Lenarz
Math 102
Practice Exam # 3
Name:
1. A 10-sided die is rolled 100 times with the following results:
Outcome Frequency
1
8
2
8
3
12
4
7
5
15
6
8
7
8
8
13
9
9
10
12
(a) What is the experimental probability of rolling a 5?
Solution:
P (5) =
15
= 0.15
100
(b) What is the theoretical probability of rolling a 5?
Solution:
P (5) =
1
= 0.1
10
(c) What is the experimental probability of rolling a multiple of 3?
Solution:
P (multiple of 3) =
12 + 8 + 9
29
=
= 0.29
100
100
(d) What is the theoretical probability of rolling a multiple of 3?
Solution:
P (multiple of 3) =
3
= 0.3
10
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2. One card is drawn from a standard 52 card deck. What is the probability of drawing:
(a) the queen of hearts?
Solution:
P (queen of hearts) =
1
52
(b) a queen?
Solution:
P (queen) =
4
1
=
52
13
P (heart) =
1
13
=
52
4
(c) a heart?
Solution:
(d) a face card (J, Q, or K)?
Solution:
P (face card) =
3
12
=
52
13
3. Two 6-sided dice are rolled. What is the probability of rolling
(a) a total of 6?
Solution: There are 5 ways to get a total of 6 − {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
so we have
5
P (total of 6) =
36
(b) not a total of 6?
Solution: This is simply the complement of the event in part (a), so we have
P (not a total of 6) = 1 − P (total of 6) = 1 −
5
31
=
36
36
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(c) total of 6 or 7?
Solution: The keyword “or” means we are looking for the probability of a
union, namely P (total of 6 ∪ total of 7). To find this we need P (total of 6),
P (total of 7), and P (total of 6 ∩ total of 7). We computed P (total of 6) in part
(a). There are 6 ways to get a total of 7 − {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
so we have
1
6
=
P (total of 7) =
36
6
Also note that
P (total of 6 ∩ total of 7) = 0
since it is impossible to have a total of 6 and a total of 7 simultaneously. Thus
we have
P (total of 6 ∪ total of 7) = P (total of 6) + P ( total of 7)
−P (total of 6 ∩ total of 7)
1
5
+ −0
=
36 6
11
=
36
(d) total of 6 or more?
Solution: The probability of getting a total of 6 or more is the complement of
the probability of getting a total of 2, 3, 4, or 5. There is only one way to get
a total of 2, 2 ways to get a total of 3, 3 ways to get a total of 4 and 4 ways to
get a total of 5. Also, the four events (getting a total of 2, a total of 3, a total
of 4 and total of 5) are mutually exclusive. Thus we have
P (total of 6 or more) = 1 − P (total of 2, 3, 4, or 5)
= 1 − P (total of 2) + P (total of 3)
=
=
=
=
+P (total of 4) + P (total of 5)
2
3
4
1
+
+
+
1−
36 36 36 36
10
1−
36
26
36
13
18
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4. A gumball machine has gumballs of five flavors. There are 10 apple, 15 berry, 12 cherry,
8 orange, and 9 mint. When a quarter is put into the machine, it dispenses 5 gumballs
at random. What is the probability that
(a) each gumball is a different flavor?
Solution: There are 10 + 15 + 12 + 8 + 9 = 54 gumballs in the machine and
54!
54 C5 = 39!5! = 3, 162, 510 ways to choose any 5 of the gumballs. There are
10 C1
× 15 C1 × 12 C1 × 8 C1 × 9 C1 = 10 × 15 × 12 × 8 × 9 = 129, 600
ways to choose one of each flavor. So the probability of selecting one gumball
of each flavor is
× 15 C1 × 12 C1 × 8 C1 × 9 C1
54 C5
129, 600
=
3, 162, 510
480
=
11, 713
≈ 0.04098
P (one of each flavor) =
10 C1
(b) at least two gumballs are the same flavor?
Solution: This is the complement of the event in part (a), so we have
P (at least 2 same flavor) = 1−P (one of each) = 1−
480
11, 233
=
≈ 0.95902
11, 713
11, 713
5. A coin is tossed ten times in a row. Find the probability that
(a) no tails show.
Solution: The probability that no tails show is the same as the probability
that all heads show. This is the probability that the first coin is heads AND the
second coin is heads AND.... so we have an intersection of independent events.
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Thus
P (no tails) = P (10 heads)
1
1 1
× × ··· ×
=
2}
|2 2 {z
10 times
1
= 10
2
1
=
1, 024
(b) exactly three tails show.
Solution: The probability of getting heads is 12 and the probability of getting
10!
= 120 ways to have exactly three tails (you
tails is 12 . There are 10 C3 = 7!3!
need to choose which 3 of the 10 tosses will be tails, the rest will be heads). So
the probability of getting exactly 3 tails is
3 7
15
1
1
120
=
≈ 0.117188
P (exactly 3 tails) =
×
× 10 C3 =
2
2
1024
128
(c) toss exactly twice as many heads as tails.
Solution: To have twice as many heads as tails we could have 1 tail and 2
heads (3 coins total), 2 tails and 4 heads (6 coins total), 3 tails and 6 heads (9
coins total), 4 tails and 8 heads (12 coins total), etc. Notice there is no way to
have twice as many heads as tails and get 10 coins. So it is impossible to do!
Thus
P (twice as many heads as tails) = 0
6. A group of 15 students is to be split into 3 groups of 5. In how many ways can this be
done?
Solution: There are 15 C5 ways to choose the first group, 10 C5 ways to choose the
second group and 5 C5 ways to choose the third group. So we multiply to get the
number of ways to choose all three groups. But that overcounts because it takes
the order of the groups into account (the first group is treated as different than the
second group, etc). So we need to divide out by the number of ways we can order
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the 3 groups which is 3!. So we have
15 C5
× 10 C5 × 5 C5
=
3!
=
=
=
=
15!
10!5!
×
10!
5!5!
×
5!
5!1!
3!
15×14×13×12×11
5×4×3×2×1
× 10×9×8×7×6
×1
5×4×3×2×1
3!
3003 × 252 × 1
3×2×1
756, 756
6
126, 126
7. Five cards are drawn from a standard 52 card deck. What is the probability of drawing
(a) 5 cards of the same color?
Solution:
52 25 24 23 22
×
×
×
×
52 51 50 49 48
253
=
4, 998
P (5 of same color) =
(b) a full house (3 cards of one rank and two cards of a different rank)?
Solution: The number of ways to draw any 5 cards is 52 C5 = 2, 598, 960. The
number of ways to get a full house is
# ways to
# of ways to
# of ways to
# of ways to
pick rank for × pick suits for ×
pick rank
×
pick suits
3 of a kind
3 of a kind
for pair
for pair
×
×
×
13 C1
4 C3
12 C1
4 C2
which is 3744. So the probability of drawing a full house is
P (full house) =
6
3744
=
≈ 0.001441
2598960
4165
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8. Suppose a jar has 4 coins: a penny, a nickel, a dime and a quarter. You remove two
coins at random without replacement. Let A be the event you remove the quarter. Let
B be the event you remove the dime. Let C be the event you remove less than 12 cents.
(a) List the sample space.
Solution:
S = {(p, n), (p, d), (p, q), (n, p), (n, d), (n, q),
(d, p), (d, n), (d, q), (q, p), (q, n), (q, d)}
(b) Draw a probability tree diagram to represent the possible scenarios.
Solution:
(c) Which pair(s) of theses events is (are) mutually exclusive?
Solution:
A = {(p, q), (n, q), (d, q), (q, p), (q, n), (q, d)}
B = {(p, d), (n, d), (d, p), (d, n), (d, q), (q, d)}
C = {(p, n), (p, d), (n, p), (d, p)}
So
A ∩ B = {(d, q), (q, d)}
A∩C = ∅
B ∩ C = {(p, d), (d, p)}
Thus only A and C are mutually exclusive.
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(d) Find P (A), P (C), and P (B).
Solution:
P (A) =
=
P (C) =
=
P (B) =
=
6
12
1
2
4
12
1
3
6
12
1
2
(e) Compute and interpret P (A ∪ B) and P (B ∪ C).
Solution: P (A ∪ B) is the probability that either A or B occurs, that is the
probability that either a quarter or dime is removed.
P (A ∪ B) = P (A) + P (B) − P (A ∩ B)
1 1 1
+ −
=
2 2 6
5
=
6
P (B ∪ C) is the probability that either B or C occurs, that is the probability
that either a dime or less than 12 cents is removed.
P (B ∪ C) = P (B) + P (C) − P (B ∩ C)
1 1 1
=
+ −
2 3 6
4
=
6
2
=
3
9. A dental assistant randomly sampled 200 patients and classified them according to
whether or not they had a least one cavity in their last checkup and according to what
type of tooth decay preventative measures they used. The information is as follows
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Brush only
Brush and floss only
Brush and tooth sealants only
Brush, floss and tooth sealants
At least one cavity
69
34
22
3
No Cavities
2
11
13
46
If a patient is picked at random from this group, find the probability that
(a) the patient had a least one cavity.
Solution: The number of patients with at least one cavity is
69 + 34 + 22 + 3 = 128
and the total number of patients is 200 so
P (at least one cavity) =
128
16
=
200
25
(b) the patient brushes only.
Solution: The number of patients that brush only is
69 + 2 = 71
and the total number of patients is 200 so
P (only brush) =
71
200
(c) the patient had no cavities, given s/he brushes, flosses and has tooth sealants.
Solution: The number of patients that have no cavities and brush, floss and
use sealants is 46. So the probability a patient has no cavities and brushes,
flosses and uses sealants is
P (no cavities ∩ brush, floss, and sealant) =
46
23
=
200
100
The number of patients that brush, floss and use sealants is 46 + 3 = 49 so the
probability a patient brushes, flosses and uses sealants is
P (brush, floss, and sealants) =
49
200
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Thus the probability that a patient has no cavities, given s/he brushes, flosses
and uses sealants is
P (no cavities|brush, floss, and sealants) =
P (no cavities ∩ brush, floss, and sealants)
P (brush, floss, and sealants)
=
23
100
49
200
=
46
49
(d) the patient brushes only, given that s/he had at least one cavity.
Solution: The number of patients that only brush and have at least one cavity
is 69. So the probability a patient only brushes and has at least one cavity is
P (only brushes ∩ at least one cavity) =
69
200
The number of patients that have at least one cavity is 69 + 34 + 22 + 3 = 128
so the probability a patient has at least one cavity is
P (at least one cavity) =
16
128
=
200
25
Thus the probability that a patient brushes only, given s/he has at least one
cavity is
P (brushes only|at least one cavity) =
P (brushes only ∩ at least one cavity)
P (at least one cavity)
=
69
200
128
200
=
69
128
10. A small college has two calculus classes. The first class has 25 students, 15 of whom are
female, and the other class has 18 students, 8 of whom are female. One of the classes is
selected at random and then two students are randomly selected from the class for an
interview. If both of the students are female, what is the probability they came from
the first class?
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Solution: Call the first class Class A and the second class Class B. Then we have
the following data
Female
15
8
Class A
Class B
Male
10
10
So we get the following probability tree
From the tree we see that the probability of choosing class A and choosing two
females is
7
7
1 3
=
P (A ∩ 2 females) = × ×
2 5 12
40
The probability of choosing two females is
P (2 females) =
7
1 4
7
1631
1 3
× ×
+ × ×
=
2 5 12 2 9 17
6120
Thus the probability of choosing class A, given that two females were chosen is
P (A|2 females) =
P (A ∩ 2 females)
P (2 females)
=
7
40
1631
6120
=
153
233
11. The probability is 0.6 that a student will study for an exam. If the student studies,
she has a 0.8 chance of getting an A on the exam. If she does not study, she has a
0.3 probability of getting an A. Make a probability tree for this situation. What is the
probability that she gets an A? If she gets an A, what is the conditional probability that
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she studied?
Solution: We get the following probability tree (where S denotes study, N denotes
not study, A denotes gets an A and NA denotes does not get an A)
From the tree we see that
P (A) = 0.6 × 0.8 + 0.4 × 0.3
= 0.6
We also know that
P (S ∩ A) = 0.6 × 0.8 = 0.48
so we have
P (S ∩ A)
P (A)
0.48
=
0.6
= 0.8
P (S|A) =
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12. If you consider the value of a roll of a single 6-sided die to be the number that is rolled,
what is the expected value of the roll of a single die?
Solution:
E =
=
=
=
=
1
1
1
1
1
1
(1) +
(2) +
(3) +
(4) +
(5) +
(6)
6
6
6
6
6
6
1 2 3 4 5 6
+ + + + +
6 6 6 6 6 6
21
6
7
2
3.5
13. Consider a game that consists of drawing a single card at random from a standard 52
card deck. You pay $3 to play the game and the $3 is not returned. If you draw an ace
you win $10. If you draw a king or a queen, you win $5. How much should you expect
to win or lose on average if you play this game?
Solution:
E =
=
=
=
=
4
8
40
(10 − 3) +
(5 − 3) +
(0 − 3)
52
52
52
1
2
10
(7) +
(2) +
(−3)
13
13
13
7
4
30
+
−
13 13 13
19
−
13
−1.46
You would expect to loose $1.46.
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14. Suppose you just inherited $100, 000 and you are trying to decide how to invest it for
the next year. You have narrowed it down to two choices. The first choice is to invest
it in a bond with a guaranteed return of 5% interest at the end of the year. The second
choice is to bet it all on the Super Bowl with a 51% chance of doubling your money and
a 49% chance of losing it all. Which investment option has the highest expected value?
Solution: The bond has an expected value of
E = (1)(100, 000 + 100, 000 × 0.05)
= 105, 000
Betting on the Super Bowl has an expected value of
E = (0.51)(200, 000) + (0.49)(0)
= 102, 000
So the bond has the higher expected value.