# Section 3.6: The Multiplicative Rule and Independent Events Now

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```					Section 3.6                                           MATH 2600/Dietrich
Page 1                                       Georgia College & State University

Section 3.6: The Multiplicative Rule and Independent Events

Now that we know about independent and dependent events, we
can work with multistage experiments where we are interested
in the probability that two or more events occur in sequence.

Consider the following spinner that is divided into thirds:

In an experiment, suppose that a single trial consists
of the spinner being spun two times. What is the
probability of spinning a 1 on both spins?

If we let A = “get a 1 on the first spin” and B = “get a 1 on the
second spin” we can then determine P(A and B), or equivalently,
P(A ∩ B).

One way that we can determine this probability is by using the
General Multiplication Rule for Probability.

General Multiplication Rule:
Let A and B be two events defined in a sample space S.

Then P(A ∩ B) = P(A) · P(B|A)

Note: How to recognize situations that result in the
compound event “A ∩ B.”
1. A followed by B.
2. A and B occurred simultaneously.
3. The intersection of A and B.
Section 3.6                                            MATH 2600/Dietrich
Page 2                                        Georgia College & State University

By the General Multiplication Rule, P(A ∩ B) = P(A) · P(B|A).
We know that P("get a 1 on the first spin") = P(A) = 1/3.

The interesting part of this problem is determining P(B|A). What
is the chance of spinning a 1 on the second spin if a 1 was the
result of the first spin?

We need to ask the following question: Does the result of the first
spin affect the results of the second spin? Since they do not, we
can say that the spins are independent events and, therefore, that
P(B|A) will equal P(B).

Special Multiplication Rule:
Let A and B be two events defined in a sample space S. If A and B
are independent events, then
P(A ∩ B) = P(A) · P(B)

For each spin, P(“1”) = 1/3 so

P(A ∩ B) = P(A)·P(B) = 1/3 · 1/3 = 1/9.

1
There is a        chance that a 1 will be spun twice.
9
Section 3.6                                               MATH 2600/Dietrich
Page 3                                           Georgia College & State University

Example:
In an experiment, suppose that a fair coin is tossed and a fair die
is rolled. Recall that a listing of all possible outcomes for a
probability experiment is called the sample space. For multistage
experiments, a useful way to determine the outcomes in a sample
space is to draw a tree diagram.

Suppose we are interested in performing
the experiment and observing heads on
the coin and a 4 on the die.

We know that P(“Heads”) = ½ since the
coin is fair.

We want to consider P(“Rolling a 4 after
Heads”). Does the chance that a 4 is
rolled on the die depend on the outcome
on the coin toss?

Since the probability of rolling a 4 is not
affected by the outcome on the coin
toss, we know that P(“Rolling a 4 after
Heads”) is the same as P(“Rolling a 4
after Tails”). Therefore, we are dealing
with independent events. We can then
conclude that P(“Rolling a 4 after
Heads”) = P(“Rolling a 4”) = 1/6.

Using the Special Multiplication Rule, there is a 1 2 ⋅ 1 6 = 112 probability
that the sequence of events (head followed by 4) occurs.

NOTE:
It is important to recognize that each of the twelve branches of this
particular tree (and, hence, the twelve possible outcomes in this sample
space) are equally likely and have probability 112 . Note also that each
outcome is mutually exclusive of every other outcome.
Section 3.6                                                  MATH 2600/Dietrich
Page 4                                              Georgia College & State University

Example:

Suppose you draw two marbles without replacement from a
bag that contains 3 purple and 3 orange marbles.

3 1
On the first draw, P(“purple”) =    = .
6 2

The sample space for the second draw depends on the first
draw. Since the first draw was purple, then the probability
2
of the second draw being purple is P(“purple”) = .
5
(If the first draw had been orange, then the probability of
drawing purple the second time would have been different
since there would be a different sample space.)

Note that these events are dependent events. By the General
Multiplication Rule, the probability of drawing two purple marbles is
1 2     1
P(“purple followed by purple”) = ⋅ = .
2 5     5

Extend:
This problem could also be considered by looking at a tree diagram.
Complete the tree diagram below by assigning appropriate probabilities at
each stage and calculating probabilities for all possible outcomes in the
sample space. (Note in this case that all four outcomes are NOT equally
likely.)
2
5   Purple          P(“purple followed by purple”) =
1        Purple
2
3
Orange          P(“purple followed by orange”) =
5

Purple
1
2
Orange

Orange
Section 3.6                                              MATH 2600/Dietrich
Page 5                                          Georgia College & State University

Example:
Four blue socks, four white socks, and four gray socks are mixed in a
drawer. You pull out two socks, one at a time, without looking and without
replacement.

a. Draw a tree diagram along with the possible outcomes and the
probabilities of each branch.

b. What is the probability of getting a pair of socks of the same color?

c.   What is the probability of getting two gray socks?

d. Suppose that, instead of pulling out two socks, you pull out four
socks. What is now the probability of getting two socks of the same
color?
Section 3.6                                                      MATH 2600/Dietrich
Page 6                                                  Georgia College & State University

1. An experiment consists of drawing two cards from a well-shuffled, standard deck
of cards.
a. If the first card is drawn and then replaced before the second card is
drawn, what is the probability of drawing a pair of aces?
b. Repeat the experiment above but do not replace the first card that was
drawn before the second card is drawn. Has the probability of drawing a
pair of aces changed? Explain.

2. An experiment consists of rolling a fair die and tossing a fair coin.
a. Find the probability of getting a 5 on the die and tails on the coin.
b. Find the probability of getting an even number on the die and heads on
the coin.
c. Find the probability of getting a 2 or 3 on the die and heads on the coin.

3. The table shows the Scrabble tiles available at the start of a game. There are
100 tiles: 42 vowels, 56 consonants, and 2 blanks. To begin play, each player
draws a tile. The player with the tile closest to the beginning of the alphabet goes
first. A blank tile beats any letter.
a. If you draw first, what is the probability that
you will select an A?

b. If you draw first and do not replace the tile,
what is the probability that you will select an E
and your opponent will select an I?

c. If you draw first and do not replace the tile,
what is the probability that you will select an E
and your opponent will win the first turn?

4. A box contains the 11 letters shown. Note that some of the letters are repeated.
Suppose that 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?

PROBABILITY
Section 3.6                                                    MATH 2600/Dietrich
Page 7                                                Georgia College & State University

5. On a quiz show, a contestant stands at the entrance to a maze that opens into
two rooms, labeled A and B as shown in the figure. The master of ceremonies’
assistant is to place a new car in one room and a donkey in the other. The
contestant must walk through the maze into one of the rooms and will win
whatever is in that room. If the contestant makes each decision in the maze at
random, in which room should the assistant place the car to give the contestant
the best chance to win?

6. Consider the three boxes shown below. 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?

AAB                 AB                    ABBB

7. There are two identical bottles. One bottle contains 2 green balls and 1 red ball.
The other contains 2 red balls. A bottle is selected at random and a single ball is
drawn. What is the probability that the ball is red?
Section 3.6                                                        MATH 2600/Dietrich
Page 8                                                    Georgia College & State University

Supplemental Topic: Computing Odds

People talk about the odds in favor of and the odds against a particular event’s
happening. When the odds in favor of the president’s being reelected are 4 to 1, this
refers to how likely the president is to win the election relative to how likely the president
is to lose. The probability of the president’s winning is 4 times the probability of losing.
If W represents the event the president wins the election and L represents the event the
P(W ) 4
president loses, then P (W ) = 4 P( L) or as a proportion, we have          = , or 4:1.
P ( L) 1

Note that W and L are complements of each other, so P ( L) = P(W c ) = 1 − P(W ) . Hence,
P(W )   4
the proportion above may also be written as          = , or 4:1.
1 − P(W ) 1

The odds against the president’s winning are how likely the president is to lose relative
to how likely the president is to win. Using the preceding information, we have
P ( L) 1
= , or 1:4.
P(W ) 4

P( L)   1
Because W and L are complements, we also may write                   = , or 1:4.
1 − P( L) 4

Formally, odds are defined as follows:

The odds in favor of an event A is the ratio of the likelihood it will happen to the
c
likelihood that it will not happen, that is P ( A) : P ( A ) .

The odds against an event A is the ratio of the probability that it will not occur to the
probability that it will occur, that is P ( Ac ) : P ( A) .

NOTE: When odds are calculated for equally likely outcomes, we have the following:
P( A) n( A) n( A c ) n( A)
Odds in favor of an event A:          =      ÷       =
P ( A c ) n( S ) n( S )   n( A c )
n( A c )
Odds against an event A:
n( A)

For example, when you roll a fair die, the number of favorable ways of rolling a 4 in one
throw of a die is 1, and the number of unfavorable ways is 5. Thus the odds in favor of
rolling a 4 are 1 to 5 or 1:5. The odds against rolling a 4 are 5:1.
Section 3.6                                                MATH 2600/Dietrich
Page 9                                            Georgia College & State University

Activity: Complete the following challenge exercise from lesson 10-7 of Holt
Mathematics Course 3.
Section 3.6                                                      MATH 2600/Dietrich
Page 10                                                 Georgia College & State University

Problems:
1. If a fair six-sided die is tossed, what are the odds in favor of the following events?
a. Getting a 3
b. Getting a number greater than 0
c. Getting a prime
d. Getting a number greater than 6

2. On an American roulette wheel, half of the slots numbered 1
through 36 are red and half are black. Two slots, numbered 0 and
00 are green. What are the odds against a red slot’s coming up on
any spin of the wheel?

3. Find the probability of an event E given that the odds against E are 9:2.

4. On a tote board at a racetrack, the odds for Royal Star are
listed as 26:1. Tote boards list the odds that the horse will
lose the race. If this is the case, what is the probability of
Royal Star’s winning the race?

5. If the probability of an event is a/b, what are
a. the odds in favor of the event?
b. the odds against it?

6. Suppose you are in two contests that are independent of each other. You are
given the odds of winning one at 1:4 and the odds of winning the other at 3:20.
How would you find the odds of winning both?

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