Problem 1.
Sample space
S
=
{
all possible combinations of the two teams
that can be formed
}
.

S

=
10
5
·
15
5
.
We assume all possible team combinations are equally likely.
1. Define
E
=
{
teams chosen such that Lebron is on company A’s team
AND
Larry is on company B’s team
}
. We want to find
P
(
E
) and we
know that
P
(
E
) =

E


S

.
A team for company A with Lebron in it can be chosen
(
9
4
)
ways. A
team for company B with Larry in it can be chosen
(
14
4
)
ways. So

E

=
9
4
·
14
4
.
Therefore
P
(
E
) =

E


S

=
(
9
4
)
·
(
14
4
)
(
10
5
)
·
(
15
5
)
.
=
1
6
=
.
167
.
2. Define
F
=
{
teams chosen such that Lebron does not play and Larry
does play for company B
}
. We want
P
(
F
).
A team for company A with Lebron not playing can be chosen
(
9
5
)
ways.
A team for company B with Larry playing in it can be chosen
(
14
4
)
ways
(as before). Then

F

=
9
5
·
14
4
.
Therefore
P
(
E
) =

F


S

=
(
9
5
)
·
(
14
4
)
(
10
5
)
·
(
15
5
)
.
=
1
6
=
.
167
.
Problem 2.
Define
F
i
=
{
firm
i
is involved
}
,
i
= 1
,
2
,
3
, C
=
{
a cost overrun
occurs
}
.
We are given
P
(
F
1
) =
.
5
,
P
(
F
2
) =
.
2
.
P
(
C

F
1
) =
.
05
,
P
(
C

F
2
) =
.
1
,
P
(
C

F
3
) =
.
15
.
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 Spring '10
 RohiniKumar
 Probability theory, The ATeam, Bayesian probability, Lebron

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