Module II – Probability and Random VariablesNormal distributionA normal curveNormally distributed variableDiscrete random variables and probability distributionsInterpretation of probability distributionsFormula for the mean of a discrete random variable:Binomial distributionBinomial DistributionBinomial Probability Formula:Module II – Probability and Random VariablesMendelian genetics and matingNormal distributionHeights(inches) Frequency(f) Rel. Freq56 - 57 3 0.000957 - 58 6 0.001858 - 59 26 0.00859 - 60 74 0.022760 - 61 147 0.04561 - 62 247 0.075762 - 63 382 0.11763 - 64 483 0.14864 - 65 559 0.171365 - 66 514 0.157566 - 67 359 0.1167 - 68 240 0.073568 - 69 122 0.037469 - 70 65 0.019970 - 71 24 0.007471 - 72 7 0.002172 - 73 5 0.001573 - 74 1 0.00033264 1The normal curve associated with a normal distribution is - Bell-shaped- Centered at - Close to the horizontal axis outside the range from 3 to 3A normal curveNormally distributed variableA variable is said to be a normally distributed variable or to have a normal distribution if its distribution has the shape of a normal curve.010020030040050060056 - 5758 - 5960 - 6162 - 6364 - 6566 - 6768 - 6970 - 7172 - 73Series1Series2For a normally distributed variable, the percentage of all possible observations that lie within any specified range equals the corresponding area under its associated normal curve, expressed as a percentageBinomial DistributionQuantitative skills: Sample space and events Probability and some rules of probability1. The Sample Space (S) associated with any experiment is the set of all possible outcomes that can occur as a result of the experiment. So naturally, we will call each element of the sample space an outcome.EXAMPLE 1: Consider the experiment of rolling a pair of fair dice. The figure below gives a representation of all the 36 equally likely outcomes of the sample space associated with this experiment.(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)(1,4) (2,4) (3,4) (4,4) (5,4) (6,4)(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)(1,6) (2,6) (3,6) (4,6) (5,6) (6,6)2. An event (E) is any subset of the sample space.3. The probability of an event E (written as P(E))in a sample space (S) with equally likely outcomes is given byP(E) = Sin outcomes ofnumber Ein outcomes ofnumber EXAMPLE 2: For the sample space in example 1, consider the eventE = {the sum of the faces is 7 or 3}.Then E = { (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), (1,2), (2,1) }Thus P(E) = Sin outcomes ofnumber Ein outcomes ofnumber = 92368Alternatively: If we letA = {sum of faces is 7} and B = {sum of faces is 3}Then, E = AB, and P(E) = P(AB) = P(A) +P(B) - P(AB) [ Additive Rule]= 920362366Observe: Let F = {the sum of faces is neither 7 nor 3}, i.e., F is the complement of E.Then, P(F) = 1 - P(E) [ Complement Rule] = 1 - 92 = 97Properties:1.1)(0  EP;2.1)( & ,0)(  SPP3. Odds for an event E = )()(EPEP = EinelementsofnumberEinelementsofnumber 4. Odds against E = )()(EPEP EinelementsofnumberEinelementsofnumber Conditional ProbabilityEXAMPLE 3: A regular deck of playing cards consists of 52 cards:13 clubs (black), 13 diamonds (red), 13 spades (black), 13 hearts (red).The 13 cards are labeled: Ace (A), 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack (J), Queen (Q), King (K).Consider the experiment of drawing a single card from the deck. The sample space associated with the experiment has 52 equally likely outcomes. Consider the event E = {a black ace is drawn}.Then we have,P(E) = 2/52.i.e., the probability of drawing a black ace is 1/26.However, suppose a card is drawn and we are informed that it is a club, then the question would be, ' what is the probability of drawing a black ace, given the information that the card drawn is a club' ? If F = {a club is drawn}, the question can be rephrased as ' what is the probability of E given F' ? This is symbolically written as: Find P(E | F)i.e., P(E | F) represents - the probability of the event E given the condition F.Clearly, the given condition reduces the size of the event E to 1 outcome, since there is only one black ace that is a club; the given condition also reduces the size of the sample space to 13 outcomes since there are 13 clubs. Thus, P(E | F) = 1/13Using the Formula:0)( ,)()()|(  FPFPFEPFEP 1315213521Note: 0)( ),()|()(  FPFPFEPFEP [ Product Rule]Independent EventsDefinition: Let E and F be two events of a sample space S with P(F) > 0.The event E is independent of the event F iff P(E | F) = P(E).Theorem: Let E, F be events for which P(E) > 0 and P(F) > 0. If E is independent of F, then F is independent of E.Test for Independence: Two events of a sample space S are independent iff0)( ),()()(  FPFPEPFEPEXAMPLE 4:A fair coin is tossed twice. Define the events E and F to beE: A head turns up on the first throw of a fair coin;F: A tail turns up on the second throw of a fair coin.Show that E and F are independent.Solution:E = {HH, HT}, and F = {HT, TT}..4/1)( },{  FEPthereforeHTFE, Also, 4/1)4/2).(4/2()()(  FPEPThus, events E and F are independent.Warning!: Mutually exclusive events are generally not independent.Discrete random variables and probability distributionsRandom Variables: Suppose a pair of dice are rolled. The value of the sum of thenumbers on the dice depends on chance. The ‘sum of the numbers on the dice’ istherefore called a random variable. A random variable is a quantitative variable whosevalue depends on chance. Another example is the number of siblings each student has in aclass. Discrete variable: A discrete variable is a variable whose possible values forms a finiteset or a countably infinite set of numbers. The variable ‘sum of the numbers on the dice’is a discrete variable. What are its possible value?Discrete random variables:A discrete random variable is a random variable whose possible values form a finite orcountably infinite set.Note: We usually use uppercase letters to denote random variables.Probability distribution: A listing of all the possible