these problems have all appeared in the power points
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These problems have all appeared in the power points. Now, here are the answers. Let’s review. Probability: I throw a six-sided die once and then flip a coin twice. Event? Possible outcomes? Total possible events? P(2 heads) P(odd, 2 heads) - PowerPoint PPT PresentationTRANSCRIPT
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These problems have all These problems have all appeared in the power appeared in the power
points.points.
Now, here are the answers.Now, here are the answers.
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Let’s review• Probability: • I throw a six-sided die once and then flip a coin twice.
– Event?– Possible outcomes?– Total possible events?– P(2 heads)– P(odd, 2 heads)– Can you make a tree diagram?
Can you use the Fundamental Counting Principle to find the number of
outcomes?
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• Probability: • I throw a six-sided die once and then flip a
coin twice. – Event? What we want. Ex: even, at least 1 head– Possible outcomes? 6, H, T 5, T, T 3, T, H– Total possible events? Either a tree diagram or
Fundamental Counting Principle: – P(2 heads) 1/4 (1/2 • 1/2)– P(odd, 2 heads) 1/8 (3/6 • 1/2 • 1/2)
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• Probability:• I have a die: its faces are 1, 2, 7, 8, 9, 12.• P(2, 2)--is this with or without replacement?• P(even, even) =• P(odd, 7) = • Are the events odd and 7 disjoint? Are they
complementary?
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• Probability:• I have a die: its faces are 1, 2, 7, 8, 9, 12.• P(2, 2)--is this with or without replacement? With
replacement. Each number has a chance to come up in the second roll. 1/36
• P(even, even) = 1/2 • 1/2 = 1/4• P(odd, 7) = 3/6 • 1/6 = 3/36• Are the events odd and 7 disjoint? Are they
complementary? Not disjoint--the roll of “7” satisfies both events. Not complementary--does not complete the whole.
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I play the lottery…• I pay a dollar, and then I pick any 3-digit number:
and 0 can be a leading digit.• If I pick the exact order of the numbers, I get $100.• If I pick the numbers, but one or more are out of
order, I get $50.• Who wins over time--me or the lottery?
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• I pay a dollar, and then I pick any 3-digit number: and 0 can be a leading digit.
• If I pick the exact order of the numbers, I get $100.• If I pick the numbers, but one or more are out of order, I get
$50.• Who wins over time--me or the lottery?• Pay $1. There are 1000 3-digit combinations (10•10•10). So,
P(exact order) = 1/1000. P(not in exact order) = 5/1000.Expected Value = -1 + (1/1000) • 100 + (5/1000) • 50.
= -1 + 100/1000 + 250/1000 = -650/1000I lose about 65 cents each time I play, on average.
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• Most days, you will teach Language Arts, Math, Social Studies, and Science. If Language Arts has to come first, how many different schedules can you make?
• 1 • 3 • 2 • 1• Permutation: the order of the schedule matters.
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Deal or no Deal• You are a contestant on Deal or No
Deal. There are four amounts showing: $5, $50, $1000, and $200,000. The banker offers $50,000.
• Should you take the deal? Explain.• How did the banker come up with
$50,000 as an offer?
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• You are a contestant on Deal or No Deal. There are four amounts showing: $5, $50, $1000, and $200,000. The banker offers $50,000.
• Should you take the deal? Explain. Each amount has an equally likely chance of being chosen. So, 3/4 of the time I will pick a case less than $50,000. Take the deal!
• How did the banker come up with $50,000 as an offer? Expected Value: (1/4)•5 + (1/4)•50 + (1/4)•1000 + (1/4)•200,000 = $50,256.25. Round down so that the banker wins over time.
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Given m // n.• T or F: 7 and 4
are vertical.• T or F: 1 4• T or F: 2 3• T or F: m 7 + m 6 = m 1• T or F: m 7 = m 6 + m 5• If m 5 = 35˚, find all the angles you can.• If m 5 = 35˚, label each angle as acute, right, obtuse.• Describe at least one reflex angle.
7 65
43
21
mn
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• T or F: 7 and 4are vertical. F
• T or F: 1 4 T, corresponding• T or F: 2 3 T, both supplementary to 1 and 4• T or F: m 7 + m 6 = m 1 T, 7 and 6 together are vertical to 4,
and 1 is congruent to 4• T or F: m 7 = m 6 + m 5 F, we can’t assume 7 is a right angle• If m 5 = 35˚, find all the angles you can. 5, 3, 2 = 35 degrees, 1, 4
= 145 degrees.• If m 5 = 35˚, label each angle as acute, right, obtuse. 5, 3, 2 =
acute; 1,4 = obtuse.• Describe at least one reflex angle. Combine 7, 3, 4.
7 65
43
21
mn
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Combinations and Permutations
• These are special cases of probability!• I have a set of like objects, and I want to
have a small group of these objects.• I have 12 different worksheets on probability.
Each student gets one:– If I give one worksheet to each of 5 students, how
many ways can I do this? – If I give one worksheet to each of the 12 students,
how many ways can I do this?
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• I have a set of like objects, and I want to have a small group of these objects.
• I have 12 different worksheets on probability. Each student gets one:– If I give one worksheet to each of 5 students, how many
ways can I do this? Permutation: If Student 1 gets A and Student 2 gets B, this is different from Student 1 gets B and Student 2 gets A. So, 12 • 11 • 10 • 9 • 8
– If I give one worksheet to each of the 12 students, how many ways can I do this? 12!
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More on permutations and combinations
• I have 15 french fries left. I like to dip them in ketchup, 3 at a time. How may ways can I do this? Assuming that the french fries are all about the same, this is a combination. (15 • 14 • 13)/(3 • 2 • 1)
• I am making hamburgers: I can put 3 condiments: ketchup, mustard, and relish, I can put 4 veggies: lettuce, tomato, onion, pickle, and I can use use 2 types of buns: plain or sesame seed. How many different hamburgers can I make? 3 • 4 • 2 = 24
• Why isn’t this an example of a permutation or combination? See comment on slide 13.
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When dependence matters
• If I have 14 chocolates in my box: 3 have fruit, 8 have caramel, 2 have nuts, one is just solid chocolate!
• P(nut, nut)• P(caramel, chocolate)• P(caramel, nut)• If I plan to eat one each day, how many
different ways can I do this?
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• If I have 14 chocolates in my box: 3 have fruit, 8 have caramel, 2 have nuts, one is just solid chocolate!
• P(nut, nut) = 2/14 • 1/13• P(caramel, chocolate) = 8/14 • 1/13• P(caramel, nut) = 8/14 • 2/13• If I plan to eat one each day, how many
different ways can I do this? If each candy is unique, then 14!
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Geometry• Sketch a diagram with 4 concurrent lines.• Now sketch a line that is parallel to one of
these lines.• Extend the concurrent lines so that the
intersections are obvious.• Identify: two supplementary angles, two
vertical angles, two adjacent angles.• Which of these are congruent?
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• Sketch a diagram with 4 concurrent lines.• Now sketch a line that is parallel to one of these
lines.• Extend the concurrent lines so that the intersections
are obvious.• Identify: two supplementary angles (•), two vertical
angles (•), two adjacent angles (•).• Which of these are congruent?
••••
•
•
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Geometry• Sketch 3 parallel lines segments.• Sketch a line that intersects all 3 of these line
segments. • Now, sketch a ray that is perpendicular to one
of the parallel line segments, but does not intersect the other two parallel line segments.
• Identify corresponding angles, supplementary angles, complementary angles, vertical angles, adjacent angles.
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• Sketch 3 parallel lines segments.• Sketch a line that intersects all
3 of these line segments. • Now, sketch a ray that is
perpendicular to one of the parallel line segments, but does not intersect the other two parallel line segments.
• Identify corresponding angles, supplementary angles, complementary angles, vertical angles, adjacent angles.
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Name attributes• Kite and square• Rectangle and trapezoid• Equilateral triangle and equilateral
quadrilateral• Equilateral quadrilateral and equiangular
quadrilateral• Convex hexagon and non-convex hexagon.
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• Kite and square adjacent congruent sides• Rectangle and trapezoid a pair of // sides• Equilateral triangle and equilateral
quadrilateral all congruent sides• Equilateral quadrilateral and equiangular
quadrilateral a square satisfies both--nothing else.
• Convex hexagon and non-convex hexagon. Both have 6 sides, 6 vertices
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Consider these trianglesacute scalene, right scalene, obtuse scalene, acute isosceles, right isosceles, obtuse isosceles, equilateral– Name all that have:– At least one right angle– At least two congruent angles– No congruent sides
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Consider these figures:Triangles: acute scalene, right scalene,
obtuse scalene, acute isosceles, right isosceles, obtuse isosceles, equilateral
Quadrilaterals: kite, trapezoid, parallelogram, rhombus, rectangle, square
Name all that have:At least 1 right angleAt least 2 congruent sidesAt least 1 pair parallel sidesAt least 1 obtuse angle and 2 congruent sidesAt least 1 right angle and 2 congruent sides
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Try theseName 3 rays. Not GE.
Name 4 different angles.
Name 2 supplementary angles. CBE and DBE
Name a pair of vertical angles. ABD and CBE
Name a pair of adjacent angles. BEG and GEF
Name 3 collinear points. A, B, E, F
D
C
BA
F
EG•
••
••
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Try theseName 2 right angles.
HDF and FDCName 2 complementary angles.
FDG and GDCName 2 supplementary angles.
HDE and EDCName 2 vertical angles.
EDH and GDCTrue or false: AD = DA. FIf m EDH = 48˚, find m GDC. 48
H FE
D
C
A
B
G
••
•
•
•
•
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Try these• Assume lines l, m, n
are parallel.
• Copy this diagram.
• Find the value of each angle.
• Angles will have measure 63˚, 117˚,27˚, or 90˚
l
63˚
tn
m
u
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Make this game fair• I am going to flip a coin 4 times.
– Make a tree diagram or make an organized list of the outcomes.
• There are four players.• These are the outcomes that can win.
– Exactly 1 head. P(1 head) = 1/4– Getting a head on the second and third flips (note, other heads
can still appear). P(X,H,H,X) = 4/16– 4 tails. P(T,T,T,T) = 1/16– Getting exactly 2 consecutive tails. 5/16 (tricky! TTHH, HTTH,
HHTT, THTT, TTHT)– One fair game: Players 1 and 2 get 5 points, Player 3 gets 20
points, and Player 4 gets 4 points.
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Let’s try some more• 5 cards, numbered 1 - 5, are shuffled and placed face down.
Your friend offers you $5 if the first 3 cards are in descending order (not necessarily consecutive) when you turn them over. Otherwise, you pay him $6. – Should you play this game?
– Make new rules so that the game is fair.
– Make new rules so that you almost always win.
– You win if (5,4,3, 5,4,2 5,4,1 5,3,2, 5,3,1, 5,2,1 4,3,2 4,3,1 4,2,1 3,2,1) So, 10 out of 5 • 4 • 3 = 60.
– So, 10/60 * (5) ≠ 50/60 • (6) Your friend wins a lot!
– One way to make it fair: You win $5 and your friend wins $1.
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Let’s try some more• One bag of marbles contains 4 blue, 2
yellow, and 5 red. Another bag contains 8 red, 3 blue, and 2 yellow. Is it possible to rearrange the marbles so that your chance of drawing a red one from both bags is greater than 1/2?
• 8/15 + 5/9; 7/13 + 6/11
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A few practice problems• A drawer contains 6 red socks and 3
blue socks.P(pull 2, get a match)P(pull 3, get 2 of a kind)P(pull 4, all 4 same color)
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• A drawer contains 6 red socks and 3 blue socks.P(pull 2, get a match) =6/9 • 5/8 + 3/9 • 2/8
• P(pull 3, get 2 of a kind)= 6/9 • 5/8 • 3/7 + 3/9 • 2/8 • 6/7
• P(pull 4, all 4 same color) = 6/9 • 5/8 • 4/7 • 3/6
• Consider the red socks first, and then the blue.
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Some basic probabilities• I have 40 marbles: 30 are green, 2 are blue, and 8
are black.• P(not green) = 10/40• P(red, red, black) with no replacement = 0 (no reds)• P(green, not green, green) with replacement.
30/40 • 10/40 • 30/40• P(5 blues in a row) = not possible!