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Introduction to Statistics

by: Maximilian Turcotte

Introduction to Statistics MATH 102

Marketplace > Colgate University > Mathematics (M) > MATH 102 > Introduction to Statistics
Maximilian Turcotte

GPA 3.7

Aaron Robertson

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Aaron Robertson
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This 4 page Class Notes was uploaded by Maximilian Turcotte on Monday October 5, 2015. The Class Notes belongs to MATH 102 at Colgate University taught by Aaron Robertson in Fall. Since its upload, it has received 28 views. For similar materials see /class/219585/math-102-colgate-university in Mathematics (M) at Colgate University.


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Date Created: 10/05/15
BASIC PROBABILITY Stats 102 A Robertson Throughout the lecture below let E and F be two sets of outcomes of an experiment We will develop 3 rules that will aid in solving basic probability problems These rules are 0 The NOT Rule ProbE 17 ProbNOT E o The Multiplication Rule ProbE and F ProbE gtlt ProbFlE o The Addition Rule ProbE or F ProbE ProbF 7 ProbE and THE NOT RULE Consider the simple statement Either E happens or E does not happen This statement is always true regardless of what E is Hence ProbE happens ProbE does NOT happen 1 Rearranging this equation gives us the NOT Rule ProbE 1 7 ProbNOT This is a very simple statement but can make all the difference when it comes to calculating some probabilities since sometimes it is easier to calculate what you don7t want to happen than it is to calculate what you do want to happen CALCULATING BASIC PROBABILITY We will typically assume that all possible outcomes of an experiment are equally likely That is no particular outcome is more likely to occur than others ln this situation we calculate the probability of E by number of outcomes in E Prob E number of possible outcomes of the experiment We are focusing on this probability since the basic situations of statistics involve sampling and surveying These are done by picking samples at random which means that all samples have an equally likely chance of being picked CONDITIONAL PROBABILITY Probabilities can not must change if additional information is known For example the probability that Colgate will win its next football game will obviously change if we nd out that every player on Colgate7s football team is sick with the u The probabilities we are interested here are called conditional probabilities These are probabilities given some information ln math these look like ProbElF and we say the probability of E given F What this is asking for is the probability that E occurs if we know that F has occurred Let7s use what we know to nd a formula for this probability Since we are given that E has occurred the possible outcomes are the number of outcomes in E And we want of those outcomes in E how many of them are in F In other words how many are in E and F Thus ProbltFE number of outcomes in E AND F number of outcomes in E by the way we calculate basic probability Now I can divide the numerator and denominator of this fraction by the same number and not change anything So we have 0 outcomes in tota o poss1 e outcome 0 t e experiment f E AND F l f bl f h of outcomes in Etotal of possible outcome of the experiment ProbElF But this is in math ProbE and F ProbFl E ProbltEgt MULTIPLICATION RULE We are now ready to present our second rule We use this rule to calculate the probability that both E and F occur Rearranging the preVious equation we have ProbE and F ProbE gtlt ProbFlE This is the multiplication rule Before doing some examples we consider a special case of the Multiplication Rule lf ProbFlE ProbE and hence ProbElF ProbE we say that E and F are independent In this case the multiplication rule becomes ProbE and F ProbE gtlt ProbF A few examples are in order Eccample 1 Roll a fair die and ip a fair coin What is the probability that your roll a Eland ip a head Let E be the outcome that we roll a El and let F be the outcome that we flip a head Then ProbE and F ProbEgtlt ProbFlE gtlt Example 2 Shuf e a deck of 52 cards What is the probability that the rst two cards are aces Let E be the outcome that the rst card is an ace and F the outcome that the second card is an ace Then ProbE and F ProbEgtlt ProbFlE gtlt Remark The outcomes in Example 1 are independent since ProbFlE ProbF but the out comes in Example 2 are not independent since ProbFlE while ProbF THE ADDITION RULE Another common probability is in the case when we want either E or F possibly both to occur le we want ProbE or 2 To do this7 we use a Venn Diagram The rectangle encloses all possible outcomes of the experiment and the circles labeled E and F enclose all outcomes in E and F7 respectively What we want to count are all the outcomes in E or F This is in the shaded area be ow E F Now7 there are three regions to count These are broken down below Q We start by counting the outcomes in E plus the outcomes in F and then writing each in terms of the three desired regions Substracting Q from each side we get Translating this into math7 we get the addition rule ProbE or F ProbE ProbF 7 ProbE and Before ending our lecture7 we have one more topic to coVer We say that two outcomes are mutually exclusive if ProbE and F 0 In other words7 E and F cannot both occur at the same time In this case7 the addition rule reduces to ProbE or F ProbE ProbF Remark A Very common mistake is to use the aboVe form of the addition rule when E and F are not mutually eXclusiVe By doing this7 the resulting answer will always be wrong SOME EXAMPLES 1 Pick three cards without replacement from a standard deck What is the probability of getting at least one ace Solution Our outcome E is getting at least one ace Hence this consists of getting exactly one exactly two or exactly three aces Since we have three seperate cases to count lets look and see if the outcome not E has fewer cases Here we see that not E is not getting at least one ace which is the same as getting no ace This is easier to count From here we use the multiplication principle with the experiments A is the rst card is not an ace B is the second card is not an ace 0 is the third card is not an ace So we want the prob of A and B and C Using the multiplication rule twice we have ProbA and B and C ProbA gtlt Prob B given A gtlt ProbC given A and B 4852 gtlt 4751 gtlt 4650 But we have counted the experiment not E so the answer is 1 7 4852 gtlt 4751 gtlt 4650 2 Roll two twenty sided dice one blue and one red Find the probability that you roll a blue 5 or that the sum of the two dice is even Solution Let E be the outcome get a blue 5 and let F be the outcome the sum is even We want ProbE or F so we use the addition rule First ProbE 120 and ProbF 12 Next we need to nd ProbE and This means we need the blue die to show 5 and the sum ofthe two dice to be even Using the multiplication rule we have ProbE and F ProbE gtlt ProbF given E 120 gtlt 12 where the 12 comes from the fact that only the numbers 135 7 9 11 13 15 17 and 19 on the red die will give an even surn when added to the blue 5 Hence by the addition rule the answer is 12012 7 12120 2140 3 A drawer contains 7 different pairs of socks the pairs being colored red blue green yellow black purple and brown with the socks all mixed up You reach in and pick two socks at random What is the probability that you get a matching pair What is the probability that you get the brown pair Solution Let E be the outcome of your rst pick and let F be the outcome of your second pick We rst answer the question What is the probability that you get a matching pair Since we just want a pair any pair it does not matter what E is What matters is that F rnatches E So reach in and pick a sock We now want to reach in and get the same color sock from those rernaining Since there are 13 socks remaining with only one that matches the probability of a match is 113 Next we answer the question What is the probability that you get the brown pair Here we have E being the outcome the rst sock is brown and F being the outcome the second sock is brown We want ProbE and F so we use the multiplication principle We have ProbE 214 17 since two of the fourteen socks are brown Next we have ProbF given E 113 Hence the answer is 17 gtlt 113 191 You could have answered this in a different way using your answer to the rst question Since the probability of getting a pair is 113 and we have seven pairs then the probability of getting the brown pair given that you have a pair is 17 Again by the multiplication principle we have 17 gtlt 113 191


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