d., we would
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IEOR 4701
Assignment 1
Proposed Solutions
Summer 2011
expect this probability to be
P
{
T
A
> T
B
} ·
P
{
T
A
> T
C
}
=
1
2
·
1
2
=
1
4
.
We verify this intuition
below:
P
{
T
A
> T
B
+
T
C
}
=
P
{
T
A
> T
B
+
T
C

T
A
> T
B
}
P
{
T
A
> T
B
}
+
P
{
T
A
> T
B
+
T
C

T
A
≤
T
B
}
P
{
T
A
≤
T
B
}
=
P
{
T
A
> T
B
+
T
C

T
A
> T
B
}
P
{
T
A
> T
B
}
+ 0
=
=
1
2
·
1
2
=
1
4
.
P
{
T
A
> T
C
}
=
P
{
T
A
> T
B
}
= 1
/
2 is due to the fact that
T
A
, T
B
, T
C
are i.i.d. You can also
verify it formally:
P
{
T
A
> T
B
}
=
Z
∞
0
P
{
T
A
> t
} ·
f
T
B
(
t
)
dt
=
Z
∞
0
e

t
·
e

t
dt
=
Z
∞
0
e

2
t
dt
=
1
2
.
You were expected to know how to do the above.
For those who are curious, we present
yet another way of thinking about this.
We use the fact that min(Exp(
λ
1
)
,
Exp(
λ
2
))
∼
Exp(
λ
1
+
λ
2
), and that
P
{
Exp(
λ
1
) = min(Exp(
λ
1
)
,
Exp(
λ
2
))
}
=
λ
1
λ
1
+
λ
2
.
Using this and the
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