Math 135, Fall 2011: HW 5
This homework is due at the beginning of class on
Friday October 14th, 2011
. You are free to
talk with each other and get help. However, you should write up your own solutions and understand
everything that you write.
Problem 1
(p.182 #4)
.
Suppose all the numbers in a list of 100 numbers are nonnegative, and
that the average of the numbers in the list is 2. Prove that at most 25 of the numbers in the list are
greater than 8.
SOLUTION. We use Markov’s inequality.
Let
X
be a number drawn at random from the list.
We know
E
[
X
] = 2. Hence
P
(
X >
8)
≤
E
[
X
]
8
≤
1
4
so at most 1/4 of the 100 numbers, i.e. 25 of the numbers in the list, can be bigger than 8.
Problem 2
(p.182 #6)
.
Let
X
be the number of spades in 7 cards dealt from a wellshuffled deck
of 52 cards containing 13 spades. Find
E
[
X
].
SOLUTION. The probability distribution of
X
,
P
(
X
=
i
), where
i
is an integer between 0 and
13, can be computed using the hypergeometric distribution:
P
(
X
=
i
) =
(
13
i
)(
39
13

i
)
(
52
13
)
⇒
E
[
X
] =
13
X
i
=0
iP
(
X
=
i
)
Alternatively, one can solve this with the method of indicator functions:
let
I
j
be the indicator
function of the event
A
j
=
{
jth card is a spade
}
. Note that
P
(
A
j
) =
13
52
.
If you are concerned about
the fact that, say, the probability of the second card being a spade depends on the previous card,
observe that you can condition:
P
(
A
2
) =
P
(
A
2

A
1
)
P
(
A
1
) +
P
(
A
2

A
c
1
)
P
(
A
c
1
)
=
12
51
13
52
+
13
51
39
52
=
13
52
(Indeed, this was a homework problem on a previous homework set). Now, if
X
is the total number
of spades in the hand,
X
=
7
X
j
=1
I
j
⇒
E
(
X
) =
E
7
X
j
=1
I
j
=
7
X
j
=1
E
[
I
j
] =
7
X
j
=1
13
52
=
7
4
1
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Problem 3
(p.182 #7)
.
In a circuit containing
n
switches, the
i
th switch is closed with probability
p
i
,
i
= 1
,
· · ·
, n
. Let
X
be the total number of switches that are closed. Is it possible to calculate
E
[
X
] without further assumptions? If so, what is
E
[
X
]?
SOLUTION. It is
not
possible to know the
probability distribution
of
X
without further assumptions.
However, we can still compute the expected value of
X
by using the method of indicator functions.
Let
A
i
=
{
Switch
i
is closed
}
.
Let
I
A
i
be the indicator function of
A
i
, so
I
A
i
= 1 if switch
i
is
closed, and 0 if switch
i
is open. We note that
X
=
n
X
i
=1
I
A
i
because
X
is simply the total number of switches that are closed. Hence by the addition rule for
expectation,
E
[
X
] =
E
"
n
X
i
=1
I
A
i
#
=
n
X
i
=1
E
[
I
A
i
]
=
n
X
i
=1
[1
P
(
A
i
) + 0
P
(
A
c
i
)]
=
n
X
i
=1
P
(
A
i
)
Problem 4
(p.182#10)
.
Let
A
and
B
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 Fall '08
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 Math, Probability theory, probability density function, x1 x2

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