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Chapter
9
Probability
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9-2 Multistage Experiments with Tree
Diagrams and Geometric Probabilities
 More Multistage Experiments
 Independent Events
 Modeling Games
 Geometric Probability (Area
Models) Revisited
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Tree Diagrams
Suppose the spinner is
spun twice. The tree
diagram shows the
possible outcomes.
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Tree Diagrams
The sample space can
be written
{BB, BR, BY, RB, RR,
RY, YB, YR, YY}.
The probability of each
outcome is
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Tree Diagrams
Alternatively, we can generate the sample space
using a table.
YB
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Example 9-5
Suppose we toss a fair coin 3 times and record the
results. Find each of the following:
a. The sample space for this experiment.
S = {HHH, HHT, HTH, THH, TTT, TTH, THT, HTT}
b. The event A of tossing 2 heads and 1 tail
A = {HHT, HTH, THH}
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Example 9-5
(continued)
c. The event B of tossing no tails.
B = {HHH}
d. The event C of tossing a head on the last toss
C = {HHH, HTH, THH, TTH}
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Example 9-6a
When rolling a pair of fair dice, find the probability
of the event A = rolling double sixes.
There are 36
possible rolls.
There is only
one way to roll
double sixes.
1
P ( A) 
36
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Example 9-6b
Find the probability of rolling a sum of 7 or 11
when rolling a pair of fair dice.
There are 36
possible rolls.
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Example 9-6b
(continued)
There are 6 ways to
form a sum of “7”, so
There are 6 ways to
form a sum of “11”, so
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Example 9-6b
(continued)
The sample space for the experiment is {2, 3, 4, 5,
6, 7, 8, 9, 10, 11, 12}, but the sample space is not
uniform; i.e., the probabilities of the given sums
are not equal.
The probability of rolling a sum of 7 or 11 is
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Example 9-7
A fair pair of dice is rolled. Let E be the event of
rolling a sum that is an even number and F the
event of rolling a sum that is a prime number. Find
the probability of rolling a sum that is even or
prime, that is, P(E U F).
E U F = {2, 4, 6, 8, 10, 12, 3, 5, 7, 11}
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Example 9-7
(continued)
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Example 9-7
(continued)
Alternate solution 1:
E = {2, 4, 6, 8, 10, 12} and F = {2, 3, 5, 7, 11}.
Thus, E and F are not mutually exclusive because
E ∩ F = {2}.
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Example 9-7
(continued)
Alternate solution 2:
E = {2, 4, 6, 8, 10, 12} and F = {2, 3, 5, 7, 11}.
Thus, E U F = {2, 3, 4, 5, 6, 7, 8, 10, 11} and
E U F = {9}.
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More Multistage Experiments
The box in figure (a) contains one colored ball and two
white balls. If a ball is drawn at random and the color
recorded, a tree diagram for the experiment might look
like the one in figure (b). Because each ball has the same
chance of being drawn, we may combine the branches
and obtain the tree diagram shown in figure (c).
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Suppose a ball is drawn at random
from the box in the figure and its
color recorded.
The ball is then replaced, and a second ball is
drawn and its color recorded. The sample space
for this two-stage experiment may be recorded
using ordered pairs as
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The tree diagram for this experiment is shown below.
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Multiplication Rule for Probabilities for
Tree Diagrams
For all multistage experiments, the probability of
the outcome along any path of a tree diagram is
equal to the product of all the probabilities along
the path.
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Suppose two balls are drawn one-by-one without
replacement.
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Independent Events
Independent events:
when the outcome of one event has no
influence on the outcome of a second event.
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Independent Events
If two coins are flipped and event E1 is obtaining
a head on the first coin and E2 is obtaining a tail
on the second coin, then E1 and E2 are
independent events because one event has no
influence on possible outcomes.
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Independent Events
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Independent Events
For any independent events E1 and E2,
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Example 9-8
The figure shows a box with 11 letters. Some
letters are repeated. Suppose 4 letters are drawn
at random from the box one-by-one without
replacement. What is the probability of the
outcome BABY, with the letters chosen in exactly
the order given?
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Example 9-8
(continued)
We are interested in only the tree branch leading
to the outcome BABY.
The probability of the first B is
are 2 B’s out of 11 letters.
because there
The probability of the second B is because there
are 9 letters left after 1 B and 1 A have been chosen.
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Example 9-9
A letter is drawn from box 1 and placed in box 2.
Then, a letter is drawn from box 2 and placed in box
3. Finally, a letter is drawn from box 3. What is the
probability that the letter drawn from box 3 is B?
(Call this event E.)
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Example 9-9
(continued)
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Modeling Games
There are two colored marbles and one white
marble in a box. Gwen mixes the marbles, and
Arthur draws two marbles at random without
replacement. If the two marbles match, Arthur
wins; otherwise, Gwen wins. Does each player
have an equal chance of winning?
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Modeling Games
The probability that the marbles are the same color
is
while the probability that they are not the
same color is
Thus, the players do not have the
same chance of winning.
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