IE 111 Fall 2007
Homework #6 Solutions
Due Friday October 19 at the beginning of class
Question 1.
To check on the security screeners at an airport, we put 4 fake bombs into a set of 52
baggage pieces.
Suppose that the screeners actually just pick 6 pieces randomly (without
replacement) to inspect.
Let X be the number of our fake bombs that they find.
a)
According to which well known distribution is X distributed?
X~ Hyper geometric ( N=52,K=4),
E[X]= np = 6*(4/52) = 0.462 ,
V[X] = np*(1p)*(Nn)/(N1) = 6*(4/52) *48/52*46/51 = 0.384
b)
Find the Probability Mass Function f(x).
Remember to specify an appropriate range for x.
P
X
(x=0) =
48
C
6
*
4
C
0
/
52
C
6
= 0.6028,
P
X
(x=1) =
48
C
5
*
4
C
1
/
52
C
6
= 0.3364,
P
X
(x=2) =
48
C
4
*
4
C
2
/
52
C
6
= 0.0573,
P
X
(x=3) =
48
C
3
*
4
C
3
/
52
C
6
= 0.0034 ,
P
X
(x=4) =
48
C
2
*
4
C
4
/
52
C
6
= 0.000055
c)
Find E(X)
E[X] = 0* P
X
(x=0) + 1* P
X
(x=1)+ 2* P
X
(x=2)+ 3* P
X
(x=3)+ 4* P
X
(x=4) = 0.4615
d)
Find V(X)
V[X] = 0* P
X
(x=0) + 1* P
X
(x=1)+ 4* P
X
(x=2)+ 9* P
X
(x=3)+ 16* P
X
(x=4) – E[X]
2
= 0.384
Question 2.
An urn has 6 red marbles and 14 green marbles.
Careful on this one, the distribution to
use keeps changing!
a)
If I pick 3 marbles
with replacement
, and let X= the number of red marbles picked,
what is the P(X=1)?
X~ Binomial (n=3, p=6/20)
P(X=1) =
3
C
1
p
1
*(1p)
2
=3* (6/20)
1
*(16/20)
2
= 0.441
b)
If I keep picking marbles
with replacement
until I get my first red marble, what is the
probability I picked exactly 7 times?
X~ Geometric (p=6/20)
P(X=7) = (1p)
6
*p
=(16/20)
6
*(6/20) = 0.0353
c)
If I pick 3 marbles
without replacement
and let X= the number of red marbles picked,
what is the P(X=1)?
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View Full DocumentX~ Hyper geometric ( N=20,K=6)
P(X=1)=
6
C
1
*
14
C
2
/
20
C
3
= 0.4789
d)
If I pick 3 marbles
without replacement
and let X= the number of red marbles picked,
what is the expected value of X?
E[X]= n*p = 3*6/20 =9/10
e)
Suppose I have picked 8 marbles
with replacement
and have not yet gotten a red
marble.
How many more picks do I expect will be required before my first red
marble?
E[X] = 1/p for geometric distribution. From the lack of memory property of
geometric distribution, no matter that I did not find red in first 8 trials, expected
number of picks until I find red after that time is 1/(6/20), which is 20/6.
f)
Suppose I have picked 9 marbles
without replacement
and have not yet gotten a red
marble.
How many more picks (without replacement) do I expect will be required
before my first red marble?
This is a tricky problem!
We can't use any of our favorite
distributions; you have to think through it from first principles.
Here's a hint to get
you started.
Let X = number of remaining picks.
Compute P(X=1), P(X=2), P(X=3),
etc. (it will help to think of the Geometric distribution, but it's not quite the same) and
then compute the mean value from there.
After 9 picks, there are 6 red and 5 green marbles left. So
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 Spring '07
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