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Probability and Sampling: Part I

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## Probability and Sampling: Part I

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**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**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.**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**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.**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**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.**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.**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.**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**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**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**The Probability Distribution of Two Six Sided Dice**.1667 .1389 .1111 Probability .0833 .0556 .0278 .0000**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**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**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**If we plot the outcomes as a histogram we begin to**get a familiar shape. 0 1 2**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**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**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**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**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?**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**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.