HWK1_sols - IEOR 4701 1 Ross 4.28 Assignment 1 Proposed...

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 memory-less 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