P B
P A
B
.
Proof:
Clearly we can write the set
B
as
1
(
)
n
i
i
B
P A
B
. Since the collection of
i
A
form a partition, the collection of sets
1
(
)
n
i
i
A
B
are disjoint sets. Rule 3 in
our definition of probability says the probability of the union of disjoint sets equals the sum of the
probabilities.
The Law of Total Probability:
To determine the probability of an event A, you sum up the probability of
all ways that A can occur (loosely speaking).
Example 2.32:
An urn contains 8 white, 5 red and 3 blue chips. A person selects 3 chips without
replacement. Determine the following probability: P(The third chip is red)
One way to solve this problem is to determine all of the ways that the third chip is red. There are 4 ways
that the third chip can be red. Thus, we have
3
1
2
3
1
2
3
1
2
3
1
2
3
(
)
(
)
(
)
(
)
(
)
5
4
3
5 11 4
11 5
4
11 10 5
16 15 14
16 15 14
16 15 14
16 15 14
1050
5
.3125
3360
16
P R
P R
R
R
P R
NR
R
P NR
R
R
P NR
NR
R
The second and third fraction in each part are determined by conditional probability. For the first term, a
red on the first chip has probability 5/16. Once the first chip is red, we are down to 15 chips of which
only 4 are red.
So
2
1
(

)
4 /15
P R
R
.
A simpler way to work THIS problem is to consider that you will pick the third chip first, set it aside and
then pick the first two. Then the answer is more simply calculated as 5/16. This simple alternative
solution should not be expected to be available on other problems.
Example 2.33:
Two cards are drawn from a standard deck of 52 cards. Determine the probability that the
first card is a Heart and the second card is an Ace.
39
The simple method does not work on this problem. Here there are two things that need to take place
and not just one thing as in Example 2.32 above (third chip red). Therefore, this problem
needs
the Law
of Total Probability approach.
The problem is written out in probability format as follows:
1
2
1
2
1
(
)
(
)
(

)
P H
A
P H
P A
H
. Determining
1
(
)
13/ 52
P H
is effortless. The second part seems to create some possible confusion. Clearly there are
51 cards left as the new whole. But, how many are Aces? Are there 3 Aces or 4 Aces. It depends if that
first card, which was a heart, is an Ace. If it was, then there are 3 Aces left for the second cars, if the first
card was not an Ace, then there are 4 Aces left.
1
2
1
2
1
2
(
)
(
)
([
]
)
1
3
12
4
51
1
52 51
52 51
52 51
52
P H
A
P A
A
P non Ace
A
w
w
Example 2.34:
An urn contains 7 white chips, 5 red chips and 3 blue chips. A chip is randomly selected
from the urn, the color is noted, then chip is returned to the urn and then the number of chips of the
noted color is doubled. Now a second chip is selected. Determine the probability that the first chip was
white given that the second chip is blue. How should we start?
1
2
1
2
1
2
2
1
2
1
2
1
2
(
)
(
)
(

)
(
)
(
)
(
)
(
)
P W
B
P W
B
P W
B
P B
P W
B
P R
B
P B
B
1
2
1
1
2
1
1
2
1
1
2
1
7
3
(
)
(

)
15 22
7
3
5
3
3
6
(
)
(

)
(
)
(

)
(
)
(

)
15 22
15 20
15 18
P W
P B
W
P W
P B
W
P R
P B
R
P B
P B
B
The first step above is the definition of conditional probability. Step 2 uses the law of total probability.