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CMU CS 15251 - Lecture

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Probability Theory: Counting in Terms of ProportionsA Probability DistributionThe Descendants Of AdamSlide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Unbiased Binomial Distribution On n+1 Elements.As the number of elements gets larger, the shape of the unbiased binomial distribution converges to a Normal (or Gaussian) distribution.Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Probabilities and countingHow many n-bit strings have an even number of 1’s?Slide 26Slide 27Slide 28Slide 29Slide 30How many n-trit strings have even number of 0’s?Some puzzlesSlide 34Actually, 6 and 7 are equally likelyAnother viewSlide 37Slide 38Slide 39So, sometimes, probabilities can be counter-intuitiveLanguage Of ProbabilityFinite Probability DistributionSlide 43Sample spaceProbabilityProbability DistributionEventsSlide 48Slide 49Uniform DistributionSlide 51Slide 52Slide 53Using the LanguageUsing the Language: visuallySlide 56Slide 57Slide 58PictureSlide 60Slide 61Slide 62Slide 63Slide 64Another way to calculate Pr(no collision)More Language Of ProbabilitySlide 67Slide 68Independence!Slide 70Slide 71Slide 72Monty Hall problemSlide 74why was this tricky?Random walks and electrical networksSlide 77Random walks come up all the timeProbability Theory:Counting in Terms of Proportions Great Theoretical Ideas In Computer ScienceSteven Rudich, Anupam Gupta CS 15-251 Spring 2004Lecture 18 March 18, 2004 Carnegie Mellon UniversityA Probability DistributionHEIGHTProportion of MALESThe Descendants Of AdamAdam was X inches tall.He had two sonsOne was X+1 inches tallOne was X-1 inches tallEach of his sons had two sons ….XX-1X+111X-2X+2XX-3X+3X-1 X+1X-4X+4X-2X+2X156655101015201X-1X+111X-2X+2XX-3X+3X-1 X+1X-4X+4X-2X+2X1566551010152011111X-2X+2XX-3X+3X-1 X+1X-4X+4X-2X+2X1566551010152011111112X-3X+3X-1 X+1X-4X+4X-2X+2X1566551010152011111112113 3X-4X+4X-2X+2X1566551010152011111112113 3114461566551010152011111112113 31144615665510101520In nth generation, there will be 2n males, each with one of n+1 different heights: h0< h1 < . . .< hn. hi = (X – n + 2i) occurs with proportionUnbiased Binomial Distribution On n+1 Elements.Let S be any set {h0, h1, …, hn} where each element hi has an associated probabilityAny such distribution is called an Unbiased Binomial Distribution or an Unbiased Bernoulli Distribution.As the number of elements gets larger, the shape of the unbiased binomial distribution converges to a Normal (or Gaussian) distribution.StandardDeviationMean11446Coin Flipping in ManhattanAt each step, we flip a coin to decide which way to go.What is the probability of ending at the intersection ofAvenue i and Street (n-i) after n steps?11446Coin Flipping in ManhattanAt each step, we flip a coin to decide which way to go.What is the probability of ending at the intersection ofAvenue i and Street (n-i) after n steps?Coin Flipping in ManhattanAt each step, we flip a coin to decide which way to go.What is the probability of ending at the intersection ofAvenue i and Street (n-i) after n steps?11446Coin Flipping in ManhattanAt each step, we flip a coin to decide which way to go.What is the probability of ending at the intersection ofAvenue i and Street (n-i) after n steps?11446Coin Flipping in ManhattanAt each step, we flip a coin to decide which way to go.What is the probability of ending at the intersection ofAvenue i and Street (n-i) after n steps?11446Coin Flipping in Manhattan2n different paths to level n, each equally likely. The probability of i heads occurring on the path we generate is: 11446n-step Random Walk on a lineStart at the origin: at each point, flip an unbiased coin to decide whether to go right or left. The probability that, in n steps, we take i steps to the right and n-i to the left (so we are at position 2i-n) is: 11446n-step Random Walk on a lineStart at the origin: at each point, flip an unbiased coin to decide whether to go right or left. The probability that, in n steps, we take i steps to the right and n-i to the left (so we are at position 2i-n) is: 11446n-step Random Walk on a lineStart at the origin: at each point, flip an unbiased coin to decide whether to go right or left. The probability that, in n steps, we take i steps to the right and n-i to the left (so we are at position 2i-n) is: 11446n-step Random Walk on a lineStart at the origin: at each point, flip an unbiased coin to decide whether to go right or left. The probability that, in n steps, we take i steps to the right and n-i to the left (so we are at position 2i-n) is: 11446n-step Random Walk on a lineStart at the origin: at each point, flip an unbiased coin to decide whether to go right or left. The probability that, in n steps, we take i steps to the right and n-i to the left (so we are at position 2i-n) is: 11446Probabilities and countingSay we want to count the number of X's with property Y One way to do it is to ask "if we pick an X at random, what is the probability it has property Y?" and then multiply by the number of X's. =Probability of X with property Y # of X with property Y# of XHow many n-bit strings have an even number of 1’s?If you flip a coin n times, what is the probability you get an even number of heads? Then multiply by 2n.Say prob was q after n-1 flips.Then, after n flips it is ½q + ½(1-q) = ½.11446Binomial distribution with bias p Start at the top. At each step, flip a coin with a bias p of heads to decide which way to go. What is the probability of ending at the intersection ofAvenue i and Street (n-i) after n steps?p1-p11446Binomial distribution with bias p p1-pStart at the top. At each step, flip a coin with a bias p of heads to decide which way to go. What is the probability of ending at the intersection ofAvenue i and Street (n-i) after n steps?11446Binomial distribution with bias p pp1-p1-ppStart at the top. At each step, flip a coin with a bias p of heads to decide which way to go. What is the probability of ending at the intersection ofAvenue i and Street (n-i) after n steps?11446Binomial distribution with bias p pp1-p1-ppStart at the top. At each step, flip a coin with a bias p of heads to decide which way to go. The probability of any fixed path with i heads (n-i tails) being chosen is: pi (1-p)n-iOverall probability we get i heads is:Bias p coin flipped n times. Probability of exactly i heads is:How many n-trit strings have even number of 0’s?If you flip a bias 1/3 coin n times, what is the probability qn you get


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CMU CS 15251 - Lecture

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