Solutions to Homework 04
1.
a.
P
[
A
∪
B
] =
P
[
A
] +
P
[
B
]

P
[
A
∩
B
] =
P
[
A
] +
P
[
B
]

P
[
A
]
P
[
B
].
b.
P
[
A
∪
B
] =
P
[
A
] +
P
[
B
]

P
[
A
∩
B
] =
P
[
A
] +
P
[
B
]. (since
A
∩
B
=
φ
⇒
P
[
A
∩
B
] = 0)
2.
a.
Using the theorem on total probability:
P
[
A
]
=
P
[
A

¯
A
]
P
[
¯
A
] +
P
[
A

A
]
P
[
A
]
=
P
[
A

¯
A
](1

P
[
A
]) +
P
[
A

A
]
P
[
A
]
=
0
.
3(1

P
[
A
]) + 0
.
5
P
[
A
]
⇒
P
[
A
] = 0
.
3750
b.
P
[
¯
A
] = 1

0
.
3750 = 0
.
6250
3.
a.
This is a case of the Binomial probability law.
P
[
k
errors] =
n
k
!
p
k
(1

p
)
n

k
b.
Type 1 errors occur with probability
pa
and do not occur with probability 1

pa
. Each
operation can now be considered a Bernoulli trial with probability of “success” as
pa
. Then,
P
[
k
1
type 1 errors] =
n
k
1
!
(
pa
)
k
1
(1

pa
)
n

k
1
c.
Similarly, type 2 errors occur with probability
p
(1

a
) and do not occur with probability
1

p
(1

a
). Then,
P
[
k
2
type 2 errors] =
n
k
2
!
(
p
(1

a
))
k
2
(1

p
(1

a
))
n

k
2
d.
This can be modeled as a multinomial distribution with three possible outcomes: no
errors, type 1 errors and type 2 errors. These three events occur with probabilities 1

p
,
pa
and
p
(1

a
) respectively. We are interested in the case where out of
n
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 Fall '08
 GELFAND
 Probability, Probability theory, k2, black balls

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