Larson/Farber Ch. 3

Section 3.3

The Addition Rule

StatisticsMrs. SpitzFall 2008

Larson/Farber Ch. 3

Check in assignment 3.2

1. Two events are independent if the occurrence of one of the events does not affect the probability of the occurrence of the other event. If P(B|A) = P(B) orP(A|B) = P(A), then Events A and B are independent.

2a. Roll a die twice. The outcome of the 2nd toss is independent of the outcome of the 1st toss.

2b. Draw two cards(without replacement) from a standard 52 card deck. The outcome of the 2nd card is dependent upon the outcome of the 1st card.

3. False. If two events are independent, P(A|B) = P(A)4. False. If events A and B are independent, then P(A and

B) = P(A)●P(B)

Larson/Farber Ch. 3

Check in assignment 3.2

5. Independent6. Dependent7. Dependent8. Independent9a. 0.89b. 0.00329c. Dependent10a. 0.74010b. 0.82210c. Dependent

11a. 0.016811b. 0.9312a. 0.2412b. 0.613a. 0.10913b. 0.38213c. 0.61814. 0.08315a. 0.83915b. 0.16715c. 0.50615d. Dependent

Larson/Farber Ch. 3

Check in assignment 3.2

16a. 0.55616b. 0.52516c. 0.16716d. Dependent17a. 0.000000024317b. 0.85917c. 0.14118a. 0.05518b. 0.23818c. 0.76219a. 0.219b. 0.04

19c. 0.00819d. 0.48820a. 0.98520b. 0.01520c. 0.00000012521. 0.95422. 0.93323a. 0.44423b. 0.424a. 0.07424b. 0.99925a. 0.46225b. 0.53825c. Yes25d. Answers will vary

Larson/Farber Ch. 3

Objectives/Assignment

• How to determine if two events are mutually exclusive

• How to use the addition rule to find the probability of two events.

• Assignment: 129-131 #1-18 all

Larson/Farber Ch. 3

What is different?

• In probability and statistics, the word “or” is usually used as an “inclusive or” rather than an “exclusive or.” For instance, there are three ways for “Event A or B” to occur.– A occurs and B does not occur– B occurs and A does not occur– A and B both occur

Larson/Farber Ch. 3

Independent does not mean mutually exclusive

• Students often confuse the concept of independent events with the concept of mutually exclusive events.

Larson/Farber Ch. 3

Study Tip

• By subtracting P(A and B), you avoid double counting the probability of outcomes that occur in both A and B.

Larson/Farber Ch. 3

Compare “A and B” to “A or B”

The compound event “A and B” means that A and B both occur in the same trial. Use the multiplication rule to find P(A and B).

The compound event “A or B” means either A can occur without B, B can occur without A or both A and B can occur. Use the addition rule to find P(A or B).

A B

A or BA and B

A B

Larson/Farber Ch. 3

Mutually Exclusive Events

Two events, A and B, are mutually exclusive if they cannot occur in the same trial.

A = A person is under 21 years old B = A person is running for the U.S. Senate

A = A person was born in PhiladelphiaB = A person was born in Houston

A B Mutually exclusiveP(A and B) = 0

When event A occurs it excludes event B in the same trial.

Larson/Farber Ch. 3

Non-Mutually Exclusive Events

If two events can occur in the same trial, they are non-mutually exclusive.

A = A person is under 25 years oldB = A person is a lawyer

A = A person was born in PhiladelphiaB = A person watches West Wing on TV

A BNon-mutually exclusiveP(A and B) ≠ 0

A and B

Larson/Farber Ch. 3

The Addition Rule

The probability that one or the other of two events will occur is: P(A) + P(B) – P(A and B)

A card is drawn from a deck. Find the probability it is a king or it is red.A = the card is a king B = the card is red.

P(A) = 4/52 and P(B) = 26/52 but P(A and B) = 2/52P(A or B) = 4/52 + 26/52 – 2/52

= 28/52 = 0.538

Larson/Farber Ch. 3

The Addition Rule

A card is drawn from a deck. Find the probability the card is a king or a 10.A = the card is a king B = the card is a 10.

P(A) = 4/52 and P(B) = 4/52 and P(A and B) = 0/52

P(A or B) = 4/52 + 4/52 – 0/52 = 8/52 = 0.054

When events are mutually exclusive, P(A or B) = P(A) + P(B)

Larson/Farber Ch. 3

The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is:

Contingency Table

3. P(Miami or Yes)

4. P(Miami or Seattle)

Omaha Seattle Miami TotalYes 100 150 150 400No 125 130 95 350Undecided 75 170 5 250Total 300 450 250 1000

One of the responses is selected at random. Find:

1. P(Miami and Yes)

2. P(Miami and Seattle)

Larson/Farber Ch. 3

Contingency Table

1. P(Miami and Yes)

2. P(Miami and Seattle)

= 250/1000 • 150/250 = 150/1000 = 0.15

= 0

Omaha Seattle Miami TotalYes 100 150 150 400No 125 130 95 350Undecided 75 170 5 250Total 300 450 250 1000

One of the responses is selected at random. Find:

Larson/Farber Ch. 3

Contingency Table

3 P(Miami or Yes)

4. P(Miami or Seattle)

250/1000 + 450/1000 – 0/1000= 700/1000 = 0.7

Omaha Seattle Miami TotalYes 100 150 150 400No 125 130 95 350Undecided 75 170 5 250Total 300 450 250 1000

250/1000 + 400/1000 – 150/1000= 500/1000 = 0.5

Larson/Farber Ch. 3

Probability at least one of two events occur

P(A or B) = P(A) + P(B) - P(A and B)

Add the simple probabilities, but to prevent double counting, don’t

forget to subtract the probability of both occurring.

For complementary events P(E') = 1 - P(E)Subtract the probability of the event from one.

The probability both of two events occurP(A and B) = P(A) • P(B|A)

Multiply the probability of the first event by the conditional probability the second event occurs, given the first occurred.

Summary