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Chapter 3
Probability
PROBABILITY AND STATISTICS
CHAPTER 3 NOTES
SECTION 2
I. SOME PROBABILITY RULES -- COMPOUND EVENTS
We have seen how to calculate the probability of simple events for a statistical
experiment. Compound events are those which combine two or more simple events.
Two connectives with which simple events are combined to produce compound events
are AND and OR.
It is essential to differentiate between them and to recognize special cases for each.
A. The AND combination P(A and B).
1. Used when we are required to compute the probability of two events which occur
together or in sequence.
2. Look for words like AND, BOTH, TOGETHER, and phrases which combine two traits,
like KING OF SPADES (king AND spade), AGGRESSIVE SALESPERSON (aggressive
AND salesperson), MALE NURSE (male AND nurse), overworked teacher,
paranoid schizophrenic, etc.
3. To select the proper formula, we must first determine whether events A and B are
dependent or independent.
a) The two events are independent if P(A) does not affect P(B).
1) Roll a die and flip a coin. No matter what the die roll is, the probability
that the coin will come up heads is still ½. Likewise, no matter what
the coin does, the probability of rolling a 5 is still 1/6.
PROBABILITY AND STATISTICS
CHAPTER 3 NOTES
SECTION 2
a When the events are independent, use the multiplication rule.
1 P(A and B) = P(A) x P(B).
a). P(head and 5) = P(head) x P(5) = (1/2)(1/6) = 1/12
b) The two events are dependent if P(A) affects P(B).
1) In the rule for dependent events, we must examine the probability
that event B will occur under the condition that event A has already
occurred. This is called conditional probability and will be denoted
P(B, given A)
a. Example: A jar contains 5 red marbles, 3 green, and 2 blue. Draw
two marbles without replacing the first one before you draw the
second. Find the probability the first marble is red and the
second is green.
1. A = first marble is red
2. B = second marble is green. P(green) depends on the color
of the first marble.
b. When the events are dependent, use the following
multiplication rule:
1. P(A and B) = P(A) x P(B given A)
PROBABILITY AND STATISTICS
CHAPTER 3 NOTES
SECTION 2
a) Calculate P(A)
b) Assume that A has occurred.
c) Make the necessary adjustments to the sample
space, and calculate P(B given A).
d) Multiply these two values.
1) P(red) = 5/10, or ½.
a Once a red marble has been drawn,
only 9 marbles left, 3 of which are
green.
2) P(B given A) = 3/9 or 1/3.
3) P(A and B) = P(A) x P(B given A) = (1/2)(1/3)
or 1/6.
4. For most AND situations, we will have to use intuition to determine whether
two events are independent or dependent (and consequently which
multiplication rule to use). However, with enough information, we can figure
out whether two events are independent.
a) We know that if two events are independent, then P(A and B) = P(A) x
P(B). The converse of this is also true. In other words, if P(A) x P(B) = P(A
and B), then the two events are independent.
PROBABILITY AND STATISTICS
CHAPTER 3 NOTES
SECTION 2
5. When the two events are independent, P(B) = P(B given A).
a) The chance that B will occur is not influenced by whether or not A has
occurred.
1) To decide whether two events are independent:
a. Evaluate P(B).
b. Evaluate P(B given A)
c. When the two results are equal, the events are independent
d. When the two results are not equal, the events are
dependent.
PROBABILITY AND STATISTICS
CHAPTER 3 NOTES
SECTION 2
EXAMPLE:
Look at #15 on page 155.
Summer Vacation This Year
Own a
Computer
YES
NO
TOTAL
YES
46
11
57
NO
55
34
89
TOTAL
101
45
146
a) Find the probability that a randomly selected family is not taking a
summer vacation this year.
45/146 = .308
PROBABILITY AND STATISTICS
CHAPTER 3 NOTES
SECTION 2
EXAMPLE:
Look at #15 on page 155.
Summer Vacation This Year
Own a
Computer
YES
NO
TOTAL
YES
46
11
57
NO
55
34
89
TOTAL
101
45
146
b) Find the probability that a randomly selected family owns a computer.
57/146 = .390
PROBABILITY AND STATISTICS
CHAPTER 3 NOTES
SECTION 2
EXAMPLE:
Look at #15 on page 155.
Summer Vacation This Year
Own a
Computer
YES
NO
TOTAL
YES
46
11
57
NO
55
34
89
TOTAL
101
45
146
c) Find the probability that a randomly selected family is taking a summer
vacation given that they own a computer.
46/57 = .807
PROBABILITY AND STATISTICS
CHAPTER 3 NOTES
SECTION 2
EXAMPLE:
Look at #15 on page 155.
Summer Vacation This Year
Own a
Computer
YES
NO
TOTAL
YES
46
11
57
NO
55
34
89
TOTAL
101
45
146
d) Find the probability a randomly selected family is taking a summer
vacation this year AND owns a computer.
46/146 = .315
PROBABILITY AND STATISTICS
CHAPTER 3 NOTES
SECTION 2
EXAMPLE:
Look at #15 on page 155.
Summer Vacation This Year
Own a
Computer
YES
NO
TOTAL
YES
46
11
57
NO
55
34
89
TOTAL
101
45
146
e) Are the events of owning a computer and taking a summer vacation this
year independent or dependent? Explain.
The probability of taking a summer vacation is 101/146 = .692.
The probability of taking a summer vacation given that a computer
is owned is 46/57 = .807
These are NOT equal, so the events are DEPENDENT.
PROBABILITY AND STATISTICS
CHAPTER 3 NOTES
SECTION 2
Assignments:
Classwork: Pages 154-155 #1-12 All
Homework:Pages 155-158 #14-30 Evens