Page 59 - Probability Demystified

48 CHAPTER 3 The Addition Rules

EXAMPLE: A card is selected at random from a deck of 52 cards. Find the
probability that it is a 6 or a diamond.

SOLUTION:

Let A ¼ the event of getting a 6; then PðAÞ¼ 4 since there are four 6s.
52
Let B ¼ the event of getting a diamond; then PðBÞ¼ 13 since there are
52
13 diamonds. Since there is one card that is both a 6 and a diamond (i.e., the
1
6 of diamonds), PðA and BÞ¼ . Hence,
52
4 13 1 16 4
PðA or BÞ¼ PðAÞþ PðBÞ PðA and BÞ¼ þ ¼ ¼
52 52 52 52 13

EXAMPLE: A die is rolled. Find the probability of getting an even number
or a number less than 4.

SOLUTION:
3
Let A ¼ an even number; then PðAÞ¼ since there are 3 even numbers—2, 4,
6
3
and 6. Let B ¼ a number less than 4; then PðBÞ¼ since there are 3 numbers
6
less than 4—1, 2, and 3. Let (A and B) ¼ even numbers less than 4 and
1
PðA and BÞ¼ since there is one even number less than 4—namely 2. Hence,
6
3 3 1 5
PðA or BÞ¼ PðAÞþ PðBÞ PðA and BÞ¼ þ ¼
6 6 6 6
The results of both these examples can be verified by using sample spaces
and classical probability.

EXAMPLE: Two dice are rolled; find the probability of getting doubles or a
sum of 8.

SOLUTION:

Let A ¼ getting doubles; then PðAÞ¼ 6 since there are 6 ways to get doubles
36
and let B ¼ getting a sum of 8. Then PðBÞ¼ 5 since there are 5 ways to get a
36
sum of 8—(6, 2), (5, 3), (4, 4), (3, 5), and (2, 6). Let (A and B) ¼ the number of
ways to get a double and a sum of 8. There is only one way for this event to
1
occur—(4, 4); then PðA and BÞ¼ . Hence,
36
6 5 1 10 5
PðA or BÞ¼ PðAÞþ PðBÞ PðA and BÞ¼ þ ¼ ¼
36 36 36 36 18