Chapter 2Trials2.1 Independent, Identically Distributed TrialsIn Chapter 1 we considered the operation of a CM. Many, but not all, CMs can be operated morethan once. For example, a coin can be tossed or a die cast many times. By contrast, the next NFLseason will operate only once.In this chapter we consider repeated operations of a CM.Let us return to the ‘Blood type’ CM of Section 1. Previously, I described the situation asfollows: A man with AB blood and a woman with AB blood will have a child. The outcome isthe child’s blood type. The sample space consists of A, B and AB. I stated in Chapter 1 that thesethree outcomes are not equally likely, but that the ELC is lurking in this problem. We get the ELCby viewing the problem somewhat differently, namely as two operations of a CM.The first operation is the selection of allele that Dad gives to the child. The second operationis the selection of allele that Mom gives to the child. The possible outcomes are A and B and itseems reasonable to assume that these are equally likely. Consider the following display of thepossibilities for the child’s blood type.Allele from MomAllele from DadA BA A ABB AB BI am willing to make the following assumption.• The allele contributed by Dad (Mom) has no influence on the allele contributed by Mom(Dad).Based on this assumption, and the earlier assumption of the ELC for each operation of the CM,I conclude that the four entries in the cells of the table above are equally likely. As a result,we have the following probabilities for the blood type of the child: P (A) = P (B) = 0.25 andP (AB) = 0.50.Here is another example. I cast a die twice and I am willing to make the following assumptions.11• The number obtained on the first cast is equally likely to be 1, 2, 3, 4, 5 or 6.• The number obtained on the second cast is equally likely to be 1, 2, 3, 4, 5 or 6.• The number obtained on the first (second) cast has no influence on the number obtained onthe second (first) cast.The 36 possible ordered results of the two casts are displayed below, where, for example, (5, 3)means that the first die landed 5 and the second die landed 3. This is different from (3, 5).Number fromNumber from second castfirst cast 1 2 3 4 5 61 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)Just like in the blood type example, b/c of my assumptions, I conclude that these 36 possibilitiesare equally likely. We will do a number of calculations now.For ease of presentation, define X1to be the number obtained on the first cast of the die and letX2denote the number obtained on the second cast of the die.We call X1and X2random variables, which means that to each possible outcome of the CMthey assign a number. Every random variable has a probability distribution which is simply alisting of its possible values along with the probability of each value. Note that X1and X2havethe same probability distribution; a fact we describe by saying that they are identically distributed,Below is the common probability distribution for X1and X2.Value Probability1 1/62 1/63 1/64 1/65 1/66 1/6Total 1As we have seen, X1and X2each has its own probability distribution (which happens to be thesame). It is also useful to talk about their joint probability distribution which is concerned withhow they interact. I will illustrate this idea with a number of computations.P (X1= 3 and X2= 4) = 1/36,b/c, by inspection exactly one of the 36 ordered pairs in the earlier table have a 3 in the first positionand a 4 in the second position.12Before we proceed, I want to invoke my laziness again. It is too much bother to type, say,P (X1= 3 and X2= 4).It is much easier to type,P (X1= 3, X2= 4).We will follow this practice; a comma within a probability statement represents the word ‘and.’For future reference, note thatP (X1= 3, X2= 4) = 1/36, as does P (X1= 3)P (X2= 4).In words, the word ‘and’ within a probability statement tells us to multiply.Here is another example.P (X1≤ 4, X2≥ 4) = 12/36,b/c, as you can see from the table below, exactly 12 of the 36 pairs have the required property.X2X11 2 3 4 5 61 X X X2 X X X3 X X X4 X X X56Note again thatP (X1≤ 4, X2≥ 4) = 12/36 gives the same answer as P (X1≤ 4)P (X2≥ 4) = (4/6)(3/6) = 12/36.This last property is called the multiplication rule for independent random variables. Itis a very important result. It says that if we have two random variables that are independent,then we can compute joint probabilities by using individual probabilities. In simpler words, ifwe want to know the probability of X1doing something and X2doing something, then we cancalculate two individual probabilities, one for X1and one for X2and then take the product ofthese two individual probabilities. As we shall see repeatedly in this class, the multiplication rulefor independent random variables is a great labor saving device.The above ideas for two casts of a die can be extended to any number of casts of a die. Inparticular, define• X1to be the number obtained on the first cast of the die;• X2to be the number obtained on the second cast of the die;• X3to be the number obtained on the third cast of the die;13• and so on, in general, Xkis the number obtained on the ‘kth’ cast of the die.If we assume that the casts are independent–that is, that no outcome has any influence on anotheroutcome–then we can use the multiplication rule to calculate probabilities. Some examples arebelow.You might be familiar with the popular dice game Yahtzee. In this game, a player casts fivedice. If all dice show the same number, then the player has achieved a Yahtzee. One of the firstthings you learn upon playing the game Yahtzee is that the event Yahtzee occurs only rarely. Wewill calculate its probability.Let’s find the probability of throwing five 1’s when casting five dice. We write this as:P (1, 1, 1, 1, 1) = (1/6)5.Now, the probability of a Yahtzee is:P (Y1or Y2or Y3or Y4or Y5or Y6),where ‘Yk’ means all five dice land with the side ‘k’ facing up; in words, ‘Yk’ means a Yahtzee onthe number ‘k.’ Clearly all of the ‘Yk’ have the same probability. Thus, by Rule 3, the probabilityof a Yahtzee is:6(1/6)5= 1/1296 = 0.000772.There is a slicker way to calculate the probability of a Yahtzee. Imagine that you cast the diceone-at-a-time. (Don’t
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