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Date: Click here to enter text. Section: 7.3 Dependent and

Date: Click here to enter text.
Section: 7.3 Dependent and Independent Events
Objective(s): Determine whether events are dependent or independent
: Find probabilities of dependent and independent events
Vocabulary
Independent events
Events are independent events if the occurrence of one event
does not affect the probability of the other.
Dependent events
Events are dependent events if the occurrence of one event
affects the probability of the other.
Conditional probability
The probability of event B, given that event A has occurred.
Independent events
If a coin is tossed twice, its landing heads up on the first toss and
landing heads up on the second toss are independent events. The
outcome of one toss does not affect the probability of heads on
the other toss. To find the probability of tossing heads twice,
1
1
1
multiply the individual probabilities, 2 βˆ— 2 = 4 .
A six-sided cube is labeled with the
numbers 1, 2, 2, 3, 3, and 3. Four
sides are colored red, one side is
white, and one side is yellow. Find the
probability of tossing a 2 and then
another 2.
Tossing a 2 once does not affect the probability of tossing a 2
again, so the events are independent.
𝑃(2 π‘Žπ‘›π‘‘ 2) = 𝑃(2) βˆ— 𝑃(2) =
1 1 1
βˆ— =
3 3 9
A six-sided cube is labeled with the
numbers 1, 2, 2, 3, 3, and 3. Four
sides are colored red, one side is
white, and one side is yellow. Find the
probability of tossing red, then white,
then yellow.
Find the probability of rolling a 6 on
one number cube and then a 6 again
on another number cube.
Find the probability of tossing heads,
then heads, then tails when tossing a
coin three times.
Dependent events
To find the probability of dependent
events, you can use conditional
probability P(B|A), the probability of
event B, given that event A has
occurred.
The result of any toss does not affect the probability of any other
outcome so the events are independent.
𝑃(π‘Ÿπ‘’π‘‘ π‘Žπ‘›π‘‘ π‘€β„Žπ‘–π‘‘π‘’ π‘Žπ‘›π‘‘ π‘¦π‘’π‘™π‘™π‘œπ‘€)
= 𝑃(π‘Ÿπ‘’π‘‘) βˆ— 𝑃(π‘€β„Žπ‘–π‘‘π‘’) βˆ— 𝑃(π‘¦π‘’π‘™π‘™π‘œπ‘€) =
𝑃(6) βˆ— 𝑃(6) =
4 1 1
4
1
βˆ— βˆ— =
=
6 6 6 216 54
1 1
1
βˆ— =
6 6 36
𝑃(β„Žπ‘’π‘Žπ‘‘π‘ ) βˆ— 𝑃(β„Žπ‘’π‘Žπ‘‘π‘ ) βˆ— 𝑃(π‘‘π‘Žπ‘–π‘™π‘ ) =
1 1 1 1
βˆ— βˆ— =
2 2 2 8
Events are dependent events if the occurrence of one event
affects the probability of the other. For example, suppose that
there are 2 lemons and 1 lime in a bag. If you pull out two pieces
of fruit, the probabilities change depending on the outcome of the
first.
The tree diagram shows the probabilities for choosing two pieces
of fruit from a bag containing 2 lemons and 1 lime.
The probability of a specific event can be found by multiplying
the probabilities on the branches that make up the event. For
2
1
2
1
example, the probability of drawing two lemons is 3 βˆ— 2 = 6 = 3 .
Two number cubes are rolled–one
white and one yellow. Find the
probability that the white cube shows
a 6 and the sum is greater than 9 .
The events are dependent because if the white cube does NOT
show six, then the yellow cube doesn’t matter at all. The yellow
cube need only be considered IF the white cube shows 6 (or viceversa). Reference the 36 outcomes chart for rolling 2 number
cubes.
𝑃(π‘€β„Žπ‘–π‘‘π‘’ 6) =
6
1
=
36 6
𝑃(π‘ π‘’π‘š > 9|π‘€β„Žπ‘–π‘‘π‘’ 6) =
3 1
=
6 2
𝑃(π‘€β„Žπ‘–π‘‘π‘’ 6 π‘Žπ‘›π‘‘ π‘ π‘’π‘š > 9)
= 𝑃(π‘€β„Žπ‘–π‘‘π‘’ 6) βˆ— 𝑃(π‘ π‘’π‘š > 9|π‘€β„Žπ‘–π‘‘π‘’ 6) =
Two number cubes are rolled–one
white and one yellow. Find the
probability that the yellow cube shows
an even number and the sum is 5 .
1 1
1
βˆ— =
6 2 12
Again, the events are dependent because if the yellow shows an
even number then that changes the probability of the sum being 5.
𝑃(π‘¦π‘’π‘™π‘™π‘œπ‘€ 𝑒𝑣𝑒𝑛) =
18 1
=
36 2
𝑃(π‘ π‘’π‘š = 5|π‘¦π‘’π‘™π‘™π‘œπ‘€ 𝑒𝑣𝑒𝑛) =
2
1
=
18 9
𝑃( π‘¦π‘’π‘™π‘™π‘œπ‘€ 𝑒𝑣𝑒𝑛 π‘Žπ‘›π‘‘ π‘ π‘’π‘š = 5)
= 𝑃(π‘¦π‘’π‘™π‘™π‘œπ‘€ 𝑒𝑣𝑒𝑛) βˆ— 𝑃(π‘ π‘’π‘š = 5|π‘¦π‘’π‘™π‘™π‘œπ‘€ 𝑒𝑣𝑒𝑛) =
Two number cubes are rolled–one red
and one black. Find the probability that
the red cube shows a number greater
than 4 and the sum is greater than 9 .
1 1
1
βˆ— =
2 9 18
The events are dependent.
𝑃(π‘Ÿπ‘’π‘‘ > 4) =
12 1
=
36 3
𝑃(π‘ π‘’π‘š > 9|π‘Ÿπ‘’π‘‘ > 4) =
5
12
𝑃( π‘Ÿπ‘’π‘‘ > 4 π‘Žπ‘›π‘‘ π‘ π‘’π‘š > 9)
= 𝑃(π‘Ÿπ‘’π‘‘ > 4) βˆ— 𝑃(π‘Ÿπ‘’π‘‘ > 4|π‘ π‘’π‘š > 9) =
1 5
5
βˆ—
=
3 12 36
Conditional probability
The table shows domestic migration
from 1995 to 2000. A person is
randomly selected. Find the
probability that an emigrant is from
the West.
𝑃(π‘Šπ‘’π‘ π‘‘|π‘’π‘šπ‘–π‘”π‘Ÿπ‘Žπ‘›π‘‘) =
2654
β‰ˆ 0.2277 = 22.77%
11656
Domestic Migration by Region
(thousands)
Region
Immigrants
Emigrants
Northeast
1537
2808
Midwest
2410
2951
South
5042
3243
West
2666
2654
Using the same table above, again a
person is randomly selected. Find the
probability that someone selected
from the South region is an
immigrant.
Using the same table above, again a
person is randomly selected. Find the
probability that someone selected is an
emigrant and from the Midwest
𝑃(π‘–π‘šπ‘šπ‘–π‘”π‘Ÿπ‘Žπ‘›π‘‘|π‘†π‘œπ‘’π‘‘β„Ž) =
5042
β‰ˆ 0.6086 = 60.86%
8285
𝑃(𝑀𝑖𝑑𝑀𝑒𝑠𝑑|π‘’π‘šπ‘–π‘”π‘Ÿπ‘Žπ‘›π‘‘) =
2951
11656
𝑃(π‘’π‘šπ‘–π‘”π‘Ÿπ‘Žπ‘›π‘‘ π‘Žπ‘›π‘‘ 𝑀𝑖𝑑𝑀𝑒𝑠𝑑|π‘’π‘šπ‘–π‘”π‘Ÿπ‘Žπ‘›π‘‘) =
=
2951
β‰ˆ 0.127 = 12.7%
23311
11656 2951
βˆ—
23311 11656
Below is a standard deck of 52 cards:
What is the probability of selecting two
hearts when the first card is replaced after
the first selection?
With replacement usually indicated independent events, and such
is the case here.
𝑃(β„Žπ‘’π‘Žπ‘Ÿπ‘‘) βˆ— 𝑃(β„Žπ‘’π‘Žπ‘Ÿπ‘‘) =
1 1
1
βˆ— =
4 4 16
What is the probability of selecting two
hearts when the first card is NOT replaced
after the first selection?
Without replacement usually indicated dependent events. After
the first selection the size of the deck is one card less than it was
before that selection and the number of hearts in the deck has
also changed, being reduced by one.
1 12
1
𝑃(β„Žπ‘’π‘Žπ‘Ÿπ‘‘) βˆ— 𝑃(β„Žπ‘’π‘Žπ‘Ÿπ‘‘|π‘“π‘–π‘Ÿπ‘ π‘‘ π‘π‘Žπ‘Ÿπ‘‘ π‘€π‘Žπ‘  π‘Ž β„Žπ‘’π‘Žπ‘Ÿπ‘‘) = βˆ—
=
4 51 17
What is the probability that a queen is
drawn is not replaced, and then a king is
drawn?
These are dependent events.
𝑃(π‘žπ‘’π‘’π‘’π‘›) βˆ— 𝑃(π‘˜π‘–π‘›π‘”|π‘“π‘–π‘Ÿπ‘ π‘‘ π‘π‘Žπ‘Ÿπ‘‘ π‘€π‘Žπ‘  π‘Ž π‘žπ‘’π‘’π‘’π‘›)
4 4
4
=
βˆ—
=
52 51 663
A bag contains 10 beads: 2 black, 3 white,
and 5 red. What is the probability of
selecting a white bead, replacing it, and
then selecting a red bead?
These events are independent.
A bag contains 10 beads: 2 black, 3 white,
and 5 red. What is the probability of
selecting a white bead, NOT replacing it,
and then selecting a red bead?
These events are dependent.
A bag contains 10 beads: 2 black, 3 white,
and 5 red. What is the probability of
selecting three red beads in a row?
𝑃(π‘€β„Žπ‘–π‘‘π‘’) βˆ— 𝑃(π‘Ÿπ‘’π‘‘|π‘“π‘–π‘Ÿπ‘ π‘‘ π‘π‘’π‘Žπ‘‘ π‘€π‘Žπ‘  π‘€β„Žπ‘–π‘‘π‘’) =
𝑃(π‘€β„Žπ‘–π‘‘π‘’) βˆ— 𝑃(π‘Ÿπ‘’π‘‘|π‘“π‘–π‘Ÿπ‘ π‘‘ π‘π‘’π‘Žπ‘‘ π‘€π‘Žπ‘  π‘€β„Žπ‘–π‘‘π‘’) =
3 5
3
βˆ—
=
10 10 20
3 5 1
βˆ— =
10 9 6
These events are dependent
𝑃(π‘Ÿπ‘’π‘‘) βˆ— 𝑃(π‘Ÿπ‘’π‘‘|π‘“π‘–π‘Ÿπ‘ π‘‘ π‘π‘’π‘Žπ‘‘ π‘€π‘Žπ‘  π‘Ÿπ‘’π‘‘) βˆ— 𝑃(π‘Ÿπ‘’π‘‘|π‘“π‘–π‘Ÿπ‘ π‘‘ π‘‘π‘€π‘œ π‘π‘’π‘Žπ‘‘π‘  π‘€π‘’π‘Ÿπ‘’ π‘Ÿπ‘’π‘‘)
=
5 4 3
60
1
βˆ— βˆ— =
=
10 9 8 720 12

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