Download April 12

Geometry Unit 5
Probability and Statistics
12 April 2017
Agenda 4/12/2017
● Bulletin
● Arcs and Angles Homework - correct and record grade
and turn in
● Probability - Introductory Activity
○ Getting to know what is in a deck of cards
● Homework
Practice Exercises: Page 164
Find each measure:
5. m ∠5
66°
m∠a = (66 + 50)°/ 2 = 116/2
a
5
=58°
m ∠5 = (180 - 58)° = 122°
50°
Practice Exercises: Page 164
Find each measure:
6. m ∠6 = a
6
Secant-Tangent
228°
m ∠6=132°
2
m ∠6 = 66°
132°
Students to show us
7.
8.
9.
10.
11. And
12
Grading - next ? minutes.
Page 164 - now done together CW score /10
(I will come around to record)
Meanwhile you tear out 165-6 and correct and
grade using answer sheet provided.
Tear out 167 - 168 for me to grade (if you did it).
Sort the deck of cards - we may be missing some
Some may be in the wrong deck (check color of back of card)
Use desktop to arrange as in the layout on the worksheet sorry it is not in color.
Once you have a full deck to work with answer the questions.
Deck
How many cards are in a deck?
How many of each type of card?
Suits
How many suits are there?
How many cards are in a suit?
What are the names of the suits?
Deck
How many cards are in a deck? 52
How many of each type of card? 4
Suits
How many suits are there? 4
How many cards are in a suit? 13
What are the names of the suits? Hearts, diamonds, clubs,
spades.
Suits
How many suits are there? 4
How many cards are in a suit? 13
What are the names of the suits? Hearts, diamonds, clubs,
spades.
Which suits are red? Hearts and diamonds
Which suits are black? Clubs and spades
How many red cards in the deck? (13 + 13) 26
How many black cards in the deck? 26
Types of Cards.
Face Cards (Cards with Faces)
How many face cards are there? 12
What are the names for the face cards? Kings, queens and
jacks.
How many face cards in a suit? 3
Numerical Cards (cards with numbers).
How many numerical cards are there? 36
How many numerical cards in a suit? 9
ACES (Not a face card, Not a numerical card)
How many aces in a deck? 4
Given a standard deck of cards, if we draw one
card at random, what are the
Chances of drawing a diamond?
Given a standard deck of cards, if we draw one
card at random, what are the
Chances of drawing a diamond? 13/ 52 = ¼ or 0.25
Chances of drawing a Face Card?
Given a standard deck of cards, if we draw one
card at random, what are the
Chances of drawing a diamond? 13/ 52 = ¼ or 0.25
Chances of drawing a Face Card? 12/52 = 3/13 or 0.23
Chances of selecting an Ace?
Given a standard deck of cards, if we draw one
card at random, what are the
Chances of drawing a diamond? 13/ 52 = ¼ or 0.25
Chances of drawing a Face Card? 12/52 = 3/13 or 0.23
Chances of selecting an Ace? 4/52 = 1/13 or 0.077
Chances of selecting a Numerical Card?
Given a standard deck of cards, if we draw one
card at random, what are the
Chances of drawing a diamond? 13/ 52 = ¼ or 0.25
Chances of drawing a Face Card? 12/52 = 3/13 or 0.23
Chances of selecting an Ace? 4/52 = 1/13 or 0.077
Chances of selecting a Numerical Card? 36/52 =9/13 or 0.69
Probability - deals with the possibility that an event
will happen. Our event - is drawing a card, the
sample space is all the cards it could be - 52
possible outcomes
P(event) = number of favorable outcomes
# of total possible outcomes
P(event) = # of items in event
# of items in sample space
Probability(drawing the card stated)
P(Jack) = 4/52 = 1/13 or 0.077
P(Red 8)
P(Red Queen or Black King)
P(Not a face card)
P(8)
P(Black Ace)
Probability(drawing the card stated)
P(Jack) = 4/52 = 1/13 or 0.077
P(Red 8)
P(Red Queen or Black King)
P(Not a face card)
P(8) = 4/52 = 1/13 or 0.077
P(Black Ace)
Probability(drawing the card stated)
P(Jack) = 4/52 = 1/13 or 0.07
P(Red 8) = 2/52 = 1/26 or 0.04
P(Red Queen or Black King)
P(Not a face card)
P(8) = 4/52 = 1/13 or 0.077
P(Black Ace)
Probability(drawing the card stated)
P(Jack) = 4/52 = 1/13 or 0.077
P(8) = 4/52 = 1/13 or 0.077
P(Red 8) = 2/52 = 1/26 or 0.04
P(Black Ace) = 1/26 or 0.04
P(Red Queen or Black King)
P(Not a face card)
Probability(drawing the card stated)
P(Red 8) = 2/52 = 1/26 or 0.04
P(Black Ace) = 1/26 or 0.04
P(Red Queen or Black King) = (2 + 2) / 52 = 1/13 = 0.077
P(Not a face card)
Probability(drawing the card stated)
P(Red 8) = 2/52 = 1/26 or 0.04
P(Black Ace) = 1/26 or 0.04
P(Red Queen or Black King) = (2 + 2) / 52 = 1/13 or 0.077
P(Not a face card) = (52 - 12)/ 52 = 40/52 = 10/13 or 0.77
Leave line empty for now - going to be title
______________
If a card is drawn from the pack, then replaced and a second card is drawn, what
is the probability that both cards will be Aces?
First card P(Ace) =
Second card P(Ace) =
Combined probability P(Ace and Ace) =
Did the outcome of the first event affect the outcome of the second event?
Leave line empty for now - going to be title
______________
If a card is drawn from the pack, then replaced and a second card is drawn, what
is the probability that both cards will be Aces?
First card P(Ace) = 4/ 52 = 1/13
Second card P(Ace) =
Combined probability P(Ace and Ace) =
Did the outcome of the first event affect the outcome of the second event?
Leave line empty for now - going to be title
______________
If a card is drawn from the pack, then replaced and a second card is drawn, what
is the probability that both cards will be Aces?
First card P(Ace) = 4/ 52 = 1/13
Second card P(Ace) = 1/13
Combined probability P(Ace and Ace) =
Did the outcome of the first event affect the outcome of the second event?
Leave line empty for now - going to be title
______________
If a card is drawn from the pack, then replaced and a second card is drawn, what
is the probability that both cards will be Aces?
First card P(Ace) = 4/ 52 = 1/13 or
0.077
Second card P(Ace) = 4/52 = 1/13
or
0.077
Combined probability P(Ace and Ace) = 1 x 1 = 1
13 13 169
(very unlikely event)
or 0.006
Did the outcome of the first event affect the outcome of the second event? No.
Compound Events
______________
If a card is drawn from the pack, then replaced and a second card is drawn, what
is the probability that both cards will be Aces?
First card P(Ace) = 4/ 52 = 1/13 or
0.077
Second card P(Ace) = 4/52 = 1/13
or
0.077
Combined probability P(Ace and Ace) = 1 x 1 = 1
13 13 169
(very unlikely event)
or 0.006
Did the outcome of the first event affect the outcome of the second event? No.
Compound Events
Independent Events outcome of one not affected by other
event outcome
If a card is drawn from the pack, then replaced and a second card is drawn, what is
the probability that both cards will be Aces?
First card P(Ace) = 4/ 52 = 1/13 or
0.077
Second card P(Ace) = 4/52 = 1/13
or
0.077
Combined probability P(Ace and Ace) = 1 x 1 = 1
13 13 169
(very unlikely event)
or 0.006
Did the outcome of the first event affect the outcome of the second event? No.
_________________
If a card is drawn from the pack, then a second card is drawn from the remaining
cards, what is the probability that both cards will be Aces?
First card P(Ace) =
Second card P(Ace) =
Combined probability P(Ace and Ace) =
Did the outcome of the first event affect the outcome of the second event?
_________________
If a card is drawn from the pack, then a second card is drawn from the remaining
cards, what is the probability that both cards will be Aces?
First card P(Ace) = 4/ 52 = 1/13
but keep it - so ...
Second card P(Ace) =
Combined probability P(Ace and Ace) =
Did the outcome of the first event affect the outcome of the second event?
_________________
If a card is drawn from the pack, then a second card is drawn from the remaining
cards, what is the probability that both cards will be Aces?
First card P(Ace) = 4/ 52 = 1/13
holding on to this Ace
Second card P(Ace) = 3/51
Now only 3 aces in the deck of 51 cards
Combined probability P(Ace and Ace) = 1 x 3 = 3 = 1
13 51
663 221
or 0.0045
Did the outcome of the first event affect the outcome of the second event?
Yes.
Dependent Events
If a card is drawn from the pack, then a second card is drawn from the remaining
cards, what is the probability that both cards will be Aces?
First card P(Ace) = 4/ 52 = 1/13
holding on to this Ace
Second card P(Ace) = 3/51
Now only 3 aces in the deck of 51 cards
Combined probability P(Ace and Ace) = 1 x 3 = 3 = 1
13 51
663 221
or 0.0045
Did the outcome of the first event affect the outcome of the second event?
Yes.
Homework
Probability Review Lesson 1.
Sample Space, Events and Notation
Read pages 1 and 2.
Complete the Probability Review Questions Homework Lesson 1 on pages 3 and 4.
See slides that follow for some additional input from me
when you are working on this at home, if you need some more
ideas.
Sample Space - set of all possible outcomes
Rolling 2 die
Sample Space - set of all possible outcomes
White
Red
1
2
3
4
5
6
1
2
3
4
5
6
Sample Space - set of all possible outcomes
White
Red
1
2
3
4
5
6
1
1, 1
1, 2
1, 3
1, 4
1, 5
1, 6
2
2, 1
2, 2
2, 3
2, 4
2,5
2, 6
3
3, 1
3, 2
3, 3
3, 4
3, 5
3, 6
4
4, 1
4, 2
4, 3
4, 4
4, 5
4,6
5
5, 1
5, 2
5, 3
5, 4
5, 5
5,6
6
6, 1
6, 2
6, 3
6, 4
6, 5
6,6
Sample Space - compare to set notation on page 1
White
Red
1
2
3
4
5
6
1
1, 1
1, 2
1, 3
1, 4
1, 5
1, 6
2
2, 1
2, 2
2, 3
2, 4
2,5
2, 6
3
3, 1
3, 2
3, 3
3, 4
3, 5
3, 6
4
4, 1
4, 2
4, 3
4, 4
4, 5
4,6
5
5, 1
5, 2
5, 3
5, 4
5, 5
5,6
6
6, 1
6, 2
6, 3
6, 4
6, 5
6,6
Sample Space - what notice about total outcomes?
White
Red
1
2
3
4
5
6
1
1, 1
1, 2
1, 3
1, 4
1, 5
1, 6
2
2, 1
2, 2
2, 3
2, 4
2,5
2, 6
3
3, 1
3, 2
3, 3
3, 4
3, 5
3, 6
4
4, 1
4, 2
4, 3
4, 4
4, 5
4,6
5
5, 1
5, 2
5, 3
5, 4
5, 5
5,6
6
6, 1
6, 2
6, 3
6, 4
6, 5
6,6
Sample Space
6 x 6 = 36 total outcomes
White
Red
1
2
3
4
5
6
1
1, 1
1, 2
1, 3
1, 4
1, 5
1, 6
2
2, 1
2, 2
2, 3
2, 4
2,5
2, 6
3
3, 1
3, 2
3, 3
3, 4
3, 5
3, 6
4
4, 1
4, 2
4, 3
4, 4
4, 5
4,6
5
5, 1
5, 2
5, 3
5, 4
5, 5
5,6
6
6, 1
6, 2
6, 3
6, 4
6, 5
6,6
Sample Space for flipping a coin one time
Using Set notation
S = {H, T}
S sample space
{ all possible outcomes }
Event - subset of the sample space
Using Set notation
S = {H, T}
S sample space
{ all possible outcomes }
Event B is the event of flipping a head
with a coin
B = {H}
Probability of an event
S = {H, T}
B = {H}
P(B) = ½
Probability of an event
S = {H, T} Possible outcomes 2
B = {H} favorable outcomes 1
P(B) = ½
P(event) = number of favorable outcomes
# of total possible outcomes
P(Sum of 7) =
White
Red
1
2
3
4
5
6
1
1, 1
1, 2
1, 3
1, 4
1, 5
1, 6
2
2, 1
2, 2
2, 3
2, 4
2,5
2, 6
3
3, 1
3, 2
3, 3
3, 4
3, 5
3, 6
4
4, 1
4, 2
4, 3
4, 4
4, 5
4,6
5
5, 1
5, 2
5, 3
5, 4
5, 5
5,6
6
6, 1
6, 2
6, 3
6, 4
6, 5
6,6
P(Sum of 7) = 6/ 36 = ⅙
White
Red
1
2
3
4
5
6
1
1, 1
1, 2
1, 3
1, 4
1, 5
1, 6
2
2, 1
2, 2
2, 3
2, 4
2,5
2, 6
3
3, 1
3, 2
3, 3
3, 4
3, 5
3, 6
4
4, 1
4, 2
4, 3
4, 4
4, 5
4,6
5
5, 1
5, 2
5, 3
5, 4
5, 5
5,6
6
6, 1
6, 2
6, 3
6, 4
6, 5
6,6
P(Sum of 7) = 6/ 36 = ⅙
One in six chance of rolling a
sum of seven when we roll two
dice.
P(sum of 7) = 0.16 or 17%
Probability Review Questions Homework Lesson 1
1. A box contains 4 opals, 5 garnets, and 6
pearls.
List the sample space
Below.
Sample space for which
event?
Probability Review Questions Homework Lesson 1
1. A box contains 4 opals, 5 garnets, and 6
pearls.
List the sample space
Below for selecting one
Jewel at random.
Probability Review Questions Homework Lesson 1
1. A box contains 4 opals, 5 garnets, and 6
pearls.
List the sample space below for selecting one
jewel at random.
S={
Probability Review Questions
1. A box contains 4 opals, 5 garnets, and 6
pearls.
List the sample space below for selecting one
jewel at random.
S = {O, O, O, O,
Probability Review Questions
1. A box contains 4 opals, 5 garnets, and 6
pearls.
List the sample space below for selecting one
jewel at random.
S = {O, O, O, O, G, G, G, G, G, P,P,P,P,P}
Probability Review Questions
2.A box contains 4 opals, 5 garnets, and 6 pearls.
A jewel is selected at random.
List each event as a set.
a) An Opal = {O, O, O, O}
b) Either an opal or a pearl
={O,O,O,O,P,P,P,P,P,P}
c) A garnet = { G, G, G, G, G}
d) Not a garnet = same as b).