Statistics 427: Homework #3
Solutions to Assigned Problems
2.76
[6 pts]
The probability that an error is made on any given question is 0
.
1, which means that the probability
that no error is made is 0
.
9. Let
E
i
=
{
error made on question
i
}
. Then
P
(
E
i
) = 0
.
1 and
P
(
E
0
i
) =
0
.
9. We are interested in the probability that no errors are made in ten questions, or
P
(
E
0
1
∩···∩
E
0
10
).
Because the
E
i
are assumed to be independent, the
E
0
i
are also independent, and so
P
(
E
0
1
∩ ··· ∩
E
0
10
) =
10
Y
i
=1
P
(
E
0
i
) = (0
.
9)
10
= 0
.
3487
.
The probability that at least one error is made is that same as 1

P
(no errors are made) =
1

0
.
9
10
= 0
.
6513. If we abstract this to probability
p
of an error and
n
total questions, then the
probabilities are (1

p
)
n
and 1

(1

p
)
n
.
2.80
[6 pts]
Let
E
i
=
{
component
i
works
}
. Then
P
(
E
i
) = 0
.
9 and we are assuming that all components work
independently. In order for the system to work,
either
the “top” system (1/2)
or
the “bottom
system” (3/4) must work. For the top system to work we must have either component 1 or 2
working, i.e.
E
1
∪
E
2
. For the bottom system to work, we must have both 3 and 4 working,
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 Spring '08
 Staff
 Statistics, Probability, Probability theory, EMC E4, Tōhoku Shinkansen

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