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### Statistics Quick Overview

Class #2

Conditional Probability Examples

- Given that a consumer has looked at a green shirt 5 times
- What is the probability that they will buy?
- What is the probability that they will buy if you give them 5% off?
- What is the probability that they will buy if you give them 10% off?
- What is the probability given that they are a new customer/existing customer?
- Given that a consumer has bought a flashlight
- What is the probability that they will buy batteries?
- What is the probability that they will buy a 2nd flashlight?

Car Example- Neither Will Start?

Solution space = 1

A and B

(75%)

= 90% + 80% - 75%=95%

1-95% = 5% chance neither will start

Conditional Probability:A Table Can Be Helpful

Acura

Starts Doesn’t

Normalize this row:

75% / 80%

Starts

Doesn’t

80%

75%

5%

BMW

20%

15%

5%

10%

90%

Let’s Say A Coach (Maybe Mr. Brown?) Had to Pick 3 Players for Hockey and Then Quiditch

- Let’s start with hockey…
- Here, order matters
- The person on the left must stay on the left
- The person on the right must stay on the right
- So, how many different potential line-ups does Mr. Brown have to consider?
- Choices are: Mr Blonde, Mr White, Mr Orange, Mr Pink, and Mr Blue

5 x 4 x 3 = 60

Where n is the number of choices, and k is the number picked. In Excel, this is PERMUT(n,k)

Let’s Carefully Write Out the Permutations

Note: Each column is a unique combination of players

Note: The entries within each row are different permutations of the players. This is our same problem again where n= 3 and k = 3

==6

Mr. Brown’s Choices for a Quiditch Front Line

- Here, order does not matter
- He just needs a front line
- All that matters is the number of unique combinations
- What observation from the permutation table helps us determine the unique combinations

Figuring out the Combinations

When calculating the permutations, we naturally determined the unique combinations (the columns) and then ran the permutations for each combination. If we divide out that last step, we will have just the combinations:

= 60 / 6=10

Binomial Distribution

- Sample Size of 10
- Case #1: Assume that the lot is good with 5% defectives
- When will you reject because you find 3 or more defectives
- Case #2: Assume that the lot has 40% defectives
- When will you accept because you find 2 or less defectives
- Let’s assume:
- s is the probability of success
- f is the probability of failure

Case #1 (only 5% of the lot is defective)

- Example of getting 3 Failures
- fssfsssssf
- Probability of this is (5%)3(95%)7
- Example of getting 4 Failures
- fssfsfsssf
- Probability of this is (5%)4(95%)6
- What are we missing?

The number of combinations

Bayes’ Rule

- A1 uses drugs P(A1) = 5%
- A2 does not use drugs P(A2) = 95%
- B tests shows drug use
- P(B | A1) = 98%
- P(B | A2) = 2%

What we want….

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