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.

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 = -650/1000I 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.

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

• 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

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 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

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

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•

••

••

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

••

•

•

•

•

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

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!