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UCLA STATS 100A - BEAMER-100A-lesson4

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Stat 100A. Introduction to ProbabilityIntroduction to Sample Spaces with equally likely outcomesJuana [email protected] Department of StatisticsJ. Sanchez Stat 100A. Introduction to ProbabilityOutlineI. Probabilities of events in a Sample Space with equally likelyoutcomes. Small sample spaces.II Introduction to counting.III. Fundamental Theorem of Counting.IV. Probability theorems apply to sample spaces with equallylikely outcomes.J. Sanchez Stat 100A. Introduction to ProbabilityI. Calculating probability of an event from a sample spacewith equally likely outcomes. Small spaces. (FYI, thisstarts material from Ross Chapter 1)In generalIf siis an outcome in the sample space S and A an event definedin S , then the probability of A can always be found by adding theprobabilities of all the outcomes of S that are in that event, ingeneralP(A) =XsiinAP(si∈ S | si∈ A)J. Sanchez Stat 100A. Introduction to ProbabilityWhen outcomes in S are equally likelyConsider an experiment whose sample space S is a finite set, sayS = {1, 2, · · · , N}, N the cardinality or number of outcomes in S.Then it is sometimes the caseP({1}) = P({2}) = · · · = p({N })which implies from axioms 1 and 2 of probability thatP({i}) =1Ni = 1, 2, , · · · , NBecause of that, we can use a simpler way of computing theprobability of one event AP(A) =Number of outcomes in ATotal number of possible outcomes in SJ. Sanchez Stat 100A. Introduction to ProbabilityTo see that, suppose that S = (s1, s2, ..., sN) is a finite samplespace. Saying that all the outcomes are equally likely means thatP(si) = 1/N , for every outcome si. N is the cardinality of thesample space (or number of outcomes in the Sample space). Then,applying the general definition of probability.P(A) =Xsi∈AP(si) =Xsi∈A1N=Number of outcomes in ATotal number of possible outcomesJ. Sanchez Stat 100A. Introduction to ProbabilityExampleIf a die is rolled and all six sides are assumed equally likely toappear, then we would haveP({1}) = P({2}) = P({3}) = P({4}) = P({5}) = P({6}) = 1/6From definition of probability of an event, it would thus follow thatthe probability of rolling an even number would equal.P({2, 4, 6}) = P({2}) + P({3}) + P({4}) = 3/6But from definition of probability of an event with equally likelyoutcomes we could just computeP({2, 4, 6}) = (3 outcomes in the even even number) / (6 possibleoutcomes in S)= 1/2J. Sanchez Stat 100A. Introduction to ProbabilityThus, to summarize:If we assume that all outcomes of an experiment are equally likelyto occur, then the probability of any event E equals the proportionof outcomes in the sample space that are contained in E.J. Sanchez Stat 100A. Introduction to ProbabilityWarning: Need to di stinguish sample s paces with equallylikely outcomes and sample spaces without equally likelyoutcomesThe space S may be the same, but the probabilities of eachoutcome may not. The example below has S a set of equally likelyoutcomes.ExampleYou roll a fair six-sided die. With a fair die, all the outcomes of aroll are equally likely. Let A be the event ”number of roll is lessthan or equal to 5), and B be the event ”a number larger or equalto 3.”Give S, P(A) and P(B).(a) S = {1, 2, 3, 4, 5, 6}(b) Prob(A) =(5)/(6 outcomes in S)=5/6(c) Prob(B) =(4 )/(6 outcomes in S)= 4/6.J. Sanchez Stat 100A. Introduction to ProbabilityWarning: Need to distinguish sample spaces with equallylikely outcomes and sample spaces without equally likelyoutcomesThe following example does not have equally likely outcomes, soonly the definition of probability of an event can be used.P(A) =Psi∈AP(si)ExampleYou roll a six-sided die that is not fair in the following way: P(1) =2/12; P (2) = 5/12; P(3) = 2/12; P(4) = 1/12; P(5) = P (6) = 1/12Let A be the event getting an even number, and B the event getting anodd number.Give S, P(A) and P(B)(a) S = {1, 2, 3, 4, 5, 6}(b) Prob(outcome is an even number) = 5/12 + 1/12 + 1/12 = 7/12(c) Prob(outcome is an odd number) = 1/6 + 1/6 + 1/12 = 5/12.J. Sanchez Stat 100A. Introduction to ProbabilityExampleYou roll two fair six-sided dice. The sample space has 36 equallylikely outcomes because the probability of each two rolls is1/6 × 1/6 (by product rule we will see next week).S =(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)What is the probability that the sum of the upturned faces willequal 7?Let A be the event that the sum of the upturned faces equals 7.Enumerate A/A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}P(A) = (number of outcomes with sum 7)/(total number ofoutcomes in S)=636= 1/6If the dice were not fair, we would need to know the probabilitiesof each pair of numbers to compute the probability of any event.With not equally likely outcomes, we would need to add theprobabilities of the outcomes in the event.J. Sanchez Stat 100A. Introduction to ProbabilityI I. Introduction to counting1With sample spaces like the one of the die, there are very fewequally likely outcomes to count, so we can count with ourfingers.2For large sample spaces with many possible equally likelyoutcomes use counting techniques, that help us calculate thenumerator and denominator of the probability. We get intothose now.J. Sanchez Stat 100A. Introduction to ProbabilityI II. Fundamental theorem of countingTheoremIf a job consists of k separate tasks, the ith of which can be donein niways, i = 1, · · · , k, then the entire job can be done inn1× n2× · · · × nkways.The lecture posted separately contains examples on how to applythis theorem to count under different scenarios. Lets see first someexamples of this fundamental theorem.J. Sanchez Stat 100A. Introduction to ProbabilityMathematical requirements for this topicFor a positive integer n, n! (read n factorial) is the product ofall the positive integers less than or equal to n. That is ,n! = n(n − 1).......(3)(2)(1)Furthermore, we define 0! = 1.For nonnegative integers n and r , where n ≥ r , we define thesymbolnr, read ”n choose r”, asnr=n!r!(n − r)!Notice that this is the binomial coefficient.J. Sanchez Stat 100A. Introduction to ProbabilityExampleA committee of 5 is to be selected from a group of 6 men and 9women. If the selection is made randomly, what is the probabilitythat


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