GRINNELL MAT 209 - MAT 209 CHAPTER 4 RANDOMNESS

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CHAPTER 4 RANDOMNESSCHAPTER 4.2 Probability ModelsProbability Rules {4.5}SECTION 4.4 Means and VariancesCHAPTER 4 RANDOMNESSWe use mathematics to describe regular patternsCircles, movements of planets, e = mc2, strength requirements for buildings, etc..Space Shuttle Challenger—what is the chance of failure?Engineers 1 in 100, management said 1 in 100,000--that would be 1 failure if launching a rocket every day for 300 years—clearly they were guessingThe mathematics of chance/randomness is called Probability.Chance behavior is unpredictable in the short run, but has a regular and predictable pattern in the long runToss a coin –In the long run, about ½ will e heads P(heads) = ½Each outcome of a roll of a dice is governed by chance, but after many repetitions a pattern emerges.RANDOM-if individual outcomes are uncertain, but there is a regular distribution of outcomes in a large number of repetitionsSamples are random – we are uncertain who will be chosen, but overall everyone has an equal chance of being selected.PROBABILITY of any outcome/event is a # between 0 and 1 that give the proportion of times that event would occur in a very long series of repetitions. CHAPTER 4.2 Probability ModelsSample space, S, is the set of all possible outcomesToss a coin 1 time S = {H, T}Toss a coin 2 times S = {HH, HT, TH, TT} or S = {0, 1, 2}EVENT any combination of outcomesProbability Rules {4.5}1) 0  P( A )  1 2) P(S) = P(All possible outcomes) = 13) P ( Ac ) = 1 – P (A)4) P(A  B) = P(A or B) = P(A) + P(B) – P (A and B) = P(A) + P(B) – P (A  B)Events A and B are disjoint if they have no outcomes in common, P(A and B) = 05) If A and B are independent events then P(A and B) = P(A)P(B)2 events are Independent if knowing the outcome of one does not change the probability of the 2nd event.6) P(A|B) = P(A given B) = P(A and B) / P(B)Equally likely outcomes (tossing a coin or rolling dice)Two dice are numbered 1-6, find the following1) P( getting an 1 on the first roll)2) P( getting an 1 on the second roll)3) P( getting an 1 on the first roll and a 1 on the second roll)4) P( the sum of the dice are less than 7)5) P( the sum of the dice are greater than or equal to 7)6) P( getting an 1 on the second roll given a 1 was given on the first roll )7) P( getting an 1 on the second roll given the sum of the dice are  7)8) P( getting an 1 on the second roll given the sum of the dice are > 7)9) P( getting an 1 on the second roll given the sum of the dice are  7)Venn DiagramFinite number of outcomesMarital status of women between the age of 25 and 29Marital Status Never Married Married Widowed DivorcedProbability .353 .574 .002 .071Assign prob to each outcome and any collection of itemsP(not married) Probability Model describes all possible outcome and assigns probabilities to any collection of outcomes.SECTION 4.3A random variable is a variable whose value is a numerical outcome of a random phenomenon.Discrete Random VariablesContinuous Random VariablesProbability DSN of a statistic tell us what values the statistic will take (and how often) in repeated samples from the same population.For discrete random variables this can look like a probability histogram.For a continuous random variable this is a density curve.CLT says means of large samples can use the normal dsn.When n is large Xbar is normally distributedSECTION 4.4 Means and VariancesEXPECTED VALUE multiply each outcome by its probability, then sum over all possible outcomesOutcomes a1, a2, ..an and probabilities p1, p2, …pn then the expected value is:a1 * p1 + a2* p2 + … + an* pn FAIR GAME you pay the expected value to play the game. So on average, people will come out even. So for pick 3 to be fair, they should charge $0.50X =  xi piX2 =  (xi - X)2 pi a+bX = a + bX aX+bY = aX + bY2a+bX = b22X2X+Y = 2X + 2Y + 2XY2X-Y = 2X + 2Y - 2XY2aX+bY = a22X + b22Y {when X and Y are independent}Pick 3 gameChoose any 3 digit number (000 – 999) pay $1 and win $500 if you get it rightIf you bought all 1000 possible tickets you would win exactly $500.So each ticket has an expected value of $500/10000 = $0.50Or$500*(1/1000) + 0*999/1000 = $0.50Find expected value of a straight box wager 3 digits,win $292 if exactly right (561), win $42 if you selected the same 3 digits (156,165,516,615,651)0*994/1000 + $42*5/1000 + $292*1/1000 = 0.502So on average in both games you would expect to win $0.50Law of Large Numbers: as our sample size increases (the # of trails, bets, etc) the mean (average) will approach expected value.So the state, casinos (and life insurance companies) are not gambling at all, they know that with thousands of people willing to bet, they will make profits based on the expected value of each game.Example 4) 90% survive Kidney transplant, (10% die), Transplant succeeds in 60% of those who survive)People who survive 5 years is 70% with new kidney, 50% for those who return to dialysisWhat is his P(survive 5 years)—tree diagram


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