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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)
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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) • Can you make a tree diagram? Can you use the Fundamental Counting Principle to find the number of outcomes?
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)
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?
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.
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?
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 = -650/1000 I lose about 65 cents each time I play, on average.
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.
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?
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.
6 7 3 5 4 2 n 1 m Given m // n. • T or F: 7 and 4are 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.
6 7 3 5 4 2 n 1 m • 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.
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?
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!
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.
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?
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!
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?
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? • • • • • •
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.
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.
Name attributes • Kite and square • Rectangle and trapezoid • Equilateral triangle and equilateral quadrilateral • Equilateral quadrilateral and equiangular quadrilateral • Convex hexagon and non-convex hexagon.
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
Consider these triangles acute 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
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 angle At least 2 congruent sides At least 1 pair parallel sides At least 1 obtuse angle and 2 congruent sides At least 1 right angle and 2 congruent sides
• • • • • D F G E B A C Try these Name 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
• • E H F • D • G • C B • A Try these Name 2 right angles. HDF and FDC Name 2 complementary angles.FDG and GDC Name 2 supplementary angles.HDE and EDC Name 2 vertical angles. EDH and GDC True or false: AD = DA. F If m EDH = 48˚, find m GDC. 48
63˚ u m n t Try these • Assume lines l, m, nare parallel. • Copy this diagram. • Find the value of each angle. • Angles will have measure 63˚, 117˚,27˚, or 90˚ l
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.
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.
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
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)
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.
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!