Download Think of the game rock, paper, scissors

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.
Given m // n.
• T or F:  7 and  4
are vertical.
7 6
• T or F:  1   4
3
5
4
• T or F:  2   3
2
1
• T or F: m  7 + m  6 = m  1
m
• 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.
n
3
7
6
5
• T or F:  7 and  4
4
2
1
n
are vertical. F
m
• 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
Try these
Name 3 rays. Not GE.
Name 4 different angles.
D•
F•
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
B
A•
C•
E
G
•
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
E
•
•
H
D
G•
•
C
B
A
•
F
•
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
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!