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MAT 205: FINITE MATH
MAT 205 CHECKPOINT 6
POINT VALUES
SECTION 7.1
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SECTION 7.4
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INSTRUCTOR: CARY SOHL
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SECTION 7.2
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SECTION 7.5
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SECTION 7.3
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SECTION 7.6
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WEEK 6 CHECKPOINT
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University of Phoenix Axia College
MAT 205: FINITE MATH
Section 7.1
7.1-1) If A and B are sets, and A  B   , then what set is
( A  B)  ( A  B ')
The empty set.
7.1-2) If A={1,2,3,4}, and B={9,11,13,15}, then if S={x|x  (A  B)}, what is S?
S={ 1, 2, 3, 4, 9, 11, 13, 15 }
7.1-3) Decide if the statement below is true or false, and support your conclusion
with an example.
If n(A)<n(B) Then A  B
No, this is not necessarily true. There is no specified connection between the
sets A and B, so while the number of elements in set A may indeed be fewer
than the number of elements of set B, it does not necessarily follow that A is a
subset of B.
Example:
A = {1, 2, 3}
B = {4, 5, 6, 7}
n(A) = 3, n(B) = 4, n(A) < n(B), but A is not a subset of B.
7.1-4) Decide if the statement below is true or false, and support your conclusion
with an example.
If A  B Then n(A)<n(B)
INSTRUCTOR: CARY SOHL
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University of Phoenix Axia College
MAT 205: FINITE MATH
This is true. If A is a proper subset of B, then every element of A is also in B, but
the number of elements of A is fewer than the number of elements of B, by
definition.
Example:
A = {1, 2, 3}
B = {1, 2, 3, 4}
A is a proper subset of B.
n(A) = 3, n(B) = 4, n(A) < n(B).
7.1-5) List the subsets of A={1,2,3}
Subsets of A: { {∅}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} }
INSTRUCTOR: CARY SOHL
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University of Phoenix Axia College
MAT 205: FINITE MATH
Section 7.2
7.2-1) A medical study of 100 cats shows that 22 have fleas, 54 have ticks, and
10 have both fleas and ticks. Draw a Venn Diagram which represents the
situation. The diagram needs to show the set F (cats with fleas), the set T
(cats with Ticks), and the number of elements in each. Draw the diagram
in any software you wish, but paste a screen capture of the drawing in the
work area below. How many cats do not have either fleas or ticks?
QuickTime™ and a
decompressor
are needed to see this picture.
7.2-2) Explain in words what the union rule, repeated below, actually means.
n( A  B)  n( A)  n( B)  n( A  B)
The number of elements in the union of sets A and B is equal to the number of
elements in set A, plus the number of elements in set B, minus the number of
elements that are in both sets A and B simultaneously.
INSTRUCTOR: CARY SOHL
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University of Phoenix Axia College
MAT 205: FINITE MATH
7.2-3) Draw a Venn Diagram, filling in the number of elements of each of the sets
and relevant regions, which represents the situation below. Use any
software you wish, but paste a screen capture of the Venn Diagram in the
work area below.
n(A)=26, n(B)=10, n(A  B)=30, n(A’)=17
QuickTime™ and a
decompressor
are needed to see this picture.
INSTRUCTOR: CARY SOHL
WEEK 4 CHECKPOINT
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University of Phoenix Axia College
MAT 205: FINITE MATH
Section 7.3
7.3-1) If a day in April is selected for a cookout, what is the sample space?
Sample space = { x is an integer; 1 ≤ x ≤ 30 }
7.3-2) A coin is flipped 3 times. Write the sample space. What is the relative
likelihood of getting 3 heads?
Sample space = { {HHH}, {HHT}, {HTH}, {HTT}, {THH}, {THT}, {TTH}, {TTT} }
P({TTT}) = 1/8
7.3-3) A single fair die is thrown. Write the sample space of outcomes. If the
event to be considered is getting an odd number, what is the probability of
the event?
Sample space = { {1}, {2}, {3}, {4}, {5}, {6} }
P(odd) = 3/6 = 1/2
7.3-4) A jar has 2 white, 3 orange, 5 yellow, and 10 black marbles. If a marble is
picked at random, what is the probability that it is white or orange?
P(white or orange) = (2 + 3)/(2+3+5+10) = 5/20 = 1/4
7.3-5) The table below shows a categorization of the members of a club by race.
What is the probability of randomly meeting an Asian member?
Race
White
Asian
African
INSTRUCTOR: CARY SOHL
# Of Members
10
20
10
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MAT 205: FINITE MATH
Hispanic 30
Other
10
P(Asian) = 20 / (10 + 20 + 10 + 30 + 10) = 20 / 80 = 1/4
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MAT 205: FINITE MATH
Section 7.4
7.4-1) Two dice are rolled. What is the probability that the first die is a 3 or the
sum is 8?
P(first die is a 3) = 6/36
P(sum is 8) = 5/36
P(first die is a 3 AND sum is 8) = 1/36
P(first die is a 3 OR sum is 8) = P(first die is a 3) + (P sum is 8) – P(first die is a 3
AND sum is 8) = 6/36 + 5/36 – 1/36 = 10/36 = 5/18
7.4-2) P(A)=.42, P(B)=.35, and P(A  B)=.61. What is P(A’  B’)?
P(A’∩B’) = 1 – P(A∪B) = 1 – 0.61 = 0.39
7.4-3) A jar has 3 yellow, 4 white, and 8 blue marbles. What are the odds in
favor of drawing a yellow marble?
Opportunities to draw yellow = 3
Opportunities to draw other than yellow = 12
Odds of drawing yellow = 3 to 12 = 1 to 4
7.4-4) A study on body types showed that 45% were short, 25% were short and
overweight, and 24% were tall and not overweight. Find the probability of
a person being overweight.
P(Short) = 0.45
P(Short & Overweight) = 0.25
P(Tall & Not Overweight) = 0.24
P(Tall) = 1 – P(Short) = 1 – 0.45 = 0.55
P(Tall & Overweight) = P(Tall) – P(Tall & Not Overweight) = 0.55 – 0.24 = 0.31
INSTRUCTOR: CARY SOHL
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MAT 205: FINITE MATH
P(Overweight) = P(Short & Overweight) + P(Tall & Overweight) = 0.25 + 0.31 =
0.56
INSTRUCTOR: CARY SOHL
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MAT 205: FINITE MATH
Section 7.5
7.5-1) A single die is rolled. Given that the number is even, what is the
probability that it is a 4? Do the calculation using the formula for
conditional probability.
P(4|Even) = P(4 and Even) / P(4) = (1/6) / (3/6) = (1/6)(6/3)= 1/3
7.5-2) Two cards are drawn from a normal deck, without replacement. What is
the probability that the second card drawn Is an ace, given that the first
card is not an ace?
P(second card is an ace|first card is not an ace) = 4/51
7.5-3) A and B are independent events. P(A)=.25, P(B)=.2. Find the probability
that A or B occurs. Find the probability that A and B occurs.
P(A and B) = (0.25)(0.2) = 0.05
P(A or B) = P(A) + P(B) – P(A and B) = 0.25 + 0.2 – 0.05 = 0.40
INSTRUCTOR: CARY SOHL
WEEK 6 CHECKPOINT
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University of Phoenix Axia College
MAT 205: FINITE MATH
Section 7.6
7.5-1) For 2 events M and N, P(M)=.4, P(N|M)=.3, AND P(N|M’)=.4.
P(M’|N).
Find
P(M’|N) = P(M’)P(N|M’) / [ P(M’)P(N|M’) + P(M)P(N|M) ]
P(M’) = 1 – P(M) = 1 – 0.4 = 0.6
P(M’|N) = (0.6)(0.4) / [ (0.6)(0.4) + (0.4)(0.3) ]
P(M’|N) = 0.24 / (0.24 + 0.12)
P(M’|N) = 0.24 / 0.36
P(M’|N) = 2/3
INSTRUCTOR: CARY SOHL
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