# Introducing probability BPS chapter 10 © 2006 W. H. Freeman and Company Brigitte Baldi, University of California, Irvine Ellen Gundlach, Purdue University.

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Introducing probability BPS chapter 10 © 2006 W. H. Freeman and Company Brigitte Baldi, University of California, Irvine Ellen Gundlach, Purdue University Marcus Pendergrass, Hampden-Sydney College

Objectives (BPS chapter 10) Introducing probability  The idea of probability  Probability models  Probability rules  Discrete sample space  Continuous sample space  Random variables  Personal probability

A phenomenon is random if individual outcomes are uncertain, but there is nonetheless a regular distribution of outcomes in a large number of repetitions. Randomness and probability The probability of any outcome of a random phenomenon can be defined as the ideal theoretical proportion of times the outcome would occur in a very long series of repetitions.

Coin toss The result of any single coin toss is random. But the result over many tosses is predictable, as long as the trials are independent (i.e., the outcome of a new coin toss is not influenced by the result of the previous toss). First series of tosses Second series The probability of heads is 0.5 = the proportion of times you get heads in many repeated trials.

The trials are independent only when you put the coin back each time. It is called sampling with replacement. Two events are independent if the probability that one event occurs on any given trial of an experiment is not affected or changed by the occurrence of the other event. When are trials not independent? Imagine that these coins were spread out so that half were heads up and half were tails up. Close your eyes and pick one. The probability of it being heads is 0.5. However, if you don’t put it back in the pile, the probability of picking up another coin and having it be heads is now less than 0.5.

Probability models mathematically describe the outcome of random processes. They consist of two parts: 1) S = Sample Space: This is a set, or list, of all possible outcomes of a random process. An event is a subset of the sample space. 2) A probability for each possible event in the sample space S. Probability models Example: Probability Model for a Coin Toss S = {Head, Tail} Probability of heads = 0.5 Probability of tails = 0.5

Important: It’s the question that determines the sample space. Sample space A. A basketball player shoots three free throws. What are the possible sequences of hits (H) and misses (M)? H H H - HHH M … M M - HHM H - HMH M - HMM … S = {HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM } Note: 8 elements, 2 3 B. A basketball player shoots three free throws. What is the number of baskets made? S = {0, 1, 2, 3} C. A person tosses a coin repeatedly. How long in seconds does it take before the first head appears? S = [0, ∞ ) = (all numbers ≥ 0)

Coin Toss Example: S = {Head, Tail} Probability of heads = 0.5 Probability of tails = 0.5 1) Probabilities range from 0 (no chance of the event) to 1 (the event has to happen). For any event A, 0 ≤ P(A) ≤ 1 Probability rules P(head) + P(tail) = 0.5 + 0.5 = 1 2) The probability of the complete sample space must equal 1. P(sample space) = 1 P(tail) = 1 – P(head) = 0.5 3) The probability of an event not occurring is 1 minus the probability that it does occur. P(A) = 1 – P(not A) Probability of getting a head = 0.5 We write this as: P(head) = 0.5 P(neither head nor tail) = 0 P(getting either a head or a tail) = 1

Probability rules (cont'd) 4) Two events A and B are disjoint if they have no outcomes in common and can never happen together. The probability that A or B occurs is the sum of their individual probabilities. P(A or B) = “P(A U B)” = P(A) + P(B) This is the addition rule for disjoint events. Example: If you flip two fair coins and the first flip does not affect the second flip, S = {HH, HT, TH, TT}. The probability of each of these events is 1/4, or 0.25. The probability that you obtain “only heads or only tails” is: P(HH or TT) = P(HH) + P(TT) = 0.25 + 0.25 = 0.50 A and B disjoint A and B not disjoint

Probability rules (cont'd) 5) Two events A and B are independent knowing that one event occurred doesn’t affect the probability that the other occurs. If A and B are independent, the probability that both A and B occurs is the product of their individual probabilities. P(A and B) = “P(A∩B)” = P(A) P(B) This is the multiplication rule for independent events. Example: Flip a fair coin twice. The coin has no “memory”, so the outcome of the first flip can’t affect the probabilities for the second flip. The first and second flips are independent. Let A = {toss 1 is heads}, B = {toss 2 is tails}. Then P(A and B) = P(HT) = P(H) P(T) = 0.5 + 0.5 = 0.25

Note: Discrete data contrast with continuous data that can take on any one of an infinite number of possible values over an interval. Discrete sample space Discrete sample spaces deal with data that can take on only certain values. These values are often integers or whole numbers. Dice are good examples of finite sample spaces. Finite means that there is a limited number of outcomes. Throwing 1 die: S = {1, 2, 3, 4, 5, 6}, and the probability of each event = 1/6.

Example  Random phenomenon: toss a fair coin three times, and record the outcomes.  What is the probability model?  Probability Model = Sample Space + Probability assignment  S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}  All outcomes equally likely (since the coin is fair)  So each outcome has probability 1/8.  We can represent the probability model as a table: outcome probability

This is the sample space: {(1,1), (1,2), (1,3), ……(6,6).} In some situations, we define an event as a combination of outcomes. In that case, the probabilities need to be calculated from our knowledge of the probabilities of the simpler events. There are 36 possible outcomes in S, all equally likely (given fair dice). Thus, the probability of any one of them is 1/36. Example: You toss two dice. What is the probability of the outcomes summing to five? Example: P(the roll of two dice sums to 5) = P(4,1) + P(3,2) + P(2,3) + P(1,4) = 4 * 1/36 = 1/9 = 0.111

Example  A 6-sided die has been weighted so that 1, 4, and 6 are more likely to appear than other digits. The probability model for the die is given in the table: outcome probability 1 25 4 63 1/5 1/6 1/5 1/6  Is this a legitimate (valid) probability model?  If we wanted the probabilities for 1, 4, and 6 to all be 1/5, and the other probabilities to all be equal to each other, what would they have to be? outcome probability 1 25 4 63 1/5 ?? 1/5 ??

Random Variables  A random variable is a variable whose value is a numerical outcome of a random phenomenon.  Example: flip a coin three times; let X = number of heads obtained  Example: take an SRS of 1000 Virginia voters; let Y = number who support the current economic stimulus plan.  Example: roll a pair of fair dice; let S = the sum of the spots that appear.  Throw a dart at a dartboard; let D = distance of dart from center.  A random variable describes a numerical outcome of a random phenomenon. It’s values are random, so they are described by a probability distribution.  Example; flip a fair coin three times, let X = number of heads obtained  P(X = 0) = ???  P(X = 1) = ???  P(X = 2) = ???  P(X = 3) = ???

Example  Toss a pair of fair dice, and record the outcomes  Random variable: let S = sum of the spots that appear S=2 S=3 S=4 S=5 S=6 S=7 S=8 S=9 S=10 S=11 S=12  Probability Model for S: S=2 S=3 S=4 S=5 S=6 S=7 S=8 S=9 S=10 S=11 S=12 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 outcome probability

Give the sample space and probabilities of each event in the following cases:  A couple wants three children. What are the arrangements of boys (B) and girls (G)? Genetics tells us that the probability that a baby is a boy or a girl is the same, 0.5. → Sample space: {BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG} → All eight outcomes in the sample space are equally likely. → The probability of each is thus 1/8.  A couple wants three children. What are the numbers of girls (X) they could have? The same genetic laws apply. We can use the probabilities above to calculate the probability for each possible number of girls. → Sample space {0, 1, 2, 3} → P(X = 0) = P(BBB) = 1/8 → P(X = 1) = P(BBG or BGB or GBB) = P(BBG) + P(BGB) + P(GBB) = 3/8 → etc. Probability Model:

More problems  An on-campus club has 10 members: 3 freshmen, 2 sophomores, 1 junior, and 4 seniors.  Random phenomenon: pick a club member at random (i.e. choose an SRS of size n = 1)  Random variable: let X = class rank of the chosen individual.  What is the probability model for X?  A club has 5 members: 3 men and 2 women.  Random phenomenon: choose an SRS of size 2 from the club.  Random variable: let N = number of women chosen.  What is the probability model for N?

Continuous sample spaces contain an infinite number of events. They typically are intervals of possible, continuously-distributed outcomes. Continuous sample space The probability of the uniformly-distributed variable Y to be within 0.3 and 0.7 is the area under the density curve corresponding to that interval. Thus: P(0.3 ≤ y ≤ 0.7) = (0.7 − 0.3)*1 = 0.4 This is a uniform density curve. There are a lot of other types of density curves. y Example: Pick a real number at random between 0 and 1 (e.g., 0.001, 0.4, 0.0063876). S = {interval containing all numbers between 0 and 1} How do we assign probabilities to events in an infinite sample space? We use density curves and compute probabilities for intervals.

Probability distribution for a continuous random variable Because the probability of drawing one individual at random depends on the frequency of this type of individual in the population, the probability is also the shaded area under the curve. The shaded area under the density curve shows the proportion, or percent, of individuals in the population with values of X between x 1 and x 2. % individuals with X such that x 1 < X < x 2

Intervals versus single outcomes The probability of a single outcome is meaningless for a continuous sample space. Only intervals can have a non-zero probability, represented by the area under the density curve for that interval. Suppose X is uniformly distributed on [0, 6] Density curve for X P(1 ≤ X ≤ 3) = (3 – 1)*(1/6) = 1/3 height = 1/6 0631 0631 P( X = 3) = P(3 ≤ X ≤ 3) = (3 – 3)*(1/6) = 0 P(1 ≤ X ≤ 3) = P(1 < X ≤ 3) = P(1 ≤ X < 3) = P(1 < X < 3) = 1/3

We generate two independent random variables X and Y uniformly distributed between 0 and 1. Let Z to be their sum. Z can take any value between 0 and 2. The density curve for Z is: Height = 1. We know this because the base = 2, and the area under the density curve must be 1. The area of a this triangle is ½ (base*height). What is the probability that Y is < 1? 012 Y What is the height of the triangle? What is the probability that Y < 0.5? What is the probability that 0.5 < Y < 1.5? 1/2 1/8

The probability distribution of many random variables is the normal distribution. This is the same distribution we first encountered in Chapter 3. We’re just thinking about it in a slightly different way. Probabilities with the normal distribution are calculated in exactly the same way as we computed proportions with the normal distribution. (Use normalCDF on the TI-83 or 84.) Normal probability distribution Example: Choose a woman at random from a certain population. Let X be the chosen woman’s height. Then X is a random variable.

Previously, we wanted to calculate the proportion of individuals in the population with a given characteristic. N (µ,  ) = N (64.5, 2.5) Distribution of women’s heights ≈ Example: What's the proportion of women with a height between 57" and 72"? That’s within ± 3 standard deviations  of the mean , thus that proportion is roughly 99.7%. Since about 99.7% of all women have heights between 57" and 72", the chance of picking one woman at random with a height in that range is also about 99.7%.

Assume the distribution of women’s heights in the previous slide. What is the probability, if we pick one woman at random, that her height is between 68” and 70” ? Because the woman is selected at random, X is a random variable. From the previous slide, we know X ~ N(  = 64.5,  = 2.5). So the probability that the selected woman is between 68’’ and 70’’ in height is P(68 < X < 70) = normalCDF(68, 70, 64.5, 2.5) = 0.06686 = about 6.7% Example Let X be the height of the woman who is selected.

We standardize normal data by calculating z-scores so that any Normal curve N(  ) can be transformed into the standard Normal curve N(0,1). Reminder: standardizing N (  ) N(64.5, 2.5) N(0,1) Standardized height (no units) →

Inverse problem: Your favorite chocolate bar is dark chocolate with whole hazelnuts. The weight on the wrapping indicates 8 oz. Whole hazelnuts vary in weight, so how can they guarantee you 8 oz. of your favorite treat? You are a bit skeptical... To avoid customer complaints and lawsuits, the manufacturer makes sure that 98% of all chocolate bars weight 8 oz. or more. The manufacturing process is roughly normal and has a known variability  = 0.2 oz. How should they calibrate the machines to produce bars with a mean  such that P(x < 8 oz.) = 2%?  = ? x = 8 oz. Lowest 2%  = 0.2 oz.

How should they calibrate the machines to produce bars with a mean m such that P(x < 8 oz.) = 2%?  = ? x = 8 oz. Lowest 2%  = 0.2 oz. x Plug in what we know: - 2.054 = (8 –  ) / 0.2 Solve for  :  = 8.4107 z = (x-  ) /  0 z = ??? Lowest 2% Standardize X z = invNorm(0.02, 0, 1) = - 2.054 Find z:

How should they calibrate the machines to produce bars with a mean m such that P(x < 8 oz.) = 2%?  = ? x = 8 oz. Lowest 2%  = 0.2 oz. x z = (x-  ) /  0 z = ??? Lowest 2% Thus, the chocolate bar should weigh μ = 8.41 oz on average.

Meaning of a probability We have several ways of defining a probability, and this has consequences on its intuitive meaning.  Theoretical probability  From understanding the phenomenon and symmetries in the problem  Example: Six-sided fair die: Each side has the same chance of turning up; therefore, each has a probability 1/6.  Example: Genetic laws of inheritance based on meiosis process.  Empirical probability  From our knowledge of numerous similar past events  Mendel discovered the probabilities of inheritance of a given trait from experiments on peas, without knowing about genes or DNA.  Example: Predicting the weather: A 30% chance of rain today means that it rained on 30% of all days with similar atmospheric conditions.

 Personal probability  From subjective considerations, typically about unique events  Example: Probability of a large meteorite hitting the Earth. Probability of life on Mars. These do not make sense in terms of frequency. A personal probability represents an individual’s personal degree of belief based on prior knowledge. It is also called Baysian probability for the mathematician who developed the concept.  We may say “there is a 40% chance of life on Mars.” In fact, either there is or there isn’t life on Mars. The 40% probability is our degree of belief, how confident we are about the presence of life on Mars based on what we know about life requirements, pictures of Mars, and probes we sent.  Our brains effortlessly (?) assess risks (probabilities) of all sorts, and businesses try to formalize this process for decision-making.

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