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AMS 311 (Fall, 2011)
Joe Mitchell
PROBABILITY THEORY
Homework Set # 2 – Solution Notes
(1).
(12 points)
Suppose that two fair dice have been tossed and the total of their top faces is found to be
divisible by 3. What is the probability that both of them have landed 6?
The sample space consists of 36 equally likely outcomes,
S
=
{
(1
,
1)
,...,
(6
,
6)
}
. We want to compute
P
(
E

F
), where
E
= “both are 6”=
{
(6
,
6)
}
and
F
= “sum is divisible by 3” =
{
(1,2), (1,5), (2,1), (2,4),
(3,3), (3,6), (4,2), (4,5), (5,1), (5,4), (6,3), (6,6)
}
.
P
(
E

F
) =
P
(
E
∩
F
)
P
(
F
)
=
P
(
{
(6
,
6)
}
)
P
(
{
(1
,
2)
,
(1
,
5)
,
(2
,
1)
,
(2
,
4)
,
(3
,
3)
,
(3
,
6)
,
(4
,
2)
,
(4
,
5)
,
(5
,
1)
,
(5
,
4)
,
(6
,
3)
,
(6
,
6)
}
)
=
1
12
(2).
(12 points)
Suppose for simplicity that the number of children in a family is 1, 2, or 4, with probability
1/3 each. Little Alice (a girl) has no brothers. What is the probability that she is an only child? (Set the
problem up carefully. Remember to deFne the sample space, and any events that you use!)
We can de±ne the sample space as
S
=
{
B, G, BB, BG, GB, GG, BBBB, BBBG, BBGB, BBGG,
BGBB, BGBG, BGGB, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG
}
. Let
A
=
{
G,GG,GGGG
}
be the event that there are no boys in the family (ie, Alice has no brothers). Let
B
i
be the event that the family has
i
children. (e.g.,
B
1
=
{
B,G
}
and
B
2
=
{
BB,BG,GB,GG
}
) The events
B
1
,
B
2
,
B
4
partition the sample space
S
.
Note that the outcomes of the sample space are
not
equally likely. However, the 2 outcomes within
B
1
are equally likely, the 4 outcomes within
B
2
are equally likely, and the 16 outcomes within
B
4
are equally
likely.
We want to compute the probability that Alice is an only child (event
B
1
happens),
given
that Alice has
no brothers (event
A
happens). That is, we want to compute
P
(
B
1

A
) =
P
(
B
1
∩
A
)
P
(
A
)
=
P
(
A

B
1
)
P
(
B
1
)
P
(
A

B
1
)
P
(
B
1
) +
P
(
A

B
2
)
P
(
B
2
) +
P
(
A

B
4
)
P
(
B
4
)
=
1
2
·
1
3
1
2
·
1
3
+
1
4
·
1
3
+
1
16
·
1
3
=
8
13
,
where we have used the fact that
P
(
A

B
1
) = 1
/
2,
P
(
A

B
2
) = 1
/
4, and
P
(
A

B
4
) = 1
/
16.
(3).
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