19
Probability Topi cs 1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial Probabilities

1.Addition Rule 2.Multiplication Rule 3.Compliments 4.Conditional Probability 5.Permutation 6.Combinations 7.Expected value 8.Geometric Probabilities 9.Binomial

Embed Size (px)

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%

Citation preview

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Review Packet!!!!