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Probability of Independent and
Dependent Events
CCM2 Unit 6: Probability
Independent and Dependent Events
• Independent Events: when one event has no
affect on the probability of the other event
occurring.
• Dependent Events: if the outcome or
probability of the first event affects the
outcome of the second.
Independent Events
Suppose a die is rolled and then a coin is tossed.
• Explain why these events are independent.
– They are independent because the outcome of
rolling a die does not affect the outcome of
tossing a coin, and vice versa.
• Fill in the table to represent the sample space
– how many total outcomes are there?
Roll 1
Head
Tail
Roll 2
Roll 3
Roll 4
Roll 5
Roll 6
Roll 1
Roll 2
Roll 3
Roll 4
Roll 5
Roll 6
Head
1,H
2,H
3,H
4,H
5,H
6,H
Tail
1,T
2,T
3,T
4,T
5,T
6,T
• How can you find the total outcomes of the
sample space WITHOUT writing it out?
Remember the Fundamental Counting Principle.
– Still stumped? Here is the process:
– How many outcomes are there for rolling the die?
• 6 outcomes
– How many outcomes are there for tossing the coin?
• 2 outcomes
– How many outcomes are there in the sample space of
rolling the die and tossing the coin?
6 • 2 = 12 outcomes
Roll 1
Roll 2
Roll 3
Roll 4
Roll 5
Roll 6
Head
1,H
2,H
3,H
4,H
5,H
6,H
Tail
1,T
2,T
3,T
4,T
5,T
6,T
Use the table to find the following probabilities:
1. P(rolling a 3)
2/12 = 1/6
2. P(Tails)
6/12 = ½
3. P(rolling a 3 AND
getting tails)
1/12
What do you notice about #1 and #2 to
get #3?
4. P(rolling an even)
6/12 = ½
5. P(heads)
6/12 = ½
6. P(rolling an even AND
getting heads)
3/12 or 1/4
Do you see the pattern again with #4
and #5 to get #6?
Multiplication Rule of Probability for
Independent Events
• The probability of two independent events
occurring can be found by the following
formula:
P(A and B) = P(A) x P(B)
• Let’s go back to the table
• How can we find the probabilities for this
table?
– Just the way you found the total number of
possibilities – by multiplying!
Head
P(H) =
1/2
Tail
P(T) =
1/2
Roll 1
Roll 2
Roll 3
P(1) = 1/6 P(2) = 1/6 P(3) = 1/6
Roll 4
Roll 5
P(4) = 1/6 P(5)= 1/6
Roll 6
P(6)= 1/6
1,H
2,H
3,H
4,H
5,H
6,H
P(1 and H) =
P(2 and H) =
P(3 and H) =
P(4 and H) =
P(5 and H) =
P(6 and H) =
(1/6)(1/2)
= 1/12
1/12
1/12
1/12
1/12
1/12
1,T
2,T
3,T
4,T
5,T
6,T
P(1 and T) =
P(2 and T) =
P(3 and T) =
P(4 and T) =
P(5 and T) =
P(6 and T) =
1/12
1/12
1/12
1/12
1/12
1/12
Example 1
At City High School, 30% of students have parttime jobs and 25% of students are on the honor
roll. What is the probability that a student
chosen at random has a part-time job and is on
the honor roll? Write your answer in context.
P(PT job and honor roll) = P(PT job) x P(honor
roll) = .30 x .25 = .075
There is a 7.5% probability that a student chosen
at random will have a part-time job and be on
the honor roll.
Dependent Events
• Remember, we said earlier that
• Dependent Events: two events are dependent if the
outcome or probability of the first event affects the
outcome or probability of the second.
• Let’s look at some scenarios and determine
whether the events are independent or
dependent.
Example 2: Determine whether the
events are independent or dependent:
a.) Selecting a marble from a container and
selecting a jack from a deck of cards.
•
Independent
b.) Rolling a number less than 4 on a die and rolling
a number that is even on a second die.
•
Independent
c.) Choosing a jack from a deck of cards and
choosing another jack, without replacement.
•
Dependent
d.) Winning a hockey game and scoring a goal.
•
Dependent
Multiplication Rule of Probability for
Dependent Events
• The probability of two dependent events
occurring can be found by the following
formula:
P(A and B) = P(A) x P(B following A)
Example 3: Let’s look at a different of Independent vs. Dependent Example
INDEPENDENT
A box contains 4 red marbles and 6 purple marbles. You are going
to choose 3 marbles with replacement .
a.) Draw a tree diagram and write the probabilities on each
Purple
branch.
Purple
Purple
Red
Red
Purple
Red
Red
Purple
Red
Purple
Red
Purple
Red
b.) What is the probability of drawing 2 purple marbles and 1 red
marble in succession (aka in order)?
P(1st draw purple) • P(2nd draw purple) • P(3rd draw red)
6/10
•
6/10
•
4/10
= 18/125 or .144 or 14.4%
DEPENDENT
A box contains 4 red marbles and 6 purple marbles. You are going
to choose 3 marbles without replacement .
a.) Draw a tree diagram and write the probabilities on each
Purple
branch.
Purple
Purple
Red
Red
Purple
Red
Red
Purple
Red
Purple
Red
Purple
Red
b.) What is the probability of drawing 2 purple marbles and 1 red
marble in succession (aka in order)?
P(1st draw purple) • P(2nd draw purple) • P(3rd draw red)
6/10
•
5/9
•
4/8
= 1/6 or .167 or 16.7%
In the above example, what is the probability of
first drawing all 4 red marbles in succession
and then drawing all 6 purple marbles in
succession without replacement?
• P(4 red then 6 purple) =
(4/10)(3/9)(2/8)(1/7)(6/6)(5/5)(4/4)(3/3)(2/2)(1/1) =
1/210 or .0048
• The probability of drawing 4 red then 6 purple without
replacement is 0.48%
• Explain why the last 6 probabilities above were all
equivalent to 1.
• This is because there were only purple marbles left, so
the probability for drawing a purple marble was 1.
Example 4: How does permutations and combinations
work into this type of probability???
a.) There are 32 students in the classroom and the teacher
needs to be make a seating chart. What is the probability
that Deshawn will be in the first seat, Sophie in the second
seat, and Delaney in the third seat?
P(Deshawn 1st) • P(Sophie 2nd) • P(Delaney 3rd )
1/32
• 1/31
•
1/30
= 3.36 x 10-5 = .0000336 = .00336%
You get
the same
answer!
OR Since the order matters, we call this a permutation
choosing 3 out of 32: 32P3 and we’re looking for a specific
one order, so:
1_
= 1
= .00336%
32•31•30
32P3
Example 4: How does permutations and combinations work into
this type of probability???
b.) There are 32 students in the classroom and the teacher can
only choose 2 students to go to the media center. What is the
probability that Arturo and Jon will be chosen?
P(Arturo or Jon) • P(the other one)
2/32
• 1/31
= .002 = .20%
You get
the same
answer!
OR Since the order does NOT matter, we call this a combination
choosing 2 out of 32: 32C2 and we’re looking for a specific one
group of people, so:
1_
= 1
= .20%
32 •31
32C2
2•1