The Addition Rule

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Section 3.3
The Addition Rule
Statistics
Mrs. Spitz
Fall 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) or
P(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. Independent
6. Dependent
7. Dependent
8. Independent
9a. 0.8
9b. 0.0032
9c. Dependent
10a. 0.740
10b. 0.822
10c. Dependent
Larson/Farber Ch. 3
11a. 0.0168
11b. 0.93
12a. 0.24
12b. 0.6
13a. 0.109
13b. 0.382
13c. 0.618
14. 0.083
15a. 0.839
15b. 0.167
15c. 0.506
15d. Dependent
Check in assignment 3.2
16a. 0.556
16b. 0.525
16c. 0.167
16d. Dependent
17a. 0.0000000243
17b. 0.859
17c. 0.141
18a. 0.055
18b. 0.238
18c. 0.762
19a. 0.2
19b. 0.04
Larson/Farber Ch. 3
19c. 0.008
19d. 0.488
20a. 0.985
20b. 0.015
20c. 0.000000125
21. 0.954
22. 0.933
23a. 0.444
23b. 0.4
24a. 0.074
24b. 0.999
25a. 0.462
25b. 0.538
25c. Yes
25d. Answers will vary
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).
B
A
A and B
Larson/Farber Ch. 3
B
A
A or B
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 Philadelphia
B = A person was born in Houston
A
B
Mutually exclusive
P(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 old
B = A person is a lawyer
A = A person was born in Philadelphia
B = A person watches West Wing on TV
A and B
Non-mutually exclusive
P(A and B) ≠ 0
Larson/Farber Ch. 3
A
B
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) = 2/52
P(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
Contingency Table
The results of responses when a sample of adults in
3 cities was asked if they liked a new juice is:
Omaha
Yes
100
No
125
Undecided 75
Total
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
One of the responses is selected at random. Find:
1. P(Miami and Yes)
3. P(Miami or Yes)
2. P(Miami and Seattle) 4. P(Miami or Seattle)
Larson/Farber Ch. 3
Contingency Table
Yes
No
Undecided
Total
Omaha
100
125
75
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
One of the responses is selected at random. Find:
1. P(Miami and Yes)
= 250/1000 • 150/250 = 150/1000
= 0.15
2. P(Miami and Seattle) = 0
Larson/Farber Ch. 3
Contingency Table
Yes
No
Undecided
Total
Omaha
100
125
75
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
3 P(Miami or Yes)
250/1000 + 400/1000 – 150/1000
= 500/1000 = 0.5
4. P(Miami or
Seattle)
250/1000 + 450/1000 – 0/1000
= 700/1000 = 0.7
Larson/Farber Ch. 3
Summary
For complementary events P(E') = 1 - P(E)
Subtract the probability of the event from one.
The probability both of two events occur
P(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.
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.
Larson/Farber Ch. 3
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