1.addition rule 2.multiplication rule 3.compliments 4.conditional probability 5.permutation...
DESCRIPTION
Multiplication Rule Finding the probability of more than one event. The word “AND” is always used when describing the situation. 1)P(rolling a 4 and then a 2) = 1/6 *1/6 = 2.8% 2) P(rolling 3 odd #’s) = 3/6*3/6*3/6 = 12.5% 1)P(rolling a 4 and then a 2) = 1/6 *1/6 = 2.8% 2) P(rolling 3 odd #’s) = 3/6*3/6*3/6 = 12.5%TRANSCRIPT
![Page 1: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/1.jpg)
Probability Topics
1. Addition Rule2. Multiplication Rule3. Compliments4. Conditional Probability5. Permutation6. Combinations7. Expected value8. Geometric Probabilities9. Binomial Probabilities
![Page 2: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/2.jpg)
Addition Rule for Non Mutually Exclusive Events
2
(A or B) = P(A) + P(B) – P(A and B)
One card is drawn from a standard deck of cards. What is the probability that it is red or an ace?
= P(Red) + P(Ace) – P(Both Red and Ace)= 26/52 + 4/52 – 2/52 = 28/52
![Page 3: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/3.jpg)
Multiplication Rule Finding the probability of more than one event.
The word “AND” is always used when describing the situation.
1) P(rolling a 4 and then a 2) = 1/6 *1/6 = 2.8%
2) P(rolling 3 odd #’s) = 3/6*3/6*3/6 = 12.5%
![Page 4: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/4.jpg)
4
Example 1 continued P(A1 AND A2) = P(A1)P(A2|A1)
P(A1) = 4/52
There are now 3 aces left in a 51-card pack
P(A2|A1) = 3/51
Overall: P(A1 AND A2) = (4/52) (3/51) = .0045
What’s the probability of pulling out two aces in a row from a deck of 52 cards?
![Page 5: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/5.jpg)
If A is an event within the sample space S of an activity or experiment, the complement of A (denoted A') consists of all outcomes in S that are not in A.
The complement of A is everything else in the problem that is NOT in A.
Compliment:P(A') = 1 - P(A)
![Page 6: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/6.jpg)
Conditional Probability
)()(
)|(BP
BAPBAP
and
measures the probability of an event given that another event has occurred
![Page 7: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/7.jpg)
1% of the population has disease X.
If someone has the disease and gets tested the test is positive every time.
If a healthy person gets tested for disease X they will get a false positive 10% of the time.
If the lab comes back positive what will be the probability the person actually has the disease?
![Page 8: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/8.jpg)
n = total number of itemsr = number chosen
)!(
!rn
nrpn
an arrangement of items in a particular order.
Permutations
![Page 9: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/9.jpg)
Permutations Examples
A combination lock will open when the right choice of three numbers (from 1 to 30) is selected. How many different lock combinations are possible assuming no number is repeated?
2436028*29*30)!330(!30
330
27!30! p
![Page 10: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/10.jpg)
Combinations
. 0 where nrrnr
nrCn
)!(!!
an arrangement of items in which order does not matter. There are always fewer combinations than permutations.
n = total number of itemsr = number chosen
![Page 11: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/11.jpg)
Combinations ExampleTo play a particular card game, each
player is dealt five cards from a standard deck of 52 cards. How many different hands are possible?
960,598,21*2*3*4*548*49*50*51*52)!552(!5
!52552
5!47!52! C
![Page 12: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/12.jpg)
a weighted average of all possible values where the weights are the probabilities of each outcome
1 1 2 2( ) ( ) ( ) ( ).n nE x x P x x P x x P x
Expected value
![Page 13: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/13.jpg)
Example: expected valueprobability distribution of ER arrivals x is the number of arrivals in one hour
X 10 11 12 13 14P(x) .4 .2 .2 .1 .1
The average expected number each hour
5
1
3.11)1(.14)1(.13)2(.12)2(.11)4(.10)(i
i xpx
![Page 14: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/14.jpg)
Geometric Distribution want to find the number of trials for the 1st success
p = probability of successq = 1 – p = probability of failureX = # of trials until first success occurs
p(x) = qx-1p
![Page 15: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/15.jpg)
15
Two Ways to use the Geometric Model
#1: the probability of getting your first success on the x trail
p(x) = qx-1p#2: the number of trials until the first success is certain
p(x) =
![Page 16: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/16.jpg)
The desired probability is: p(x) = qx-1p4 1p(4) (.75) (.25) 0.0117
EXAMPLE:
On Friday’s 25% of the customers at an ATM make deposits. What is the probability that it takes 4 customers at the ATM before the first one makes a deposit.
✔ Two Categories: Success: make a deposit
Failure: don’t make a deposit✔ Probability success same for each trial✔ Wish to find the probability of the first
![Page 17: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/17.jpg)
n = number of trialsx = number of successesn – x = number of failuresp = probability of success in one trialq = 1 – p = probability of failure in one trial
BINOMIAL PROBABILITYfinding the probability of a specific
number of successes
![Page 18: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/18.jpg)
EXAMPLE 2You are taking a 10 question multiple choice test. If each question has four choices and you guess on each question, what is the probability of getting exactly 7 questions correct?
• p = 0.25 = guessing the correct answer • q = 0.75 = guessing the wrong answer• n = 10• x = 7
![Page 19: 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial](https://reader036.vdocuments.site/reader036/viewer/2022062302/5a4d1b2e7f8b9ab05999a0ff/html5/thumbnails/19.jpg)
Review Packet!!!!