Monday, January 17, 2011
Lesson 74- Using the Binomial Theorem to Calculate Probabilities
Investigate:
P (exactly x correct guesses in 6 rolls) = 6Cx (1/6)^x (5/6) ^6-x
note: The # of pathways with x C's times the probability of a specific pathway with x C's
Example 1:
P(exactly 6 correct guesses in 6 rolls) = 6C6(1/6)^6= 0.0000214
Note: calculations like these that use the word exactly you may use the function call Binompdf, put in the # of plays, chances, guesses
Binompdf (6,1/6,6)= 0.0000214
If the question says at least or at most in the question you must use Binomcdf:
P(at most to heads in 12 flips)= binomcdf( 12, 0.5, 2)= 0.0192
Binomial Experiment:
A binomial experiment consists of:
- a fixed number of independent and identical trials
- only two possible outcomes ( success or failure) per trail
Binomial Probability:
The probability of getting x successes and n-x failures in n trials is
P(x)= nCx p^x q^n-x P=success Q= failure
P (exactly x correct guesses in 6 rolls) = 6Cx (1/6)^x (5/6) ^6-x
note: The # of pathways with x C's times the probability of a specific pathway with x C's
Example 1:
P(exactly 6 correct guesses in 6 rolls) = 6C6(1/6)^6= 0.0000214
Note: calculations like these that use the word exactly you may use the function call Binompdf, put in the # of plays, chances, guesses
Binompdf (6,1/6,6)= 0.0000214
If the question says at least or at most in the question you must use Binomcdf:
P(at most to heads in 12 flips)= binomcdf( 12, 0.5, 2)= 0.0192
Binomial Experiment:
A binomial experiment consists of:
- a fixed number of independent and identical trials
- only two possible outcomes ( success or failure) per trail
Binomial Probability:
The probability of getting x successes and n-x failures in n trials is
P(x)= nCx p^x q^n-x P=success Q= failure
Lesson 73- Using Permutations and Combinations to Calculate Probabilities
Using Combination's:
Sample space is list of all possible ways (equally likely) to select 2 cards from 52 cards:
Number of unordered ways to select 2 cards from 52 cards: 52C2 = 1326
Number of unordered ways to select 2 aces: 4C2 = 6
P(both cards aces) = 4C2 = 1/ 221
52C2
Using Permutations:
Number of ordered ways to select 2 cards from 52 cards: 52P2
Number of ways to select 2 aces: 4P2
P(both cards aces)= 4P2 = 1/221
52P2
Example 1: Three prizes are awarded in a raffle. One hundred people hold a ticket each
a) what is the probability that Alice, Ben, and Concetta win first, second, and third prizes respectively?
* Order Does Matter*
1 = 1.03 x 10^-6
100P3
b) What is the probability that Alice, Ben and Concetta are the 3 prize winners (though not necessarilyin that order)?
* Order Does Not Matter*
1 = 6.18 x 10 ^ -6
100C3
* Order Does Matter* = Permutation * Order Does Not Matter*= Combination
Sample space is list of all possible ways (equally likely) to select 2 cards from 52 cards:
Number of unordered ways to select 2 cards from 52 cards: 52C2 = 1326
Number of unordered ways to select 2 aces: 4C2 = 6
P(both cards aces) = 4C2 = 1/ 221
52C2
Using Permutations:
Number of ordered ways to select 2 cards from 52 cards: 52P2
Number of ways to select 2 aces: 4P2
P(both cards aces)= 4P2 = 1/221
52P2
Example 1: Three prizes are awarded in a raffle. One hundred people hold a ticket each
a) what is the probability that Alice, Ben, and Concetta win first, second, and third prizes respectively?
* Order Does Matter*
1 = 1.03 x 10^-6
100P3
b) What is the probability that Alice, Ben and Concetta are the 3 prize winners (though not necessarilyin that order)?
* Order Does Not Matter*
1 = 6.18 x 10 ^ -6
100C3
* Order Does Matter* = Permutation * Order Does Not Matter*= Combination
Lesson 72- Problems Involving Conditional Probability
Conditional Probability:
Probability of given A in the event of B
Formula: P(a | b)
Example 1:
A card is drawn from a shuffled deck for 52 cards, and not replaced. Then a second card is drawn.
a) What is the probability that the second card drawn is a king?
P( k and k ) + P( K(with line on top) and K)
4/52 x 3/51 + 48/52 x 4/51 = 1/13
* use a tree diagram
Probability of given A in the event of B
Formula: P(a | b)
Example 1:
A card is drawn from a shuffled deck for 52 cards, and not replaced. Then a second card is drawn.
a) What is the probability that the second card drawn is a king?
P( k and k ) + P( K(with line on top) and K)
4/52 x 3/51 + 48/52 x 4/51 = 1/13
* use a tree diagram
Lesson 71- The event a and b
When do you multiply probabilities?
Situation 1: Drawing two balls from the pot, without replacement.
A: the first ball is white Find: P(a) = 2/3 P(B given that the first ball drawn is white)
B: the second ball is black = P( b | a) = 1/2
* note: using a tree diagram for this question will be useful!
Multiplication Law:
General Case: P( a and b) = P(a) x P(B|A)
Equivalently: p(b | a) = P (A and B)
P(A)
Special Sub-case: Events A and B are independent if p(B | A) = P(b)
For independent events: P(a and b) = p(a) x p(b)
Two events that are not independent are called dependent
Example:
Three cards are drawn with replacement from a shuffled deck of 52 cards
What is the probability that all three are spades?
P (S and S and S) = 13/52 x 12/51 x 11/50 = 11/850
Situation 1: Drawing two balls from the pot, without replacement.
A: the first ball is white Find: P(a) = 2/3 P(B given that the first ball drawn is white)
B: the second ball is black = P( b | a) = 1/2
* note: using a tree diagram for this question will be useful!
Multiplication Law:
General Case: P( a and b) = P(a) x P(B|A)
Equivalently: p(b | a) = P (A and B)
P(A)
Special Sub-case: Events A and B are independent if p(B | A) = P(b)
For independent events: P(a and b) = p(a) x p(b)
Two events that are not independent are called dependent
Example:
Three cards are drawn with replacement from a shuffled deck of 52 cards
What is the probability that all three are spades?
P (S and S and S) = 13/52 x 12/51 x 11/50 = 11/850
Sunday, January 16, 2011
Lesson 70- The event A or B
Addition Law:
a) General Case:
P (A or B) = P(A) + P(B) - P(A and B) * Overlap
b) Special Sub-case:
If A and B are mutually exclusive then
P(A and B) = P(A) + P(B)
Example 1:
One card is randomly drawn from deck of 52 cards
Define the following events:
S: The card is a spade
R: The card is red
F: The card is face card
Which of these three are mutually exclusive?
S, R (do not overlap)
Determine the following probabilities directly by counting outcomes:
P(s) = 13/52 P(r)= 26/52 P(s and r)= O P(s or r)= 13/52 + 26/52 = 39/52
P(s) = 13/52 P(f) = 12/52 P(s and f) = 3/52 P(s or f) = 13/52 + 12/52 - 3/52 = 22/52
*note: using a Venn-diagram with these problems will make things a lot easier, as the info. is right in front of you.
a) General Case:
P (A or B) = P(A) + P(B) - P(A and B) * Overlap
b) Special Sub-case:
If A and B are mutually exclusive then
P(A and B) = P(A) + P(B)
Example 1:
One card is randomly drawn from deck of 52 cards
Define the following events:
S: The card is a spade
R: The card is red
F: The card is face card
Which of these three are mutually exclusive?
S, R (do not overlap)
Determine the following probabilities directly by counting outcomes:
P(s) = 13/52 P(r)= 26/52 P(s and r)= O P(s or r)= 13/52 + 26/52 = 39/52
P(s) = 13/52 P(f) = 12/52 P(s and f) = 3/52 P(s or f) = 13/52 + 12/52 - 3/52 = 22/52
*note: using a Venn-diagram with these problems will make things a lot easier, as the info. is right in front of you.
Lesson 67 - Related Events
A list of all possible (non-overlapping) outcomes of an experiment is called a sample space
Example 1:
One bill is randomly selected from the pot
a) Sample space:
$0, $5, $10
b) What is the probability that the $5 bill is selected?
P(5)= 1/3
Related Events:
Complement- Opposite
The event that A does not occur is the complement of event A, and is denoted A(with line on top)
*note: P(A) + P(A)*with line on top = 1
A and B
The event that both A and B occur is denoted by A and B
A or B
The event that A or B occurs (or both) is denoted by A or B
Example 2:
If each of the 13 outcomes in the sample space is equally likely, find:
Sample Space:
P(A)= 4/13 = 30% P(B)= 5/13
P(not A)= 9/13= 70% P(notB)= 8/13
P(A and B)= 1/13 P(A or B)= 4/13 + 5/13 - 1/13 = 8 /13
Example 1:
One bill is randomly selected from the pot
a) Sample space:
$0, $5, $10
b) What is the probability that the $5 bill is selected?
P(5)= 1/3
Related Events:
Complement- Opposite
The event that A does not occur is the complement of event A, and is denoted A(with line on top)
*note: P(A) + P(A)*with line on top = 1
A and B
The event that both A and B occur is denoted by A and B
A or B
The event that A or B occurs (or both) is denoted by A or B
Example 2:
If each of the 13 outcomes in the sample space is equally likely, find:
Sample Space:
P(not A)= 9/13= 70% P(notB)= 8/13
P(A and B)= 1/13 P(A or B)= 4/13 + 5/13 - 1/13 = 8 /13
Lesson 66- Experimental and Theoretical Probability
Theoretical Probability:
Example 1:
a)What is the theoretical probability of getting a 5 when rolling a die?
Outcomes: 1,2,3,4,5,6 < Total Outcomes (all)
Favorable outcome: P(5) = 1/6 < Theoretical
b) Simulate rolling a die 300 times and recording the number of times a 5 appears.
For this calculation you have to use the Randbin
Randbin (1, 1/6,300) = 56/300
Example 2:
You can also use a tree diagram to calculate your answer.
What is the probability that in a family of 3 there are exactly 2 girls?
Tree Diagram:
Determine the number of branches that contain 2 girls?
P(2G)= 3/8 = 37.5%
* find the branches that show only two girls
Probability:
If an experiment has n equally likely outcomes of witch r outcomes are favourable to event A, then the probability of event A is:
P(a) = r/n
Example 1:
a)What is the theoretical probability of getting a 5 when rolling a die?
Outcomes: 1,2,3,4,5,6 < Total Outcomes (all)
Favorable outcome: P(5) = 1/6 < Theoretical
b) Simulate rolling a die 300 times and recording the number of times a 5 appears.
For this calculation you have to use the Randbin
Randbin (1, 1/6,300) = 56/300
Example 2:
You can also use a tree diagram to calculate your answer.
What is the probability that in a family of 3 there are exactly 2 girls?
Tree Diagram:
Determine the number of branches that contain 2 girls?
P(2G)= 3/8 = 37.5%
* find the branches that show only two girls
Probability:
If an experiment has n equally likely outcomes of witch r outcomes are favourable to event A, then the probability of event A is:
P(a) = r/n
Saturday, January 15, 2011
Lesson 62- The Binomial Theorem
Expand and simplify each of the following powers of the binomial a + b:
(a + b)^1 = 1a + b^1
(a + b)^2 = 1a^2 + 2ab + 1b^2
Powers of a: decreasing
Powers of b: increasing
Example 1: (2x + 1)^4
= 4C0 (2x)^4(1)^0 + 4C1 (2x)^3 (1)^1 + 4C2 (2x)^2 (1)^2 + 4C3 (2x)^1(1)^3 + 4C4 (2x)^0 (1)^4
= 16x^4 + 32x^3 + 24x^2 + 8x + 1
Binomial Throrem using combinations:
( a + b)^n = nC0 (a)^n (b)^0
(a + b)^1 = 1a + b^1
(a + b)^2 = 1a^2 + 2ab + 1b^2
Powers of a: decreasing
Powers of b: increasing
Example 1: (2x + 1)^4
= 4C0 (2x)^4(1)^0 + 4C1 (2x)^3 (1)^1 + 4C2 (2x)^2 (1)^2 + 4C3 (2x)^1(1)^3 + 4C4 (2x)^0 (1)^4
= 16x^4 + 32x^3 + 24x^2 + 8x + 1
Binomial Throrem using combinations:
( a + b)^n = nC0 (a)^n (b)^0
Lesson 61- Pascal's Triangle
General Triangle:
The Symmetrical Pattern:
5C1 = 5C4
Justification: by symmetry choosing 1 out is choosing 4 in
Generalization: nCr = nCn-1
The Recursive Pattern:
5C3 = 4C2 + 4C3
Justification: add above 2 terms
Generalization: nCr = n-1Cr-1 + n-1Cr
5C1 = 5C4
Justification: by symmetry choosing 1 out is choosing 4 in
Generalization: nCr = nCn-1
The Recursive Pattern:
5C3 = 4C2 + 4C3
Justification: add above 2 terms
Generalization: nCr = n-1Cr-1 + n-1Cr
Lesson 60- Combinations
Combination's have no order and Permutations do have an order.
Example 1:
If 5 sprinters compete in a race, how many different ways can the medals for first, second and third place be awarded?
5 x 4 x 3 = 5p3 = 60 genine
This is an example of a permutation of 5 objects taken 3 at a time
If 5 sprinters compete in a race and the fastest 3 qualify for the relay team, how many different relay teams can be formed?
YYY NN Yes: succeed No: fail 5!/ 3! 2! = 120/ 12 = 10
This is an example of a combination of 5 objects taken 3 at a time
Combinations:
An unordered arrangement if distinct objects is called a combination
The number if combination's of n distinct objects taken r at a time is nCr
Example 1:
If 5 sprinters compete in a race, how many different ways can the medals for first, second and third place be awarded?
5 x 4 x 3 = 5p3 = 60 genine
This is an example of a permutation of 5 objects taken 3 at a time
If 5 sprinters compete in a race and the fastest 3 qualify for the relay team, how many different relay teams can be formed?
YYY NN Yes: succeed No: fail 5!/ 3! 2! = 120/ 12 = 10
This is an example of a combination of 5 objects taken 3 at a time
Combinations:
An unordered arrangement if distinct objects is called a combination
The number if combination's of n distinct objects taken r at a time is nCr
Lesson 59 - Permutations involving identical objects
What happens if two of the three letters are the same?
Take the number of total letters and divide it by the number of repeated letters
For example the word OGOPOGO:
There are 7 letters all together, and two letters which have repeats.
the equation for the permutations is 7!/ 4! 2! 4 - o and 2- g
which equals 105
The number of permutations of n objects of which there are a objects alike of one kind, b alike of another kind, c alike of another kind, and so on. N!
A!B!C!
Example 2:
A true-false test has 7 questions. How many answer keys are possible if 3 answers are T and 4 answers are F?
TTT FFFF 7! = 5040 = 35
3! 4! 144
Example 3 :
On the following grid, how many different paths can A take to get to B, assuming one can only travel east and south?
Take the number of total letters and divide it by the number of repeated letters
For example the word OGOPOGO:
There are 7 letters all together, and two letters which have repeats.
the equation for the permutations is 7!/ 4! 2! 4 - o and 2- g
which equals 105
The number of permutations of n objects of which there are a objects alike of one kind, b alike of another kind, c alike of another kind, and so on. N!
A!B!C!
Example 2:
A true-false test has 7 questions. How many answer keys are possible if 3 answers are T and 4 answers are F?
TTT FFFF 7! = 5040 = 35
3! 4! 144
Example 3 :
On the following grid, how many different paths can A take to get to B, assuming one can only travel east and south?
Thursday, January 13, 2011
Lesson 58 - Permutations Involving different objects
Investigate:
Two letters, A and B, can be written in two different orders, AB and BA, These are permutations or A and B.
a) List all of the permutations of 3 letters A, B , and C.
ABC, ACB, BAC, BCA, CAB, CBA
3 x 2 x 1 = 6
How many permutations are there?
6
b) List all of the permutations of 4 letters A, B, C, and D
ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BCAD, BCDA, BADC, BACD, BDAC, BDCA
How many permutations are there?
4 x 3 x 2 x 1= 24
c) Predict the number of permutations of 5 letters A, B, C, D, and E
5 x 4 x 3 x 2 x 1 = 5!
! : 5 4 3 2 1
Example 1:
When you press the "scramble" button on a cd player it plays a permutation of the songs on the CD. If the cd has 5 songs on it, how many permutations of the songs possible?
5! or 5p5 = 120 5p5: 5 songs , p: means how many your picking, you are picking 5
Factorial Notation:
1! = 1
2!= 2 x 1
3!= 3 x 2 x 1
n!= n (n-1)(n-2)(n-3) ......1
An ordered arrangement of distinct objects is called a permutation
The number of permutations of n distinct objects is n!
The number of permutations of n distinct objects taken r at a time is npr
Example 2:
np2= n!/ (n-2)!
n(n-1)(n-2)!/ (n-2)!
= n(n-1)
Two letters, A and B, can be written in two different orders, AB and BA, These are permutations or A and B.
a) List all of the permutations of 3 letters A, B , and C.
ABC, ACB, BAC, BCA, CAB, CBA
3 x 2 x 1 = 6
How many permutations are there?
6
b) List all of the permutations of 4 letters A, B, C, and D
ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BCAD, BCDA, BADC, BACD, BDAC, BDCA
How many permutations are there?
4 x 3 x 2 x 1= 24
c) Predict the number of permutations of 5 letters A, B, C, D, and E
5 x 4 x 3 x 2 x 1 = 5!
! : 5 4 3 2 1
Example 1:
When you press the "scramble" button on a cd player it plays a permutation of the songs on the CD. If the cd has 5 songs on it, how many permutations of the songs possible?
5! or 5p5 = 120 5p5: 5 songs , p: means how many your picking, you are picking 5
Factorial Notation:
1! = 1
2!= 2 x 1
3!= 3 x 2 x 1
n!= n (n-1)(n-2)(n-3) ......1
An ordered arrangement of distinct objects is called a permutation
The number of permutations of n distinct objects is n!
The number of permutations of n distinct objects taken r at a time is npr
Example 2:
np2= n!/ (n-2)!
n(n-1)
= n(n-1)
lesson 57 - The fundamental counting principles
Fundamental Counting Principle:
If one item can be selected in m ways, and for each way a second item can be selected in n ways then the two ways can be selected in mn ways.
Example 1:
How many different two digit numbers are there?
9 x 10 = 90
^ ^ 0,1,2,3,4,5,6,7,8,9 < these are your choices
1,2,3,4,5,6,7,8,9
Example 2:
A multiple-choice test has 7 questions, with 4 possible answers for each question. Suppose students answer each question by guessing randomly.
a) How many possible answers are there for each question?
4
b) How many different patterns are possible for the answers to the 7 questions on the test?
4 x 4 x 4 x 4 x 4 x 4 x 4 = 16384
c) What is the probability that all 7 questions will be answered correctly?
1/ 16384
If one item can be selected in m ways, and for each way a second item can be selected in n ways then the two ways can be selected in mn ways.
Example 1:
How many different two digit numbers are there?
9 x 10 = 90
^ ^ 0,1,2,3,4,5,6,7,8,9 < these are your choices
1,2,3,4,5,6,7,8,9
Example 2:
A multiple-choice test has 7 questions, with 4 possible answers for each question. Suppose students answer each question by guessing randomly.
a) How many possible answers are there for each question?
4
b) How many different patterns are possible for the answers to the 7 questions on the test?
4 x 4 x 4 x 4 x 4 x 4 x 4 = 16384
c) What is the probability that all 7 questions will be answered correctly?
1/ 16384
Wednesday, January 12, 2011
Lesson 50 - 56
Lesson 50-51 : Identity package
Lesson 52- 56: Practice Tests, Corrections and Chapter Test
Lesson 52- 56: Practice Tests, Corrections and Chapter Test
Lesson 49 - Identities and double angles
Identities:
Sin( x + y) = sinxcosy + cosxsiny Cos ( x + y)= cosxcosy - sinxsiny
Sin( x - y) = sinxcosy - cosxsiny Cos ( x - y)= cosxcosy + sinxsiny
Double Angles:
Sin(2x) = Sin (x + x)
= sinxcosx + cosxsinx (like terms)
= 2sinxcosx > same as Sin(2x)
cos(2x) = Cos ( x + x)
= cosxcosx - sinxsinx cos^2x - sin^2
= cos^2x -sin^2x = cos^2x - (1 - cos^2x)
>> = 1 - 2sin^2x = cos^2x - 1 + cos^2x
>> = 2cos^2x - 1
Sin( x + y) = sinxcosy + cosxsiny Cos ( x + y)= cosxcosy - sinxsiny
Sin( x - y) = sinxcosy - cosxsiny Cos ( x - y)= cosxcosy + sinxsiny
Double Angles:
Sin(2x) = Sin (x + x)
= sinxcosx + cosxsinx (like terms)
= 2sinxcosx > same as Sin(2x)
cos(2x) = Cos ( x + x)
= cosxcosx - sinxsinx cos^2x - sin^2
= cos^2x -sin^2x = cos^2x - (1 - cos^2x)
>> = 1 - 2sin^2x = cos^2x - 1 + cos^2x
>> = 2cos^2x - 1
Lesson 48 - Conjugate Rule
Rules:
1) Identity/manipulation
2) sin/cos
3) complex fractions
4) Factor + Cancel
5) combine fractions
6) Conjugate Rule
Prove: sinx = 1 + cosx
1 - cosx sinx
LS = RS
sinx (1 + cosx) < multiply by the conjugate
(1-cosx) (1+ cosx)
sinx (1+cosx)
1- cos^2x
sinx (1+cosx)
sin^2x
1+ cosx = 1+ cosx
sinx sinx
Is Cos(x + y) = cosx + cosy
Cos (12 + 37) = Cos(12) + Cos(37) < Verify Numerically
Cos 49 = 0.97 + 0.80
0.66 = 1.78
NOT EQUAL!!!
You can proof graphically, algebraically, or numerically.
1) Identity/manipulation
2) sin/cos
3) complex fractions
4) Factor + Cancel
5) combine fractions
6) Conjugate Rule
Prove: sinx = 1 + cosx
1 - cosx sinx
LS = RS
sinx (1 + cosx) < multiply by the conjugate
(1-cosx) (1+ cosx)
sinx (1+cosx)
1- cos^2x
sin
1+ cosx = 1+ cosx
sinx sinx
Is Cos(x + y) = cosx + cosy
Cos (12 + 37) = Cos(12) + Cos(37) < Verify Numerically
Cos 49 = 0.97 + 0.80
NOT EQUAL!!!
You can proof graphically, algebraically, or numerically.
Wednesday, January 5, 2011
Lesson 42 - Proofs
Prove: tanx( tanx + cotx) = sec^2 x
Left side = Right side
tan^2x + tanx(cotx)
tan^2 +tanx(1/tanx)
tan^2x + 1
sec^2x sec^2x
LS = RS
Steps:
1) Manipulation
2) sin/cos
3) Fraction
4) combine fraction
5) fracor and cancel
6) LS = RS
Prove:
cscx - sinx = cosx(cotx)
LS = RS
1/sinx - sinx (sinx) cosx( cosx/sinx)
1 (sinx) cos^2x
1 - sin^2x sinx
sinx
cos^2x
sinx LS=RS
Left side = Right side
tan^2x + tanx(cotx)
tan^2 +
tan^2x + 1
sec^2x sec^2x
LS = RS
Steps:
1) Manipulation
2) sin/cos
3) Fraction
4) combine fraction
5) fracor and cancel
6) LS = RS
Prove:
cscx - sinx = cosx(cotx)
LS = RS
1/sinx - sinx (sinx) cosx( cosx/sinx)
1 (sinx) cos^2x
1 - sin^2x sinx
sinx
cos^2x
sinx LS=RS
Lesson 41 - Identities
Identities: Steps:
Cos^2x + sin^2x = 1 1) Manipulation
1+ tan^2x = sec^2x 2) Put everything in terms of sin/cos
cotx + 1= csc^2 x 3) complex fraction
Simplify:
csc^2x - cot^2x cosx + sinx
sinx(1/cosx) secx cscx
1
= tanx (cosx) cosx + sinx (sinx)
(cosx)1/cosx 1/sinx (sinx)
cos^2x + sin^2x
= 1
Cos^2x + sin^2x = 1 1) Manipulation
1+ tan^2x = sec^2x 2) Put everything in terms of sin/cos
cotx + 1= csc^2 x 3) complex fraction
Simplify:
csc^2x - cot^2x cosx + sinx
sinx(1/cosx) secx cscx
1
= tanx (cosx) cosx + sinx (sinx)
(
cos^2x + sin^2x
= 1
Subscribe to:
Posts (Atom)