Transcript Probability

Algebra-2
Counting and Probability
Quiz 10-1, 10-2
n!
n Cr 
r!(n  r )!
n Pr 
n!
(n  r )!
1. Which of these are an example of a “descrete” set of data?
a. x such that : x  5, 7, 9 , 11
b. x such that : 5  x  11
2. Make a “tree diagram” showing all the ways the letters
‘x’, ‘y’, and ‘z’ can be arranged in order.
3. You are paying for groceries at the store. You have the
following bills: $100, $50, $20, $10, $5, $2, and $1.
What are number of different sums of money that you can pull
out of your wallet if you pull out 3 bills without looking?
Counting
How many ways can you arrange the letters
‘x’, ‘y’, and ‘z’ in order?
3
2
1
___ ___ ___
Why isn’t it
3
3
3
?
___
___
___
You can’t re-use a letter in this situation.
Can you think of a situation where you could
re-use a number or a letter?
Phone numbers
Social security numbers
License plates
Go to demo
Vocabulary
“arranging without replacement: when you use an item in
the arrangement, it is “used up” and can’t be used again.
Think of arranging people in a line. Once a person is in the
front of the line, he cannot also be in the back of the line
at the same time.
“arranging with replacement: when an item is used in one
position in an arrangement, it can be used again in another
position in the arrangement.
Think of arranging numbers and Letters on a license plate:
the previous number or letter can be used again.
Effect on Muliplication Principle of counting
(Product of the # of options for each step)
arranging without replacement:
Arranging 3 people in a line.
3* 2 *1
Factorial
arranging with replacement:
Arranging 3 numbers on a licence plate.
10 *10 *10
Your turn:
Which is it (with or without replacement) for:
1. Assigning 3 committee members to the positions of:
“Pres”, “Vice-Pres”, and “Secretary”
2. The total number of social security numbers with 9
digits.
Using the Multiplication Principle
If a license plate has three letters followed by three
numerical digits. Find the number of different
license plates that could be formed if there is no
restriction on the letters or digits that can be used.
L L L # # #
How many possibilities for the 1st position (letter)?
26
Using the Multiplication Principle
26 * 26
L
L
L
#
#
#
How many possibilities for the 2nd position?
Using the Multiplication Principle
26 * 26 * 26
L
L
L
#
#
#
How many possibilities for the 3rd position?
Using the Multiplication Principle
26 * 26 * 26 * 10
L
L
L
#
#
#
How many possibilities for the 4th position (number)?
Using the Multiplication Principle
26 * 26 * 26 * 10 * 10
L
L
L
#
#
#
How many possibilities for the 5th position?
Using the Multiplication Principle
26 * 26 * 26 * 10 * 10 * 10
L
L
L
#
#
#
How many possibilities for the 6th position?
Total number of distinct license plates = 676,000
Your Turn:
3. How many distinct license plates can be made using
6 digits (numerals 0 – 9)? (# # # # # #)
1,000,000
4. How many distinct license plates can be made using
2 digits (numerals 0 – 9) and 4 letters ( A – Z) ?
(# # L L L L)
45,697,600
Your Turn:
5. Count the number of different 8-letter “words” (groups of 8
letters) that can be formed using the letters in the word
COMPUTER.
Each permutation of the 8 letters forms a different
word. There are 8! = 40,320 such permutations.
What if two of the letters are the same?
Count the number of different 4-letter “words” that can be
formed using the letters in the word
“WAAG”.
WAAG”.
Let “A” be the 1st A.
Let “A” be the 2nd A.
What’s the difference between AAWG and AAWG?
There’s no difference!! They are not distinguishable
from each other. So we really have “double counted”
a bunch of words.
What if two of the letters are the same?
Count the number of different 4-letter “words” that can be
formed using the letters in the word
“WAAG”.
AAWG (AAWG) is one example of double counting.
AWAG (AWAG) is another example of double counting.
To remove the “double counting” we must
divide out the number of possible ways to
permutate A and A
We must divide by 2!.
What if three of the letters are the same?
Count the number of different 5-letter “words” that can be
formed using the letters in the word
“WAAAG”.
“WAAAG”.
AAAWG AAAWG AAAWG AAAWG AAAWG AAAWG
AAAWG AAAWG AAAWG AAAWG AAAWG AAAWG
These are all examples of the same word and have been
“double counted”.
To remove the “double counting” we must
divide out the number of ways to permutate A, A and A
We must divide by 3!
Distinguishable Permutations
If a set 12 items to be permutated has 3 objects of one kind,
and 4 objects of another kind, and 5 objects of another kind,
then the number of distinguishable ways to arrange the 12
items is:
12!

3!*4!*5!
 27,720
We must “divide out” the permutations of the same object
that result in indistinguishable arrangements.
Distinguishable Permutations
In general, we find the number of distinguishable permutations
when using some elements that are indistinguishable
as follows:
If a set N items to be permutated has A objects of one kind,
and B objects of another kind, and C objects of another kind,
and A + B + C = N then the number of distinguishable ways
to arrange the N items is:
N!

A!*B!*C!
Your Turn:
6. You have the following bills in your wallet:
three $20’s, four $10’s, five $5’s, and six $1’s
What is the number of distinct ways you could pay
out the bills one at a time?
18!
 514,594,080

3!*4!*5!*6!
Counting
How many 5 card hands are there with all face cards (king,
queen, jack).
This tells you the hands all have 5 face cards. So how
many arrangements are there when taking 12 cards and
picking 5 ?
Which is it?
Permutation: (different order  counted separately)
Combination: (different order  not counted separately)
12
C5  792
Your Turn:
7. How many 5 card hands are there with no face
cards?
 40 C5
 658,008
Counting
Sometimes there are more than one condition that must be met.
How many 5 card hands have all 5 cards the same suite
(hearts, diamonds, spades, clubs).
Combination: (different order  not counted separately)
1st we must pick the suite:
4
C1
2nd we must pick the 5 cards from that suite: 13 C5
By the multipication principle: total number hands is:
4
C1 *13 C5
 5148
Counting
How many 5 card hands have exactly 2 aces ?
Combination: (different order  not counted separately)
1st we must pick the 2 aces:
4
C2
2nd we must pick the other 3 cards: (if the hand has
exactly 2 aces, then we must not include the other two
aces as possible picks)
48
C3
By the multipication principle: total number of hands is:
4
C2 *48 C3
 103776
Your Turn:
8. How many 5 card hands are there with two fives
and two sixes?
4 C2 *4 C2 *44 C1  1584
Hint: (1) pick the 2 fives, (2) pick the 2 sixes, (3)
Pick the last card. Use the multiplication rule.
Probability
“What’s the chance of something
happening?”
“There is a 100% chance it will rain today.”
“There is less than a 5% chance you will be picked.
Your turn:
9. Can probability be equal to 50%?
10. Can there be a (– 20)% chance something will
happen?
11. What is the smallest number that a probability
can be?
12. What is the largest number that a probability
can be?
Probability
When discussing probability, you can use either
“%”, fraction, or the decimal equivalent.
 “There is a 40% chance of thunderstorms today.”
In mathematics, we convert % to the decimal
equivalent or leave it in fraction form.
 “The probability of rain today is 0.4.”
Theoretical Probability
The probability of an event occurring:
# of ways to achieve that event
P(event) 
total # of possible outcomes
There are 4 different colored marbles in a bag (red,
blue, green and yellow). What is the probability of
pulling out a red one on the first try?
# of red marbles in the bag
P(red ) 
total # of marbles in the bag
1
P (red )   0.25
4
Examples
# of ways to achieve that event
P(event) 
total # of possible outcomes
The probability of rolling a ‘5’ using one die.
1
P (5)   0.166667
6
The probability of drawing a “king” from a deck of cards.
4
1
P (king ) 

 0.0769
52 13
# of ways to achieve that event
P(event) 
total # of possible outcomes
The challenge you have is counting the ways to
achieve the event (sometimes called successful
events) and then counting the total possible outcomes.
What is the probility of pulling an A, followed
by a B, and then a C out of a bag with the
letters ‘A’, ‘B’, and ‘C’ in it ?
# of ways to draw A, B, then C
P( A, B, C ) 
total # of ways to draw the 3 letters out of the bag
1
P( A, B, C ) 
3 permutate 3
1

3!
1
  0.1 6
6
Your Turn:
# of ways to achieve that event
P(event) 
total # of possible outcomes
13. What is the probability of picking the correct number
when someone asks you to pick a number from 1 to 10.
14. There are 2 red marbles and 3 green ones in a bag.
What is the probability of picking out a red marble on
the first try?
Probability only works if the events are completely random.
Picking a committee using numbers out of a hat or a similar
random method of picking them is the only way that
probability will work.
Your Turn:
# of ways to achieve that event
P(event) 
total # of possible outcomes
15. 10 people are trying to be selected for a 3 person
committee. You don’t know any of the people. What is the
probality of you guessing who
will be on the committee?
16. What is the probability of having a 5 card hand with
a single pair of kings in it?
Geometric Probability: ratio of areas
Assumming that at an arrow randomly hits anywhere
in the four square area, what is the probability of
hitting in the #1 square?
Since all squares have
the same area,
and #1 is ¼ of the total
area  probability is ¼.
1
2
3
4
Geometric Probability: the area of each ring is given.
If an arrow will randomly hit anywhere inside of the red
circle, what is the probability of hitting the center blue circle?
light blue area
P(center ) 
total area
P(center ) 
25
25  75  125
25
P (center ) 
225
25
P(center) 
225
P(center)  0. 1
125
75
25
Geometric Probability
17.
What is the probability of hitting the pink ring?
18.
What is the probability of hitting either the pink or dark blue ring?
125
75
25
End here
Probability using combinations and permutations.
# of ways to achieve that event
P(event) 
total # of possible outcomes
At the Roy High School Talent show 7 musicians are scheduled to perform.
What is the probability that they will perform in alphabetical order of their
last names (nobody has the same last name) ?
There is only one order of performers that is in alphabetical order.
How many ways can you arrange 7 persons names in order?
Is this a permutation or combination?
1
P(alphabetic al order ) 
7 permutate 7
1
P (alphabetic al order ) 
7!
1

5040
 0.0002
Probability using combinations and permutations.
# of ways to achieve that event
P(event) 
total # of possible outcomes
At the Roy High School Talent show 7 musicians are scheduled to perform.
3 performers are girls and 4 are boys. What is the probability that all 3
girls will be first?
P(3 girls 1st ) 
# ways to arrange 3 girls in order
# of ways to arrange 3 (of 7) people in order
How many ways can you get the first 3 performers to be girls?
How many ways can you arrange 3 of 7 people in order?
Is this a permutation or combination?
1
3 choose 3
P(3 girls 1 ) 

7 choose 3
7 C3
st
1

35
 0.029
# of ways to achieve that event
P(event) 
total # of possible outcomes
Your Turn:
19.
At the Roy High School Talent show 7 musicians are scheduled to
perform. They are: Bill, Brad, Bob, and Brody (boys) and Kylee, Kaylee,
and Kyla (3 girls). What is the probability that 2 boys will be first?
P(2 boys 1st ) 
# ways to arrange 2 of 4 boys in order
# of ways to arrange 2 (of 7) people in order
4 choose 2
P(2 boys 1 ) 
7 choose 2
st
C2

7 C2
4
6

21
 0.29
Your Turn:
# of ways to achieve that event
P(event) 
total # of possible outcomes
20.
What is the probability of getting 4 aces in a randomly dealt hand of 4
cards?
# ways to get 4 aces out 4 cards
P(4 aces) 
# of ways to get 4 cards out of 52.
4 choose 4
P(r aces) 
52 choose 4
C4

52 C 4
4
1

270,725
 0.0000037
Your Turn:
# of ways to achieve that event
P(event) 
total # of possible outcomes
21. A lottery uses numbers 1 thru 46. 6 numbers are drawn
randomly. The order in which you choose the numbers
doesn’t matter. What is the probability of winning the lottery
if you buy one ticket (assume nobody else picks the winning
number) ?
# ways to get the 6 correct numbers
P (6 # ' s ) 
# of ways to pick 6 of 46 numbers
How many ways can you get the 6 out of 6 correct numbers?
How many ways can you pick 6 of 46 numbers?
Is picking 6 of 46 a permutation or a combination?
1
6 choose 6
P( win ) 

46 choose 6 46 C6
1

9,366,819
 0.00000011
Cards:
# of ways to achieve that event
P(event) 
total # of possible outcomes
What is the probability of getting 4 aces a randomly dealt hand of 5 cards?
P(4 aces) 
number of ways to have 4 aces out of 5 cards
total number of 5 card hands
C 4 *38 C1
P(4 aces) 
52 C5
4
38

2,598,960
 0.000015
# of ways to achieve that event
P(event) 
total # of possible outcomes
Your turn:
17.
What is the probability of getting 3 aces and 2 kings from randomly
dealt hand of 5 cards?
P(3 aces, 2 k' s) 
number of ways to have 3 aces and 2 k' s
total number of 5 card hands
C3 *4 C 2
P(3 aces, 2 k' s) 
52 C5
4
24

2,598,960
 0.000015