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IE 111 Fall Semester 2007
Homework #4
Due Monday, 10/1 at start of class
Question 1.
Suppose we have a biased coin such that P(head)
≠
P(tail).
Suppose I flip the coin three
times.
Let the random variable X be the number of heads in the 3 flips.
You are given
the following information:
P(X=0) = 2/27
P(X=1) = 6/27
P(X=2) = ?
P(X=3) = 9/27
a)
Find P(X=2)
2/27 + 6/27 + P(X=2) + 9/27 = 1
there fore P(X=2) = 10/27
b)
Find P(X
≤
1)
P(X
≤
1) = P(X=0) + P(X=1) = 8/27
c)
Find P( 1
≤
X < 2)
P(1
≤
X<2) = P(X=1) = 6/27
a)
Find P( X=1  X
≤
1)
P(X=1  X
≤
1) = P(X=1and X
≤
1) / P(X
≤
1)
=
P(X=1) / P(X
≤
1)
=
(6/27) / (8/27) = 6/8
Question 2.
A die has 4 sides (not 6 like a regular die).
The four sides are labeled 1, 3, 5, and 10
respectively.
It is equally likely that you will get a 1 or a 3.
It is equally likely that you
will get a 5 or a 10.
It is three times more likely that you will get a 5 or10 than a 1 or 3.
a)
Let X = the outcome of a roll of the die.
Find the probability mass function of X.
x
1
3
5
10
P
X
(x)
p
p
3p
3p
8p = 1
thus p=1/8 and the PMF is
x
1
3
5
10
P
X
(x)
1/8
1/8
3/8
3/8
b)
If Y=2X
2
+1,
find P
Y
(y)
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1
3
5
10
y
3
19
51
201
P
Y
(y)
1/8
1/8
3/8
3/8
c)
If I roll the die 10 times, what is the probability I get exactly 3 tens?
Let Z = number of tens in 10 rolls.
Z is Binomial( n=10, p=1/3)
P(Z=3) =
10
C
3
(3/8)
3
(5/8)
7
=
0.235741
Question 3.
Let the random variable X be the value of the up face of a certain
unfair
die.
For this
die, the probability of getting a specific number (1 through 6) is proportional to twice
that number.
Thus getting a “5” is 10 times as likely as getting a “1”, the probability of a
“4” is four times the probability of a “2”, etc.
a)
Find the Probability Mass Function P
X
(x).
x
1
2
3
4
5
6
P
X
(x)
p
4p
6p
8p
10p
12p
p+4p+6p+8p+10p+12p = 1
41p=1
thus p=1/41
so the PMF is
x
1
2
3
4
5
6
P
X
(x)
1/41
4/41
6/41
8/41
10/41
12/41
b)
Find P(X
≤
5)
P(X
≤
5) = 1  P(X=6) = 1  12/41 = 29/41
c)
P(X
≠
4  3 < X
≤
5)
P(X
≠
4  3<X
≤
5)
= P(X
≠
4 and 3<X
≤
5) / P(3<X
≤
5) = P(X=5) / P(X=4 or X=5)
= (10/41) / (18/41) = 5/9
d) Find the probability of rolling a number greater than or equal to “4” three times in a
row.
P(X
≥
4) = P(X=4) + P(X=5) + P(X=6) = (8+10+12)/41
=
30/41
P(three times in a row) = (30/41)
3
e)
The Cumulative Distribution Function F
X
(x)
is Defined as F
X
(x)
= P(X
≤
x)
Find the C.D.F. for the random variable X.
F
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This homework help was uploaded on 02/26/2008 for the course IE 111 taught by Professor Storer during the Spring '07 term at Lehigh University .
 Spring '07
 Storer

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