IE 111 Fall 2009
Homework #5, solutions
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~ Hypergeometric (M=52, r=4, N=6)
b)
Find the Probability Mass Function f(x).
Remember to specify an appropriate domain for x.
P(X=0) =
48
C
6
*
4
C
0
/
52
C
6
= 0.6028,
P(X=1) =
48
C
5
*
4
C
1
/
52
C
6
= 0.3364,
P(X=2) =
48
C
4
*
4
C
2
/
52
C
6
= 0.0573,
P(X=3) =
48
C
3
*
4
C
3
/
52
C
6
= 0.0034,
P(X=4) =
48
C
2
*
4
C
4
/
52
C
6
= 0.000055,
or in functional form:
P
X
=
x
=
۴۸
۶
−
x
۴
x
۵۲
۶
, x
=
۰,۱,۲,۳,۴.
c)
Find E(X)
E[X]=
0* P(X=0) + 1* P(X=1)+ 2* P(X=2)+ 3*P(X=3)+ 4* P(X=4) = 0.461538462.
Also notice that if we approximate this hypergeometric distribution with the binomial distribution with
parameters N=6, p=r/M=4/52, then the approximate expected value is Np=24/52=0.4615. Actually,
though the binomial distribution is only an approximation, the expected value is exact.
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)?
X~ Hypergeometric (M=20, N=3, r=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?
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
E[X]= N*(r/M) = 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
P(X=1)= 6/11 = 0.545
P(X=2)= 5/11*6/10 = 0.2727
P(X=3)= 5/11*4/10*6/9 = 0.1212
P(X=4)= 5/11*4/10*3/9*6/8 = 0.0455
P(X=5)= 5/11*4/10*3/9*2/8*6/7 = 0.013
P(X=6)= 5/11*4/10*3/9*2/8*1/7*6/6 = 0.0022
P(X=x) = 0 when x >=9
So E[X] = 1*P(X=1)+ 2*P(X=1)+ 3*P(X=3)+ 4*P(X=4)+ 5*P(X=5)+ 6*P(X=6)=1.7143
Question 3.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 Storer
 Probability theory, Geometric distribution

Click to edit the document details