AMS 311 (Fall, 2009)
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 4. 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
=“bothare6”=
{
(6
,
6)
}
and
F
= “sum is divisible by 4” =
{
(1,3), (2,2), (2,6), (3,1),
(3,5), (4,4), (5,3), (6,2), (6,6)
}
.
P
(
E

F
)=
P
(
E
∩
F
)
P
(
F
)
=
P
(
{
(6
,
6)
}
)
P
(
{
(1
,
3)
,
(2
,
2)
,
(2
,
6)
,
(3
,
1)
,
(3
,
5)
,
(4
,
4)
,
(5
,
3)
,
(6
,
2)
,
(6
,
6)
}
)
=
1
9
(2).
(12 points)
Suppose for simplicity that the number of children in a familyis1
,2
,or3
,w
i
thprobab
i
l
i
ty
1/3 each. Little Bobby has no sisters. What is the probabilitytha
theisanon
lych
i
ld?(Se
ttheprob
lemup
carefully. Remember to deFne the sample space, and any eventstha
tyouuse
!)
We can de±ne the sample space as
S
=
{
B, G, BB, BG, GB, GG, BBB, BBG, BGB, BGG, GBB, GBG,
GGB, GGG
}
. Let
A
=
{
B,BB,BBB
}
be the event that there are no girls in the family (ie, Bobby has no
sisters). Let
B
i
be the event that the family has
i
children. (e.g.,
B
1
=
{
B
}
and
B
2
=
{
BB,BG,GB,GG
}
)
P
(
B
1
∩
A
)=
P
(
B
1
)
·
P
(
A

B
1
)=(1
/
3)(1
/
2) = 4
/
24
P
(
B
2
∩
A
)=
P
(
B
2
)
·
P
(
A

B
2
)=(1
/
3)(1
/
4) = 2
/
24
P
(
B
3
∩
A
)=
P
(
B
3
)
·
P
(
A

B
3
)=(1
/
3)(1
/
8) = 1
/
24
We want to compute:
P
(
B
1

A
)=
P
(
B
1
∩
A
)
P
(
A
)
=
4
/
24
4
/
24 + 2
/