1 / 24

Probability and Sampling: Part I

Probability and Sampling: Part I. What are the odds?. From Reason. From a deck of cards: What are the odds of getting the king of spades? What are the odds of getting a king? What are the odds of getting a diamond suit? . 1/52 or 0.0192. 4/52 or 0.0769. 13/52 or 0.2500.

kiley
Download Presentation

Probability and Sampling: Part I

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Probability and Sampling: Part I

  2. What are the odds? From Reason From a deck of cards: What are the odds of getting the king of spades? What are the odds of getting a king? What are the odds of getting a diamond suit? 1/52 or 0.0192 4/52 or 0.0769 13/52 or 0.2500 What are the odds of finding a newspaper at the news stand? New York times Village Voice Jerusalem Post From Experience

  3. Sampling and Probability What is the probability of pulling out a red marble ? P( R) = Total number of Red marbles Total number of marbles or .5833 P( R) = 7/12 What about probabilities for MULTIPLE events? Bag of 7 red and 4 blue, 1 white marbles.

  4. Two Rules for Probabilities of Multiple Events Addition rule: “or” Probability of a red or a blue marble? • Mutually exclusive events • Non-mutually exclusive events Multiplication rule: “and” Probability of a red and a blue marble? • With replacement • Without replacement

  5. Addition Rule: (Part I) What is the probability of getting a white or a red marble? P(W or R) = P(W) + P(R) P(W or R) = 1/12 + 7/12 P(W or R) = 8/12 = .6667 Bag of 7 red and 4 blue, 1 white marbles.

  6. Addition Rule: (Part II) What is the probability of getting a red or a glossy marble? P(R or G) = P(R) + P(G) – P(R & G) P(R or G) = 7/12 + 2/12 – 1/12 Glossy marbles P(R or G) = 8/12 = .6667 Mutually exclusive events do not require subtraction

  7. Addition Rule: (Part I) What is the probability of getting a white or a red marble? P(W or R) = P(W) + P(R) P(W or R) = 1/12 + 7/12 P(W or R) = 8/12 = .6667 Bag of 7 red and 4 blue, 1 white marbles. Can a marble be both white and red at the same time? No, so these are mutually exclusive events, which do NOT require subtraction.

  8. Multiplication Rule: (Part I) What is the probability of getting a white and then a red marble? If you do put the first marble back. P(W and then R) = P(W) * P(R) P(W and then R) = 1/12 * 7/12 P(W and then R) = 7/144 = .0486 If you put the objects back after you’ve taken them out, you have sampled with replacement. Bag of 7 red and 4 blue, 1 white marbles.

  9. Multiplication Rule: (Part I) What is the probability of getting a white and then a red marble? If you do NOT put the first marble back. P(W and then R) = P(W) * P(R|W) P(W and then R) = 1/12 * 7/11 P(W and then R) = 7/132 = .0530 If you do not put the objects back after you’ve taken them out, you have sampled without replacement. Bag of 7 red and 4 blue, 1 white marbles.

  10. Multiplication Rule: (Part II) How to handle sequences of events: What is the probability of reaching into a fresh bag and getting the sequence R, W, B, R, R? Without replacement P(R) = 7/12 P(W|R) = 1/11 P(B|R, W) = 4/10 P(R|R, W, B) = 6/9 P(R|R, W, B, R) = 5/8 (7/12)(1/11)(4/10)(6/9)(5/8) = .0088

  11. There is only one way to make a “2” two dice in one toss “1” and “1” There is only one way to make a “12” two dice in one toss “6” and “6” Sum of Two Fair Dice

  12. There are 6 ways to make a “7” with two dice in one toss: Sum of Two Fair Dice Die 1 = 6, Die 2 =1 Die 1 = 5, Die 2 =2 Die 1 = 4, Die 2 =3 Die 1 = 1, Die 2 =6 Die 1 = 2, Die 2 =5 Die 1 = 3, Die 2 =4

  13. The Probability Distribution of Two Six Sided Dice .1667 .1389 .1111 Probability .0833 .0556 .0278 .0000

  14. H O.5 O.5 T Coin Toss Baseline probability A single event that can go one of two ways -- Two mutually exclusive events. Expressed as probability: Two possible outcomes with equal likelihood. P(H) = ½ = 0.5 P(T) = ½ = 0.5

  15. Outcomes H HH H O.5 O.5 HT O.5 O.5 T P(HT) = P(H) * P(T) =(0.5)(0.5) = 0.25 T H O.5 TH P(TH) = P(T) * P(H)=(0.5)(0.5) = 0.25 P(TT) = P(T) * P(T) =(0.5)(0.5) = 0.25 O.5 TT T What about multiple events? (More than one flip of the coin) Multiplication rulefor calculating the probability of a sequence of outcomes ... Independent events, Sampling with replacement P(HH) = P(H) * P(H) Probability of landing on heads twice in a row = Probability of landing on heads on the first flip Probability of landing on heads on the second flip

  16. Addition rulefor calculating the probability of outcomes that are of the same kind: Outcomes P(HH) = 0.25 H HH H O.5 O.5 P(HT) = 0.25 HT O.5 O.5 T T H P(TH) = 0.25 O.5 TH O.5 T P(TT) = 0.25 TT # of Tails P(0 T) = 0.25 } P(1 T) = 0.25 + 0.25 = 0.5 P(2 T) = 0.25

  17. If we plot the outcomes as a histogram we begin to get a familiar shape. 0 1 2

  18. This works for Sequences of any length H H T H H O.5 T … T H O.5 H T T T H 0 1 2 T

  19. A list of ALL the possible outcomes of N events when each event only has two outcomes: FAIR COIN!! Use of the Binomial Table .1250 Flip a coin 3 times, what’s the probability of 0 tails?: Flip a coin 2 times, what’s the probability of 2 tails?: .2500 Flip a coin 4 times, what’s the probability of 4 tails?: .0625 Flip a coin 3 times, what’s the probability of 1 head?: .3750 Flip a coin 4 times, what’s the probability of 3 heads?: .2500

  20. A list of ALL the possible outcomes of N events when each event only has two outcomes: FAIR COIN!! Use of the Binomial Table .5000 Flip a coin 3 times, what’s the probability of 2 or more tails?: Flip a coin 2 times, what’s the probability of 1 or less tails?: .7500 Flip a coin 4 times, what’s the probability of 2 or less tails?: .6875 Flip a coin 3 times, what’s the probability of 1 or more heads?: .8750

  21. A list of ALL the possible outcomes of N events when each event only has two outcomes: Use of the Binomial Table Ever wonder how likely you are to pass a True/False exam if you JUST GUESSED? Assume there are 20 true/false questions on the exam. You need to answer 13 or more correctly to get a 65+. P(13) + p(14) + p(15) + p(16) + p(17) + p(18) + p(19) + p(20) .0739 + .0370 + .0148 + .0046 +.0011 + .0002 + .0000 + .0000 = .1316 or 13.16% chance

  22. Use of the Binomial Table A list of ALL the possible outcomes of N events that have only two outcomes: What if the baseline probability is not .50? Suppose you are given a coin which you KNOW is weighted: 60% of the time it shows up heads, and 40% of the time it shows up tails. .1536 What is the probability of obtaining 3 tails out of 4 flips of this coin? .1296 What is the probability of obtaining 0 tails out of 4 flips of this coin? .3456 What is the probability of obtaining 3 heads out of 4 flips of this coin? .0256 What is the probability of obtaining 0 heads out of 4 flips of the coin?

  23. Use of the Binomial Table A list of ALL the possible outcomes of N events that have only two outcomes: What if the baseline probability is not .50? Suppose you are given a coin which you KNOW is weighted: 60% of the time it shows up heads, and 40% of the time it shows up tails. What is the probability of obtaining 3 or fewer heads out of 4 flips of this coin? 3 or fewer heads = 1 or more tails = .8704 What is the probability of obtaining 1 or more heads out of 3 flips of the coin? 1 or more heads = 2 or fewer tails = .9360

  24. The Critical Value of an Inferential Statistic Critical Value of the statistic is the value that demarcates the outcomes that will allow us to make conclusions about the data.

More Related