Chapter 5- Probability Review

Section 5.1

An event is the set of possible outcomes

Probability is between 0 and 1 The event A has a complement, the

event not A. Together these two probabilities sum 1.ex. At least one and none are complements

Probability of an event = number of outcomes in event / number of equally

likely outcomes

Section 5.1

Probability Distributions give all values resulting from a random process.

Sample space is the complete list of disjoint outcomes. All outcomes in a sample space must have total probability equal to 1.

Ex. The sample space for rolling a die is {1,2,3,4,5,6} The sample space for rolling two die is the table of 36 outcomes we’ve seen.

Disjoint

Disjoint and mutually exclusive mean the same thing.

Disjoint means two different outcomes can’t occur on the same opportunity

Ex. Can’t roll an extra credit and no collect on the same roll. Can’t get a heads and tail on the same flip.

These items on a Venn diagram would have no intersection

Disjoint Continued

Flipping a coin and getting a head and tail is disjoint

Flipping a coin twice and getting a head and tail is disjoint.

Law of Large Numbers

In random sampling, the larger the sample, the closer the proportion of successes in the sample tends to be to the population proportion.

The difference between a sample proportion and the population proportion must get smaller as the sample size gets larger.

Fundamental Principle of Counting Processes can be split into stages (Flip

coin once, then again) If there are k stages with different

possible outcomes for each stage, the number of total possible outcomes is

n1*n2*n3*n4… nk

Ex. How many total outcome for rolling a die and flipping a coin?

6*2 = 12

Probability Simulations

1) Assumptions

What probability are you assuming? Are events independent?

Probability Simulations

2) Model*How specifically will you use your table of random digits? *Make sure to say what to do with repeats, unallocated numbers*Fully describe what constitutes a run and what statistics you’re collecting.*Make a table to show how digits /groups are assigned

Probability Simulations

3) Repetition

Run the simulations and record the results in a frequency table.

Probability Simulations

4) Conclusion

Write the conclusion in context of the situation. Be sure to say the probability is ESTIMATED.

Addition Rule of Probability Remember first that “or” means one

or the other or both

P(A or B) = P(A) + P(B) – P(A and B)

If A and B are disjoint, there is no intersection. Therefore, P(A and B)= 0.

If A and B are disjoint: P(A or B) = P(A) + P(B)

Conditional Probability and the Multiplication Rule The probability of an event A and B

both happening isP(A and B) = P(A) * P(B|A)

P(A and B) = P(B) * P(A|B)

The probability of an event changes based on what happened before.

Conditional Probability

Rearranging means P(A|B) = P(A and B) / P(B)

Probability of both divided by the probability of the first event.

Independent Events

The occurrence of one event doesn’t change the probability of the second event occurring

Test for Independence: IS P(A|B) = P(A) or P(B|A) =

P(B) ?If yes, you have independent events.

Sometimes this isn’t obvious that one has an effect without checking

Multiplication Rule for Independent Events Remember if events are

independent,P(A|B) = P(A) or P(B|A) = P(B)

Therefore, since P(A and B) = P(A) * P(B|A)

P(A and B) = P(A) * P(B) for independent events

Review Questions

If events A and B are independent and P(A) = 0.3 and P(B) = 0.5, then which of these is true?

A. P(A and B) = 0.8B. P(A or B) = 0.15C. P(A or B) = 0.8D. P(A | B) = 0.3E. P(A | B) = 0.5

Answer

If events A and B are independent and P(A) = 0.3 and P(B) = 0.5, then which of these is true?

A. P(A and B) = 0.8B. P(A or B) = 0.15C. P(A or B) = 0.8D. P(A | B) = 0.3 Probability of A is not changed

based on the occurrence of event B

E. P(A | B) = 0.5

Two Way Table Questions

Crash Type

Single Vehicle Multiple Vehicles

Total

Alcohol Related 10,741 4,887 15,628

Not Alcohol Related

11,345 11,336 22,681

Total 22,086 16,223 38,309

If a fatal auto crash is chosen at random, what is the approximate probability that the crash was alcohol related, given that it involved a single vehicle?A. 0.28B. 0.49C. 0.58D. 0.69E. The answer cannot be determined from the information given.

AnswerCrash Type

Single Vehicle Multiple Vehicles

Total

Alcohol Related 10,741 4,887 15,628

Not Alcohol Related

11,345 11,336 22,681

Total 22,086 16,223 38,309

If a fatal auto crash is chosen at random, what is the approximate probability that the crash was alcohol related, given that it involved a single vehicle?

B. 0.4910,741/22,086 = 0.49

Two Way Table Questions

Crash Type

Single Vehicle Multiple Vehicles

Total

Alcohol Related 10,741 4,887 15,628

Not Alcohol Related

11,345 11,336 22,681

Total 22,086 16,223 38,309

What is the approximate probability that a randomly chosen fatal auto crash involves a single vehicle and is alcohol related?A. 0.28B. 0.49C. 0.58D. 0.69E. The answer cannot be determined from the information given

AnswerCrash Type

Single Vehicle Multiple Vehicles

Total

Alcohol Related 10,741 4,887 15,628

Not Alcohol Related

11,345 11,336 22,681

Total 22,086 16,223 38,309

What is the approximate probability that a randomly chosen fatal auto crash involves a single vehicle and is alcohol related?A. 0.28Because 10,741/38,309 = 0.28

Review Question

For all events A and B, P(A and B) =A. P(A) · P(B)B. P(B | A)C. P(A | B)D. P(A) + P(B)E. P(B) · P(A | B)

Answer

For all events A and B, P(A and B) =A. P(A) · P(B)B. P(B | A)C. P(A | B)D. P(A) + P(B)E. P(B) · P(A | B)This is the multiplication rule.If they are independent, P(A|B) = P(A)

Review Question

The mathematics department at a school has twenty instructors. Six are easy graders. Twelve are considered to be good teachers. Seven are neither. If a student is assigned randomly to one of the easy graders, what is the probability that the instructor will also be good?

A. 7/20B. 5/12C. 7/12D. 5/6E. The answer cannot be determined from the

information given.

Answer

D. 5/6

Easy Not Easy

Total

Good 5 7 12Not Good

1 7 8

Total 6 14 20

Review Question The management of Young & Sons Sporting

Supply, Inc., is responding to a claim of discrimination. If the company has employed the 60 people in this table, how many females over the age of 40 must the company hire so that the age and sex of its employees are independent? 40 Years > 40 Years Total

Male 25 15

Female 20

Total

AnswerFirst, let N be the number of females

older than 40 to be hired.

Set up a proportion so P(male |<40) = P(Female|< 40)N=12

40 Years > 40 Years TotalMale 25 15 40Female 20 N 20+NTotal 45 15+N 60+N

Review Question

If P(A) = 0.4, P(B) = 0.2, and P(A and B) = 0.08, which of these is true?

A. Events A and B are independent and mutually exclusive.

B. Events A and B are independent but not mutually exclusive.

C. Events A and B are mutually exclusive but not independent.

D. Events A and B are neither independent nor mutually exclusive.

E. Events A and B are independent, but whether A and B are mutually exclusive cannot be determined from the given information.

AnswerIf P(A) = 0.4, P(B) = 0.2, and P(A and B) = 0.08, which of

these is true?A. Events A and B are independent and mutually

exclusive.B. Events A and B are independent but not

mutually exclusive.C. Events A and B are mutually exclusive but not

independent.D. Events A and B are neither independent nor mutually

exclusive.E. Events A and B are independent, but whether A and

B are mutually exclusive cannot be determined from the given information.