The sample mean is an unbiased estimator of the population mean. What do we expect the sample mean to be equal to when the population mean is equal to: 1. = 8? 2. = 0? 3. = 20? 4. = "DC 11 5. = - 40? 11 6. = 0.03?

in B and C obtaining incorrect pairs of bmts. Thus there are 5 allocations with the property that only A obtains the correct boots; similarly there are 5 allocations such that only B obtains the correct boots, and 5 such that only C. does. Adding to these the single allocation such that they all obtain the correct boots, we find that there are 16 allocations which result in at least one child obtaining the correct boots, so that the required probability is iii = 33?. (d) We may consider the red and yellow boots separately. The probability that the child with the yellow boots puts them on correctly is 3,. For the red boots, we consider a sample space .. consisting of all 4! russible allocations of the boots to the feet of the children concerned (in this part of the question the order is important). Only one such allocation is correct, so that the probability that the red-booted children are correctly shod is :1: 3;. Finally, assuming independence between the allocation of red and yellow boots, we obtain the probability that all three children are correctly shod as ix 1' = iii Notes (1} The Words 'at random' are frequently encountered in questions such as this. Strictly speaking, they imply only that the outcome cannot be predicted in advance. The interpretation we have chosen here is muoh stronger, implying, for parts (a) and (b), that each of the 90 possible outcomes has the same chance of occurring. Such an assumption of 'cqually likely outcomes' is in line with everyday usage in statements such as 'Winning Premium Bond numbers are chosen at random.' in solving probability problems, it is the natural one in the absence of any further information. and we shall frequently make it in our solutions to the problems which follow. (2) It is possible to solve parts (oi-(e) on the basis of a sample space in which all 6! possible ordered allocations of the hrmts to the children are equally likely. Then, for example. there are 3 {= 23) outcomes in which all three children obtain the correct boots. since each child can obtain his [or her) boots in two possible orders. (3) Some readers may prefer to tackle this problem by dividing each part into stages and using the multiplication laws of probability. We now present a solution along these lines. (a) We may imagine A having first choice of boots, followed by B and finally C. If we adopt a notation in which, for example. A1 denotes the event that the first boot chosen by it belongs to A. we require the probability of the event Aydglf'lgnfl'g. This probability may he expressed as Pflfi thfir'lz id 0131131 !.-t lnA alpriz id irm 2ml 1) (we have omitted We further conditional probabilities in order to shorten the above expression: they relate to CS 'choice', and are both equal to 1 since C has no choice at all}. Now Pr(d1)= 1, since A has initially 6 horns to choose from, of which 2 are correct; and Pr(.-t-2'A1 = ;, since at this stage A has 5 boots to choose from, of which 1 is correct. Similarly. Pr(Bl;AlA-_.) = I and FILE: nynaznal) = 3', Multiplying these probabilities. we obtain the solution 51-} , as before. (h) [f we let MA stand for the event that A. obtains a matching pair of boots. and So on, we require Pf[MhnMnnMc) = Pr(MA)Pr(MH:MA) (as above, the third component PrUll'c MAHMB) is equal to 1, and has therefore been omitted). Now Frill-{,0 is just the probability that the second hoot selected by A matches the first, i.e. ; similarly PrmeMA) = %. Multiplying these probabilities, we obtain the solution 4] 15 ' {c} We non.r let A denote the event that A obtains the correct pair of boots, and define a and C' similarly. We then require Prlrl UB UC}. Using the general addition law for three events, this probability may be expanded as PrfA) + P113) + Pr(C) Pr(A {"13} Pr(li' Flt?) H Frpt me} + Pr(.'t n3 C). Now, using the notation used in solving part (a), since A = Almtg, we obtain no) = Prtdill'rtzldt) = M a is Similarly, FILE) and Pr{C) have the same value. The four remaining terms that we require in order to obtain Prfrt U3 UC ) are all the same, and equal to PrUt BC) (since, for example, if A and B obtain the correct boots, so also must C). This probability has already been shown, in part (a), to be 315. Thus Pr(ALJBUC) = 3x115r 3x910 + 6'6 = %. (d) We shall use a notation in which, for example, LA will stand for the event that the boot worn by A on his (or her) left foot is the correct one. We may. without loss of generality. take it that A is the owner of the yellow boots. Dealing rst with these, we obtain Fromm = Proorrotlto = gxi, = %- Turning to the red btmts, we obtain Pr(L HnRBndeRc) = PT(LB)X PrUt'Bl Lax Pr{Lc|L3FlRB) X PffciLgnRBnLd _1 I 1 I_ 1 _ _x_x..xs - __ I. 1 2 l 2* Multiplying these two probabilities, we obtain the solution :5, as before. 121.2 Dealing four cards Four cards are dealt from a standard pack of 52 cards. Find (i) the probability that all four are spades; (ii) the probability that two or fewer are spades; (iii) the probability that all four are spades, giVen that the first two are spades; (iv) the probability that spades and hearts alternate. Solution We shall solve this problem using three different approaches. Of these, Method 1 is probably the most straightforward and appealing. Method 1 Let 5,. denote the event that the ith card dealt is a spade, and H,- the event that it is a heart. The required probabilities are then obtained as follows: (i) Ff(3]n31n53n34) = Pr(31)Pr(SgiSl)Pr(33 s,nsz)Pr(s,|s.n33nsn I3 12 11 10 52 x 51 x so x 49 11 4165' ll ll (ii) Rather than sum the probabilities of the three events 'no spades'. 'one spade', and 'two spades', it will be simpler to adopt the standard device of obtaining the probability of the event complementary to that specified in the problem, and then subtracting from 1. Adopting this approach, we have to calculate the probabilities of only two events; there is, indeed, a further reduction in effort, since we have already found one of these probabilities in the solution to part (i). Thus Pr(two or fewer are spades) = 1 - Pr(three are spades) - Pr(all four are spades), and, since the second of the probabilities is known, we need only the following: Pr(three are spades) = Pr(S,nSans;n54) + Pr(Sins205;n54) + Pr (Sins 205;n54) + Pr( Sinszns 354). The first of these four probabilities is 13 12 11 39 Pr(S,)Pr($2|S,)Pr($3|S,nS2)Pr( 54|Sins2n$3) = 57* 51 50 49 Similarly, the remaining three probabilities are 13 12 39 11 13 39 12 11 39 , 13 12 11 52 51 50 49 52 51 50 49 52 51 50 49 respectively. Combining these results with the solution to part (i), we obtain Pr(two or fewer are spades) = 1 _ (13x12x11x10) + (4x13x12x11x39) 52x51x50X49 19 912 20 825 (iii) Pr(SInSanS ;nS4|Sins2) = Pr($3|Sins2)Pr($4|Sins2nS;) 11 10 50 *49 11 245 (iv) The required probability is composed of two parts, Pr(S,nH2nS;(H4) Pr(HinS2nH3054). The first of these probabilities is Pr($1)Pr(H2)$1)Pr($3|SinH2)Pr(H4 \\SinH2nS3) = 13 13 12 12 52 51 50 49 and the second is found, similarly, to have the same value. The required probability is thus 2x132x 122 156 52x51 x50X49 20 825 Method 2 In this method we base our solutions on the enumeration of appropriate outcomes in a sa space consisting of all possible hands of four cards, taking the order in which the cards are into account. This sample space consists of 52x51 x50x49 outcomes - all equally likely. (i) The number of different hands consisting of four spades is 13x12x11 x10. Hence probability that a hand consists entirely of spades is 2x51x50x49 , as before. (ii) As in Method 1, it is simpler to count the number of hands with three or four spades subtract. The number of different hands consisting of three spades followed by one card w is not a spade is 13X12x11 x39; considering the three other positions in which the card w is not a spade can appear, we see that the number of relevant hands is the same for Hence the total number of hands containing three or four spades is (13x12x11x10 (4x13x12x11x39), from which we can deduce the same answer as that obtained Method 1. (iii) In order to obtain the required conditional probability, we must find the probability the first two cards are spades. Since the number of hands with this property is 13x 12 x50 it follows that the corresponding probability is 13x12. $ 52x51 . Dividing the probability obtaineanswer to (i) by this quantity. we obtain the required result. (iv) The number of possible hands in which spades and hearts alternate. first position, is ixlxllxll; the number of hands in which spades a. with a heart in the first position, is the same. We thus obtain the same probability as was obtained by Method 1. Method .'l In this method (which can he used only for parts (i) and (ii) of the problem) we again consider a sample space of equally likely fourcard hands, but do not take the order in which the cards are dealt into account. There are thus [542] outcomes in the sample space. {i} There are [143] possible hands consisting of four spades. The probability of such a hand is thus which, after appropriate cancellations, reduces to the solution already obtained. (ii) To count the number of hands containing three spades and one card of another suit, we note that there are 39 possibilities for the latter card, and that each of these may be combined with [1;] possible combinations of three spades. The total number of hands is thus 39x [[33] and, dividing by (542] and cancelling, we obtain the same answer as before. Notes (1) In Method 1, the introduction of an appropriate notatiou for the \"cuts of interest makes it possible to write out a fairly concise, yet clear, solution. While one might prefer a solution where the events are described verbally, this becomes somewhat lengthy. Consider, for example, part (i): Pr(all four cards are spades) = PrUirst card is a spade) x Pr(sccond card is a spade first card is a spade) x Pr(third card is a spade i first two cards are spades) x Pr(fourth card is a spade l first three cards are spades) \"HACKS. On the other hand, writing down Pr(SSSS}, without explaining what this means, is not enough; and the inadequacy of such a notation becomes more evident when we start writing out a solution, involving meaningless terms like Pr{S 55}. (See Note 5 to Problem 1A.\") and Note 1 to Problem 18.3 for a discussion of similar points.) (2) [n solving part (ii) by Method 1 it is tempting, in finding the probability of three spades, to obtain Pr(31flszS3I'iS4) and then multiply by four, on the grounds that there are four possible positions for the card which is not a spade. What is more, this argument works! However, although the argument is valid for similar questions involving the rolling of dice, or the random selection of cards with replacement [and is fundamental to the derivation of the binomial distribution), it is not valid in situations where selection is performed without reptiles-merit. In the present case note that, although the four components which are summed to obtain the probability of three spades are equal, the probabilities which are multiplied to obtain them are not. A slightly more complicated argument, such as that we have given, is therefore necessary. (See Note 2 to Problem 13.3 for a similar point, and for mention of a problem for which the simpler argument does not giVe the correct answer.) (3) When we are dealing with a sample space in which all outcomes are equally likely, we may 1154: the fact that Pr{A on = 1%?)11, where n denotes the number of outcomes in the indicated event. Thus, in solving part (iii) by Method 2, we have u(3152r'133n54}=13x12x11xl, and \"(Sl32}=13X12X50X3-19, leading to the same result as before. In (4) As a further variant on the Method 2 solution to part (iii). we may consider a reduced sample space in which, given that the first two cards dealt are spades, there are 50x49 possibilities for the next two cards dealt all of them equally likely. Of these, llxl consist of two spades, so that the required conditional probability is 35; , as before. This argument is rather less straightfonyard than it may seem. We are, in fact, conditioning on the first two cards being a particular pair of spades (the three and the queen, say). Since the probability obtained does not depend on which pair we are conditioning on, it is also the probability conditional on the first two cards being any pair of spades. t (5] While the approach of Method 1 is probably the simplest for this problem, that of Method 3 can be more easily extended to cover similar problems where larger numbers are involved. For example, the probability that, in a hand consisting of ten cards, six are spades is, by an argument similar to that used for part (ii), 13 39 ls l (4 l 52 ' [10] We are dealing here with the hypergeometric distribution, which is discussed in greater detail in Problem ZAJ l . 1&3 The school assembly At the morning assembly, five schoolchildren Alan, Barbara, Clare, Daniel and Edward sit down in a row along with five other children (whose names need not concern us here). If the children arrange themselves at random, find the probabilities of the following events. (a) Alan and Barbara sit together. (b) Clare, Daniel and Edward sit together. {c} Clare, Daniel and Edward sit together but Alan and Barbara sit apart. (d) Daniel sits between Clare and Edward (but not necessarily adjacent to either of them). The sample mean is an unbiased estimator ofthe population mean. What do we expect the sample mean to be equal to when the population mean is equal to: #- l. = B? J\". 2. = D? J\". 3. = 20? J\". 4. = ,1}. oc ? J\". 5. = - 4:)? p. 6. = {1333? \f