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Geometry – Probability Unit
Common Core State Standard
S.CP.1 – Describe events as subsets of sample space using characteristics of the outcome, or as unions, intersections, or
complements of other events (“or”, “and”, “not”)
S.CP.2 – Understand that two events A and B are independent if the probability of A and B occurring together is the
product of their probabilities, and use this characterization to determine if they are independent.
S.CP.3 – Understand the conditional probability of A given B as P(A and B)/P(B), and interpret independence of A and B
as saying that the conditional probability of A given B is the same as the probability of A, and the conditional probability
of B given A is the same as the probability of B.
S.CP.4 – Construct and interpret two-way frequency tables of data when two categories are associated with each object
being classified. Use the two-way table as a sample space to decide if events are independent and to approximate
conditional probabilities.
S.CP.5 – Recognize and explain the concepts of conditional probability and independence in everyday language and
everyday situations.
S.CP.6 – Find the conditional probability of A given B as the fraction of B’s outcomes that also belong to A, and interpret
the answer in terms of the model.
S.CP.7 – Apply the Addition Rule, P(A or B) = P(A) +P(B) – P(A and B), and interpret the answer in terms of the model.
Supplies Needed: Enough Coins (pennies) for every student in the class on Day 1. For Days 1 & 2, a deck of cards
would be useful for demonstration purposes.
Day(s)
Content
1 & 2 Define sample space. Define event as a subset of the sample
space. Use dice and cards to illustrate product rule and
independence. (S.CP.1, S.CP.2)
3 & 4 Use tables to illustrate conditional probabilities,
P  A | B
5
6
7
P  A  B
. Use two-way tables with P(A|B) =
P  B
P(A) to illustrate independence. (S.CP.3, S.CP.4, S.CP.5,
S.CP.6)
Use tables to illustrate general addition rule. (S.CP.7)
Review
Test
1
Activities
Notes + Worksheet (Summary) Note that the
Summary Questions for all days could be used
for homework, classwork, group work, etc.
Notes + Worksheet (Summary)
Notes + Worksheet (Summary)
Review Sheet – Pending
Test – Pending, but, note that it will be based
directly off the notes/activities. Students should
not use notes on this test.
Probability Days 1 & 2 Introduction to Probability & Multiplication Rule (S.CP.1, S.CP.2)
What is the probability of an event?
Probability of an event is its long-run frequency.
What happens in the “long-run”?
Complete coin tossing experiment… have each student toss a coin 10 times each. Then, create a frequency histogram (as
seen below) that displays the class results. Discuss with students why it didn’t come out “50-50”… because the long run is
forever… it’s what will happen in an infinite number of tosses of the coin.
60
Frequency
50
40
30
20
10
0
Heads
Tails
Define the sample space for the outcome of the flip of a coin.
{H, T}
What is a sample space?
A sample space contains all possible outcomes for an event. In the case of the flipping of a coin, the possible outcomes
are heads or tails. So, the sample space is {H, T}.
What does it mean for two events to be independent of each other?
For two events to be independent of one another, the probability of one event does not affect another.
Is the flipping of a coin independent of any other flip of the coin?
Yes… the probability of the outcome being, say, heads is ½. This does not affect the next flip of the coin. Hence, the
flipping of a coin is an independent event!
Define the sample space for the outcome of the roll of a die.
{1, 2, 3, 4, 5, 6}
What are the outcomes that make up the sample space?
The possible outcomes of the roll of the die… {1, 2, 3, 4, 5, 6}
2
What is the probability of rolling a 6 on a single die? Rolling a 4? Rolling a 3? Rolling a 7?
1/6, 1/6, 1/6, 0 Follow-up question… Why is the probability of rolling a seven zero? Answer: There is no seven in the
sample space.
Is rolling a die an independent event? Explain.
Yes… the probability of any particular side coming face up is 1/6. This is true for any roll of the die as one roll does not
influence the next. Therefore, the rolling of a die is independent.
Define the sample space for the sum of two dice.
+
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4 5 6
5 6 7
6 7 8
7 8 9
8 9 10
9 10 11
10 11 12
Don’t give the kids the
table… let them
struggle and try to
figure it out on their
own… when they can’t
figure it out, give them
the table.
{2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7,8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 11, 11, 12}
How many outcomes make up this space? 36
What is the probability of rolling a sum of 9? A sum of 4? A sum of 11? A sum of 13?
4/36, 5/36, 2/36, 0
At this point, students should have a basic understanding of probabilities of equally likely events. Now, to move on to
multiplying probabilities.
How many outcomes are there if you were to flip three coins?
Have students list out all the possible ways… there are eight ways… HHH, HHT, THH, HTH, HTT, THT, TTH, TTT.
Now, put three blanks on the board like this…
there are for each coin
and ask them how many possible outcomes
Coin 1 Coin 2 Coin 3
The finished blanks will look like this… 2 2 2 = 8
Just ask the students what operation needs to go between the
2s… multiplication… and then students have the multiplication rule down!
Use this idea to answer the question below.
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What is the probability of flipping a coin three times in a row and getting a head each time?
Relate the ideas of the above question… the students wrote out the eight ways this could occur… there is only one way
for there to be three heads in a row, so, the answer is 1/8. You must relate this to the multiplication principle as well…
students need to see that they can multiply probabilities together when the events are independent… we’ll get to nonindependent events later. Students need to see that the probability of flipping a coin and getting a head each time (three
heads) is 1/ 2 1/ 2 1/ 2  1/ 8 .
What is the probability of flipping a coin 5 times and getting a head one each flip?
1/ 2 
5
 1/ 32
Make sure students write out ½ five times to see the multiplication rule working.
What is the probability that the first head observed is on the 4th flip of the coin?
Here, it’s going to be (not a head)(not a head)(not a head)(head) = (1/2)(1/2)(1/2)(1/2) = 1/16… note that the
probabilities don’t change here because getting a heads or tails are equally likely… moving on to something more
interesting next…
How many cards are in a standard deck of cards? How many suits? How many face cards? How many colors?
52 cards. 4 suits – clubs, spades, hearts, diamonds. 12 face cards – for each suit, Queen, Jack, King. 2 Colors – red and
black.
What does it mean to sample with and without replacement?
Sampling with replacement means to pull a card from the deck and then return it. Sampling without replacement means
to pull a card from the deck and not return it.
What is the probability of pulling a face card from the deck?
12/52
What is the probability of pulling a Jack from the deck?
4/52
What is the probability of pulling a red card from the deck?
26/52
What is the probability of pulling a Queen from the deck and then a Seven from the deck with replacement? Are these
independent events?
Ok… here’s where it might hit the fan with the kids… but this is where we’ve been leading up to… because we’re dealing
with replacement, no worries.  P(Queen and then Seven) = (4/52)(4/52) = 1/169. Yes, these are independent events…
because the card was replaced after it was drawn, the probability of pulling a Seven from the deck didn’t change…
therefore, these are independent events!
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What is the probability of pulling a Queen from the deck and then a Seven from the deck without replacement? Are
these independent events?
P(Queen and then Seven) = (4/52)(4/51) = 4/663… these are not independent because the probability of pulling a Seven
changed because we didn’t replace the card we first drew!
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Summary Questions for Probability Days 1 & 2
1. What is a sample space? The sample space contains all the possible outcomes that could take place.
2. What is probability? Probability is the long-run frequency of an event… what will happen in the long-run over an
infinite number of trials.
3. How do you determine the probability of an event? Probability is determined by calculating
# successes
possible outcomes
4. An icosahedron is a polyhedron with 20 triangular faces. If the sides are numbered 1 through 20, what is the
probability that the sum of two icosahedral dice is 19?
Students will need to make a table with 20 sides each and use it to answer the question. 18/400.
5. What does it mean for two events to be independent?
Two events are independent if the outcome of one event does not affect the chances/probability of another event.
For problems six through fourteen, assume a standard 52-card deck is used.
6. What is the probability of drawing a red card? 26/52
7. What is the probability of drawing a purple card? 0/52
8. What is the probability of drawing a queen and then drawing a king without replacement? Are these independent?
(4/52)(4/51) = 4/663
Not independent.
9. What is the probability of drawing a queen and then drawing a king with replacement? Are these independent?
(4/52)(4/52) = 1/169
Independent.
10. What is the probability of drawing an ace and then another ace with replacement? Are these independent?
(4/52)(4/52) = 1/169
Independent.
11. What is the probability of drawing an ace and then another ace without replacement? Are these independent?
(4/52)(3/51) = 1/221
Not independent.
12. What is the probability of not drawing a face card? 40/52
13. What is the probability of not drawing a face card and then drawing a seven without replacement? Are these
independent? (40/52)(4/51) = 40/663
Not independent.
14. What is the probability of drawing a non-face card and then flipping a coin and it landing heads? Are these
independent? (40/52)(1/2) = 5/13 Independent.
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15. Are the following events independent?
a. In a jar of marbles, there are 10 marbles. 4 are blue, 3 are orange, and the rest are purple. You draw a
marble from the jar, replace it, and then draw another marble. Yes.
b. A card is drawn from a deck of cards. It is not replaced. Another card is drawn. No.
c. A die is rolled. The number that is face up is recorded. The die is rolled again and the number that is face up
is recorded. Yes.
d. While driving, you come to an intersection where you must turn right or left. You turn left. No. When you
turn left, this affects your chances of turning right… which are then zero… so when one action influences
another, the events are not independent.
e. It is snowing in Denver, Colorado. It is snowing in Dubuque. Yes. The weather in Colorado is probably not
affecting the weather here in Dubuque.
f.
It is snowing in Dyersville. It is snowing in Dubuque. No. Because Dyersville is west of Dubuque, it is more
than likely that since weather tends to travel in an easterly direction, the weather in Dyersville will affect the
weather here in Dubuque.
g. A roulette wheel is spun. 19 is the winning number. The wheel is spun again. 33 is the winning number this
time. Yes. Because each spin of the roulette wheel is independent of the next, these are independent. You
may need to briefly discuss a roulette wheel with the students.
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Days 3 & 4 Two-Way Tables, Conditional Probabilities, and Independence (S.CP.3, S.CP.4, S.CP.5, S.CP.6)
Use the table below about athletes in Dubuque to answer the questions that follow. The goal here is to get students to
be able to read and interpret two-way tables.
Hempstead
Senior
Total
Golf
13
11
24
Basketball
25
22
47
Soccer
18
22
40
Total
56
55
111
a. How many total athletes are there? 111
b. How many athletes attend Hempstead? 56
c. How many athletes play basketball? 47
d. What is the probability that an athlete plays basketball and attends Hempstead? 25/111
e. What is the probability that an athlete is a golfer and attends Senior? 11/111
f.
What is the probability that an athlete is a golfer? 24/111
g. What is the probability that a student attends Hempstead? 56/111
h. Given that an athlete is a golfer, what is the probability that the athlete attends Hempstead? Ok… here is where
we get into conditional probabilities… not too bad… but it can be tricky if the kids don’t read carefully. Here, the
only athletes we’re interested in are the golfers… so, the denominator will be all the golfers… OF THE GOLFERS,
we want to know how many attend Hempstead… so, we want the numerator to be the golfers who attend
Hempstead… resulting in 13/24.
i.
Given that an athlete attends Senior, what is the probability that the athlete plays soccer? Similar to part H
above… we’re interested in only athletes who attend Senior and of those, we want to know how many play
soccer… resulting in 22/55.
j.
Are golf and the high school an athlete attends independent of each other? What does this mean? Ok… this
where students need to think… remember, independence means that the probability of one event does not
impact the probability of another event. Students will need to be introduced to some formal probability notation
and language here. In short, students need to be shown that for events A and B, P(B|A) = P(B). What does this
mean? Ok… first of all, P(B|A) reads “the probability of B given A”. This is a conditional probability… students
have already done this… it is just like parts H and I above. Using part H as an example… the technical way to
write out part H is P(Hempstead|Golfer). Now, if we can show that P(Hempstead|Golfer) = P(Hempstead), then
we will have proven that being a golfer had no impact/influence on the student attending Hempstead. If this is
true, then choosing golf and high school choice are independent of one another. P(Hempstead|Golf) = 13/24 =
0.54. P(Hempstead) = 56/111 = 0.50. Since these are not equal, we may state that golfing and high school choice
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are not independent of each other. What does this mean? This means that there is some sort of relationship
between golfing and high school choice. We don’t care what that relationship is… we just know that it exists.
Examples
1. Of voters in a recent election, 57% were male, 64% were Democrat, and 35% were both male and Democrat. Fill
in the table and answer the questions below:
Democrat
Not
Democrat
Total
Male
Not Male
Total
0.35
0.29
0.64
0.22
0.14
0.36
0.57
0.43
1
a) What is the probability that a voter chosen at random is a male if they are a Democrat? 0.35/0.64 = 0.547
b) What is the probability that a voter chosen at random is a Democrat if they are not male? 0.29/0.43 = 0.674
c) Are the two events, being male and being a Democrat independent? If these two events were independent,
then P(Male|Democrat) = P(Male). From part B above, P(Male|Democrat) = 0.547 and P(Male) = 0.57.
Because these are not equal, we cannot say that being male and being a Democrat are independent events…
that is, there is some relationship between gender and political preference.
2. The American Red Cross says that about 40% of people have blood Type A and 11% have Type B. Fill in the table
and answer the questions below:
B
Not B
Total
A
0
0.4
0.4
Not A
0.11
0.49
0.6
Total
0.11
0.89
1
a) What is the probability that a randomly selected blood donor has neither Type A nor Type B? 0.49
b) What is the probability that a randomly selected blood donor has Type A if they have Type B? 0/0.11 = 0
c) Are having Type A blood and Type B blood disjoint? Yes. Disjoint simply means that neither can occur at the
same time. If you have Type A, you don’t have Type B… you can’t have both types. So, they are disjoint.
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3. At a particular toy store, 15% of the toys are hand-made. Of the hand-made toys, 10% of them will break within
the first 6 months of use. Of the machine-made toys, only 2% of them will break within the first 6 months of
use. Fill in the table and answer the questions below:
Break
Doesn’t
Break
Total
Handmade
0.10
Not Handmade
0.02
0.05
0.83
0.88
0.15
0.85
1
Total
0.12
a) What is the probability that a toy will break within the first 6 months of use? 0.12
b) What is the probability that Kayla’s toy (a hand-made toy) will not break? This is asking for the probability
that the toy breaks given that it is hand-made. 0.05/0.15 = 0.333.
c) What is the probability that Nick’s broken toy is a machine-made toy? This is asking for the probability that
the toy is machine-made given that it is broken. 0.02/0.12 = 0.167.
d) What is the probability that Allison’s two hand-made toys will not break? (.05/.15)(.05/.15) = 1/9
Note that the 0.6 comes from P(B) = 0.6.
4. P(A) = 0.5, P(B) = 0.6, 𝑃(𝐴 ∪ 𝐵) = 0.85. Fill in the table and answer the questions below:
A
Not A
Total
B
0.25
0.35
0.6
Not B
0.25
0.15
0.4
Total
0.5
0.5
1
Note that the 0.5 comes from P(A) = 0.5.
The tricky part comes when you’re trying to
determine P  A  B  . We know that
P  A  B   0.85 … usually, this is P(A) + P(B),
but in this case P(A)+P(B)=0.5+0.6=1.1. This can’t
be. So, we need to subtract the intersection so
that P(A)+P(B)=0.85. The difference is 1.10.85=0.25. This is the intersection of A and B.
After this, finishing the table just involves simple
arithmetic.
a) What is 𝑃(𝐴 ∩ 𝐵)? 0.25
b) What is P(A|B)?
Again, a conditional probability…
0.25/0.6 = 0.417.
c) What is P(B|A)?
Another conditional probability…
0.25/0.5 = 0.5.
d) Are A and B independent? Need to show P(A|B) = P(A). P(A) = 0.5. The two are not equal. Not independent.
10
5. Research shows that only 30% of those using a particular home pregnancy kit are actually pregnant. When a
pregnant woman uses this kit, it correctly indicates pregnancy 96% of the time. A woman who is not pregnant
gets a correct negative indication 90% of the time. Fill in the table and answer the questions below:
Positive
Negative
Total
Pregnant
Not
Pregnant
Total
0.288 =
(0.96)(0.3)
0.07
0.358
0.012
0.3
0.63 =
(0.9)(0.7)
0.7
0.642
1
a) What is the probability that a woman will get a positive test result? 0.358
b) What is the probability that a woman who gets a positive test result is actually pregnant? 0.288/0.358 =
0.804
c) What is the probability that a woman who is not pregnant gets a positive test result? 0.07/0.7 = 0.1
11
Summary Questions For Probability Days 3 & 4
1. Two friends, Mimi and Najeeha, have cars in desperate need of repair. On any given day, the probability that
Mimi’s car will break down is 0.5, the probability that Najeeha’s car will break down in 0.6, and the probability
that both their cars will break down is 0.3. Fill in the table and answer the questions below:
Najeeha’s Car Breaks
Down
Najeeha’s Car Doesn’t
Break Down
Total
Mimi’s Car Breaks Down
Mimi’s Car Doesn’t
Break Down
Total
0.3
0.3
0.6
0.2
0.2
0.4
0.5
0.5
1
a) What is the probability that Najeeha’s car will break down if Mimi’s car breaks down? 0.3/0.5 = 0.6
b) Are the two cars breaking down independent? If the cars breaking down are independent, then P(Najeeha’s
Car Breaks Down|Mimi’s Car Breaks Down) = P(Najeeha’s Car Breaks Down). We find 0.3/0.5 = 0.6 = 0.6.
Since the probabilities are the same, the two cars breaking down are independent.
2. The probability that a football player weighs more than 230 pounds is 0.68, that he is at least 75 inches tall 0.55,
and that he weighs more than 230 pounds and is at least 75 inches tall is 0.43. Fill in the table and answer the
questions below:
More than 230 pounds
Less than 230 pounds
Total
At least 75 inches tall
0.43
0.12
0.55
Less than 75 inches tall
0.25
0.20
0.45
Total
0.68
0.32
1
a) What is the probability that a football player is at least 75 inches tall if he weighs more than 230 pounds?
This is a conditional probability: 0.43/0.68 = 0.632
b) What is the probability that a football player is at least 75 inches tall if he does not weigh more than 230
pounds? This is also a conditional probability: 0.12/0.32 = 0.375
c) Are the events weighing more than 230 pounds and being at least 75 inches tall independent? Need to show
P(more than 230 pounds|at least 75 inches tall) = P(more than 230 pounds). We find 0.43/0.55 = 0.782 
0.68. Because these are not equal, we find that the two events are not independent.
12
3. One card is randomly selected from a standard 52-card deck. Lizzie wants to examine the probability of finding
an ace and the probability of finding a black card. Fill in the table and answer the questions below:
Ace
Not Ace
Total
Black Card
2/52
24/52
26/52
Not Black Card
2/52
24/52
26/52
Total
4/52
48/52
52/52 = 1
a) What is the probability that Lizzie will get a black card if she gets an ace? (2/52)/(4/52) = 1/2
b) What is the probability that Lizzie will not get an ace if she gets a black card? (24/52)/(26/52) = 12/13
c) Are getting an ace and getting a black card independent? Need to show P(Ace/Black) = P(Ace).
(2/52)/(26/52) = 1/13 = 4/52 = 1/13. The events are independent.
4. Employment records indicate that the probability a worker is married if he/she is a college graduate is 0.55. The
probability that a randomly selected worker is a college graduate is 0.58. The probability that a randomly
selected worker is not married is 0.35. Fill in the table and answer the questions below:
Married
Not Married
Total
College Graduate
0.319 = (0.55)(0.58)
0.261
0.58
Not College Graduate
0.331
0.089
0.42
Total
0.65
0.35
1
a) What is the probability that a randomly selected worker is not married but is a college graduate? 0.261
b) What is the probability that a randomly selected worker is married and a college graduate? 0.319
c) What is the probability that a randomly selected worker is married if they are not a college graduate?
0.331/0.42 = 0.788
d) What is the probability that a randomly selected worker is a college graduate if they are not married?
0.261/0.35 = 0.746
e) Are being married and being a college graduate independent? Need to show P(Married|College Grad) =
P(Married). So, 0.319/0.58 = 0.55  0.65. Therefore, they are not independent.
13
Day 5 The General Addition Rule (S.CP.1, S.CP.3, S.CP.5)
In a group of 30 students, 19 are taking geometry only, 10 are taking algebra 2 only, and the rest are taking both.
Geometry
Not Geometry
Total
Algebra 2
1/30
10/30
11/30
Not Algebra 2
19/30
0/30
19/30
Total
20/30
10/30
1
a. How many are taking neither? 0
b. What is the probability that a student is taking algebra 2 or geometry? This is the General Addition Rule… it
states P  A  B   P  A  P  B   P  A  B  . This rule allows us to add probabilities when we are asked if
this OR that outcome occurred. We subtract the intersection because it gets counted twice. In this case, we get
P(algebra 2) + P(geometry) – P(algebra 2 and geometry) = 11/30 + 20/30 – 1/30 = 30/30 = 1.
Use the table below about athletes in Dubuque to answer the questions that follow.
Hempstead
Senior
Total
Golf
13
11
24
Basketball
25
22
47
Soccer
18
22
40
Total
56
55
111
a. What is the probability that a student plays golf or attends Senior? 24/111 + 55/111 – 11/111 = 68/111.
b. What is the probability that a student plays basketball or attends Hempstead? 47/111 + 56/111 – 25/111 =
78/111.
c. What is the probability that a student attends Senior or plays soccer? 55/111 + 40/111 – 22/111 = 73/111.
14
Examples
1. Of voters in a recent election, 57% were male, 64% were Democrat, and 35% were both male and Democrat.
Democrat
Not
Democrat
Total
Male
Not Male
Total
0.35
0.29
0.64
0.22
0.14
0.36
0.57
0.43
1
a) What is the probability that a voter chosen at random is either male or Democrat? 0.57 + 0.64 – 0.35 = 0.86
b) What is the probability that a voter chosen at random is not a male or a not a Democrat? 0.43 + 0.36 – 0.14 =
0.65
2. At a particular toy store, 15% of the toys are hand-made. Of the hand-made toys, 10% of them will break within the
first 6 months of use. Of the machine-made toys, only 2% of them will break within the first 6 months of use.
Break
Doesn’t
Break
Total
Handmade
0.10
Not Handmade
0.02
0.05
0.83
0.88
0.15
0.85
1
Total
0.12
a) What is the probability a toy is hand-made or doesn’t break? 0.15 + 0.88 – 0.05 = 0.98
b) What is the probability that a toy is hand-made or breaks? 0.15 + 0.12 – 0.10 = 0.17
c) What is the probability that a toy is not hand-made or breaks? 0.85 + 0.12 – 0.02 = 0.95
d) What is the probability that a toy is not hand-made or doesn’t break? 0.85 + 0.88 – 0.83 = 0.90
15
Summary Questions For Probability Day 5
1. Two friends, Mimi and Najeeha, have cars in desperate need of repair. On any given day, the probability that Mimi’s
car will break down is 0.5, the probability that Najeeha’s car will break down in 0.6, and the probability that both
their cars will break down is 0.3
Najeeha’s Car Breaks
Down
Najeeha’s Car Doesn’t
Break Down
Total
Mimi’s Car Breaks Down
Mimi’s Car Doesn’t
Break Down
Total
0.3
0.3
0.6
0.2
0.2
0.4
0.5
0.5
1
a) What is the probability that Mimi or Najeeha’s car will break down? 0.5 + 0.6 – 0.3 = 0.8
b) What is the probability that Mimi’s car breaks down or Najeeha’s doesn’t break down? 0.5 + 0.4 – 0.2 = 0.5
c) What is the probability that Mimi’s car doesn’t break down or Najeeha’s car doesn’t break down? 0.5 + 0.4 – 0.2
= 0.7
2. The probability that a football player weighs more than 230 pounds is 0.68, that he is at least 75 inches tall 0.55, and
that he weighs more than 230 pounds and is at least 75 inches tall is 0.43.
More than 230 pounds
Less than 230 pounds
Total
At least 75 inches tall
0.43
0.12
0.55
Less than 75 inches tall
0.25
0.20
0.45
Total
0.68
0.32
1
a) What is the probability that a football player weighs more than 230 pounds or is less than 75 inches tall?
0.68 + 0.45 – 0.25 = 0.88
b) What is the probability that a football player weighs less than 230 pounds or is at least 75 inches tall?
0.32 + 0.55 – 0.12 = 0.75
c) What is the probability that a football player weights more than 230 pounds or is more than 75 inches tall?
0.68 + 0.55 – 0.43 = 0.80
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3. Employment records indicate that the probability a worker is married if he/she is a college graduate is 0.55. The
probability that a randomly selected worker is a college graduate is 0.58. The probability that a randomly selected
worker is not married is 0.35.
Married
Not Married
Total
College Graduate
0.319
0.261
0.58
Not College Graduate
0.331
0.089
0.42
Total
0.65
0.35
1
a) What is the probability of being married or being a college graduate? 0.65 + 0.58 – 0.319 = 0.911
b) What is the probability of being married or not being a college graduate? 0.65 + 0.42 – 0.331 = 0.739
c) What is the probability of not being a college graduate or not being married? 0.42 + 0.35 – 0.089 = 0.681
4. One card is randomly selected from a standard 52-card deck. Lizzie wants to examine the probability of finding an
ace and the probability of finding a black card.
Ace
Not Ace
Total
Black Card
2/52
24/52
26/52
Not Black Card
2/52
24/52
26/52
Total
4/52
48/52
52/52 = 1
a) What is the probability that Lizzie will select an ace or a black card? 4/52 + 26/52 – 2/52 = 28/52
b) What is the probability that Lizzie will select neither an ace nor a black card? 48/52 + 26/52 – 24/52 = 50/52
c) What is the probability that Lizzie will select an ace or not a black card? 4/52 + 26/52 – 2/52 = 28/52
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