Solutions to Homework 4  3C
12.3.10
The question is asking us to find
P
(
A

B
) where we let
A
= the event that at least two heads occurred,
B
= the event that the second toss was a head.
We can find this using the formula
P
(
A

B
) =
P
(
A
∩
B
)
/P
(
B
). The sample
space is Ω =
{
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
}
, and thus
we can see that the sets we need are
A
∩
B
=
{
HHH, HHT, THH
}
and
B
=
{
HHH, HHT, THH, THT
}
. From these we can deduce that
P
(
A
∩
B
) =

A
∩
B


Ω

=
3
8
, P
(
B
) =

B


Ω

=
1
2
,
and hence
P
(
A

B
) = 3
/
4.
2.3.14
Let the
A
be the event that that the ball we pick is blue, and let
B
i
be the
event that we pick the ball out of bag
i
. e.g.
B
1
is the event that we pick the
ball out of bag 1. To find
P
(
A
) we will use the Law of Total Probability
P
(
A
) =
6
i
=1
P
(
A

B
i
)
P
(
B
i
)
.
Therefore we need to find
P
(
A

B
i
) and
P
(
B
i
).
Here’s another way of thinking of the
B
i
: its the event that the value of the
dieroll is
i
. Therefore we can deduce that
P
(
B
i
) =
1
6
since the probability that we roll each value on a die is equal, and there are 6
sides.
P
(
A

B
i
) represents the probability that we pick a blue ball out of bag
i
.
Bag
i
contains
i
blue ball and 2 green balls (so
i
+ 2 in total). Therefore we can
deduce that
P
(
A

B
i
) =
i
i
+ 2
.
Putting this all together, we get
P
(
A
) =
1
6
1
3
+
1
2
+
3
5
+
2
3
+
5
7
+
3
4
= 0
.
59
12.3.23
1
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Since no replacement is allowed after you take the first ball, we can intuitively
guess that the two events are not independent. To show that this we need to
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 Spring '07
 SCHONMANN
 Probability theory, Probability mass function

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