Chapter 2 Trials 2 1 Independent Identically Distributed Trials In Chapter 1 we considered the operation of a CM Many but not all CMs can be operated more than once For example a coin can be tossed or a die cast many times By contrast the next NFL season 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 as follows A man with AB blood and a woman with AB blood will have a child The outcome is the child s blood type The sample space consists of A B and AB I stated in Chapter 1 that these three outcomes are not equally likely but that the ELC is lurking in this problem We get the ELC by 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 operation is the selection of allele that Mom gives to the child The possible outcomes are A and B and it seems reasonable to assume that these are equally likely Consider the following display of the possibilities for the child s blood type Allele from Dad A B Allele from Mom A B A AB AB B I 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 and P 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 on the 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 from first cast 1 2 3 4 5 6 1 1 1 2 1 3 1 4 1 5 1 6 1 Number from second cast 2 3 4 5 1 2 1 3 1 4 1 5 2 2 2 3 2 4 2 5 3 2 3 3 3 4 3 5 4 2 4 3 4 4 4 5 5 2 5 3 5 4 5 5 6 2 6 3 6 4 6 5 6 1 6 2 6 3 6 4 6 5 6 6 6 Just like in the blood type example b c of my assumptions I conclude that these 36 possibilities are equally likely We will do a number of calculations now For ease of presentation define X1 to be the number obtained on the first cast of the die and let X2 denote the number obtained on the second cast of the die We call X1 and X2 random variables which means that to each possible outcome of the CM they assign a number Every random variable has a probability distribution which is simply a listing of its possible values along with the probability of each value Note that X1 and X2 have the same probability distribution a fact we describe by saying that they are identically distributed Below is the common probability distribution for X1 and X2 Value 1 2 3 4 5 6 Total Probability 1 6 1 6 1 6 1 6 1 6 1 6 1 As we have seen X1 and X2 each has its own probability distribution which happens to be the same It is also useful to talk about their joint probability distribution which is concerned with how 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 position and a 4 in the second position 12 Before 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 that P 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 X1 1 2 3 4 5 6 1 X2 2 3 4 X X X X 5 X X X X 6 X X X X Note again that P 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 It is 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 if we want to know the probability of X1 doing something and X2 doing something then we can calculate two individual probabilities one for X1 and one for X2 and then take the product of these two individual probabilities As we shall see repeatedly in this class the multiplication rule for 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 In particular define X1 to be the number obtained on the first cast of the die X2 to be the number obtained on the second cast of the die X3 to be the number obtained on the third cast of the die 13 and so on in general Xk is 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 another outcome then we can use the multiplication rule to calculate probabilities Some examples are below You might be familiar with the popular dice game Yahtzee In this game a player casts five dice If all dice show the same number then the player has achieved a Yahtzee One of the first things you learn upon playing the game Yahtzee is that the event Yahtzee occurs only rarely We will calculate its probability Let s find the …
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