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
(1-p)
2
=3* (6/20)
1
*(1-6/20)
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) = (1-p)
6
p
=(1-6/20)
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