Solution Manual For Probability, Statistics, and Random Processes for Engineers, 4th Edition
- Statistics
- 343 pages
- Shared May 2025
Get quick and accurate answers to your textbook questions with Solution Manual For Probability, Statistics, and Random Processes for Engineers, 4th Edition, a detailed guide to solving every exercise.
Preview 16 of 343 pages
See the first 16 pages exactly as they appear in the document.
Page 1 Read page 1 as text
1 Solutions to Chapter 1 1. The intent of this rather vague problem is to get you to compare the two notions, probability as intuition and relative frequency theory. There are many possible answers to how to make the statement "Ralph is probably guilty of theft" have a numerical value in the relative frequency theory. First step is to de fi ne a repeatable experiment along with its outcomes. The favorable outcome in this case would be ’guilty.’ Repeating this experiment a large number of times would then give the desired probability in a relative frequency sense. We thus see that it may entail a lot of work to attach an objective numerical value to such a subjective statement, if in fact it can be done at all. One possible approach would be to look through courthouse statistics for cases similar to Ralph’s, similar both in terms of the case itself and the defendant. If we found a su ffi ciently large number of these cases, ten at least, we could then form the probability = , where is the number of favorable (guilty) verdicts, and is the total number of found cases. Here we e ff ectively assume that the judge and jury are omniscient. Another possibility is to fi nd a large number of people with personalities and backgrounds similar to Ralph’s, and to expose them to a very similar situation in which theft is possible. The fraction of these people that then steal in relation to the total number of people, would then give an objective meaning to the phrase "Ralph is probably guilty of theft." 2. Note that → 3 , but 3 6 → , i.e., implies 3 but not the other way around. Thus if we turn over card 2 and fi nd a 3 . So what? It was never stated that a 3 → . Likewise, with card 3. On the other hand, if we turn over card 4 and fi nd a , then the rule is violated. Hence, we must turn over card 4 and card 1, of course. 3. First step here is to decide which kind of probability to use. Since no probabilities are explicitly given, it is reasonable to assume that all numbers are equally likely. E ff ectively we assume that the wheel is “fair." This then allows us to use the classical theory along with the axiomatic theory to solve this problem. Now we must fi nd the corresponding prob- ability model. We are told in the problem statement that the experiment is “spinning the wheel." We identify the pointed-to numbers as the outcomes . The sample space is thus Ω = { 1 2 3 4 5 6 7 8 9 } The total number of outcomes is then 9. The probability of each elemental event { } is then taken as [ { } ] , = 1 9 , as in the classical theory. We are also told in the problem statement that the contestant wins if an even number shows. The set of even numbers in Ω is { 2 4 6 8 } We can write this event as a disjoint union of four singleton (atomic) events { 2 4 6 8 } = { 2 } ∪ { 4 } ∪ { 6 } ∪ { 8 } Now we can apply axiom 3 of probability to write [ { 2 4 6 8 } ] = [ { 2 } ] + [ { 4 } ] + [ { 6 } ] + [ { 8 } ] = 1 9 + 1 9 + 1 9 + 1 9 = 4 9 We have seen that some ’reasonable’ assumptions are necessary to transform the given word problem into something that exactly corresponds to a probability model. It turns out that this is a general problem for such word problems, i.e. problems given in natural English.
Page 2 Page 3 Read page 3 as text
2 4. The experiment involves fl ipping a fair coin 3 times. The outcome of each coin toss is either a head or a tail. Therefore, the sample space of the combined experiment that contains all the possible outcomes of the 3 tosses, is given by Ω = { } Since all the coins are fair, all the outcomes of the experiment are equally likely. The proba- bility of each singleton event, i.e. an event with a single outcome, is then 1 8 . We are interested in fi nding the probability of the event , which is the event of obtaining 2 heads and 1 tail. There are 3 favorable outcomes for this event given by = { } . Therefore, [ ] = [ { } ∪ { } ∪ { } ] = [ ] + [ ] + [ ] = 3 8 . Note that we are able to write the probability of the event as the sum of probability of the singleton events (from Axiom 3) because the singlteon events of any experiment are mutually exclusive. Why? 5. The experiment contains drawing two balls (with replacement) from an urn containing balls numbered 1 , 2 , and 3 . The sample space of the experiment is given by Ω = { 11 12 13 21 22 23 31 32 33 } The event of drawing a ball twice is said to occur when one of the outcomes 11 , 22 , or 33 occurs. Therefore, the event of drawing 2 equal balls , is given by = { 11 22 33 } and [ ] = [ { 11 } ]+ [ { 22 } ]+ [ { 33 } ] Since the balls are drawn at random, it can assumed that drawing each ball is equally likely. Therefore, the singleton events, or equivalently outcomes of the experiment, are equally likely. Hence, [ ] = 3( 1 9 ) = 1 3 . 6. Let 1 2 6 represent the six balls. Each outcome will be represented by the two balls that were drawn. In the fi rst experiment, the balls are drawn without replacement; hence, the two balls drawn cannot have the same index. Then the sample space containing all the outcomes is given by Ω 1 = { 1 2 1 3 1 4 1 5 1 6 2 1 2 3 2 4 2 5 2 6 3 1 3 2 3 4 3 5 3 6 4 1 4 2 4 3 4 5 4 6 5 1 5 2 5 3 5 4 5 6 6 1 6 2 6 3 6 4 6 5 } This can be written compactly as Ω 1 = { ( ) | 1 ≤ ≤ 6 1 ≤ ≤ 6 6 = } If the fi rst ball is replaced before the second draw, then in addition to the outcomes in the earlier part, there are outcomes where both the two balls drawn are the same. The sample space for the new experiment is given by Ω 2 = Ω 1 ∪ { 1 1 2 2 3 3 4 4 5 5 6 6 } This can also be written as Ω 2 = { ( ) | 1 ≤ ≤ 6 1 ≤ ≤ 6 } .
Page 4 Read page 4 as text
3 7. Let be the height of the man and be the height of the woman. Each outcome of the experiment can be expressed as a two-tuple ( ) . Thus (a) The sample space Ω is the set of all possible pairs of heights for the man and woman. This is given as Ω = { ( ) : 0 0 } (b) The event , which is a subset of Ω is given by = { ( ) : 0 0 } 8. The word problem describes the physical experiment of drawing numbered balls from an urn. We need to fi nd a corresponding mathematical model. First we form an appropriate event space with meaningful outcomes. Here the physical experiment is ’draw ball from urn,’ so the outcome in words is ’particular labeled ball drawn,’ which we can identify with its label. So we select as outcome in our mathematical model, the number on the drawn ball’s face, i.e. the particular label. The outcomes are thus the integers 1,2,3,4,5,6,7,8, 9, and 10. The sample space is then Ω = { 1 2 3 4 5 6 7 8 9 10 } and is the set of all ten outcomes. We are told that is ’the event of drawing a ball numbered no greater than 5.’ Thus we de fi ne in our event fi eld = { 1 2 3 4 5 } The other event speci fi ed in the word problem is ’the event of drawing a ball greater than 3 but less than 9.’ In our mathematical event fi eld this corresponds to = { 4 5 6 7 8 } Having constructed our sample space with indicated events, we can use elementary set theory to determine the following answers: = { 6 7 8 9 10 } = { 1 2 3 9 10 } = { 4 5 } ∪ = { 1 2 3 4 5 6 7 8 } = { 1 2 3 } = { 6 7 8 } ∪ = { 1 2 3 6 7 8 9 10 } ( ) ∪ ( ) = { 1 2 3 6 7 8 } ( ) ∪ ( ) = { 4 5 9 10 } ( ∪ ) = { 9 10 } ( ) = { 1 2 3 6 7 8 9 10 } The last part of the problem asks us to ’express these events in words.’ Since we have a mathematical model, we should really more precisely ask what each of these events corresponds to in words . We know of course that corresponds to ’drawing a ball numbered no greater than 5.’ We can thus loosely write = { 0 drawing a ball numbered no greater than 5’}, although in our mathematical model is just the set of integers { 1 2 3 4 5 } . So when we write = { ’drawing a ball numbered no greater than 5’}, what we really mean is that the event in our mathematical model corresponds to the physical event ’drawing a ball numbered no greater than 5’ mentioned in the word problem. With this caveat in mind, we can then write: = { 0 drawing a ball greater than 5 0 } = { 0 drawing a ball not in the range 4-8 inclusive 0 } = { 0 drawing a ball greater than 3 and no greater than 5 0 } etc. 9. The sample space containing four equally likely outcomes is given by Ω = { 1 2 3 4 } . Two events = { 1 2 } and = { 2 3 } are given. The required events can be easily obtained by observation.
Page 5 Read page 5 as text
4 = set of outcomes in and not in = { 1 } . = set of outcomes in and not in = { 3 } . = set of outcomes in and = { 2 } . ∪ = set of outcomes in or in = { 1 2 3 } . 10. = ∪ This can be proved using the distributive law on = Ω = ( ∪ ) = ∪ ∪ = ( ) ∪ ( ) ∪ ( ) Here we fi rst write = ( ∪ ) and = ( ∪ ) Then we can write ∪ = ( ( ∪ )) ∪ ( ( ∪ )) = ( ∪ ) ∪ ( ∪ ) = ∪ ∪ ∪ = ∪ ∪ using the above laws and formulas. Notice that the above two decompositions are into disjoint sets. From the third axiom of probability, we know that the probability of union of disjoint sets is the sum of the probabilities of the disjoint sets. Therefore, we can add the probabilities over the unions. 11. In a given random experiment there are four equally likely outcomes 1 2 3 and 4 Let the event , { 1 2 } [ ] = [ { 1 2 } ] = [ { 1 } ] + [ { 2 } ] = 1 4 + 1 4 = 1 2 = { 3 4 } [ ] = [ { 3 4 } ] = [ { 3 } ] + [ { 4 } ] = 1 4 + 1 4 = 1 2 Note that we are told that the four outcomes are equally likely. This means that the four singleton (atomic) events have equal probability. [ ] = 1 2 = 1 − [ ] = 1 − 1 2 12. (a) The three axioms of probability are given below (a) [label=()] (b) For any event , the probability of the even occuring is always non-negative. [ ] ≥ 0 This ensures that probability is never negative. (c) The probability of occurence of the sample space event Ω is one. [ Ω ] = 1 This ensures that probability of no event exceeds one. The fi rst two axioms ensures that the probability is a quantity between 0 and 1 , inclusive. (d) For any two events that are disjoint, the probability of the union of the events is the sum of the probabilities of the two events. [ ∪ ] = [ ] + [ ] when = This axiom tells us that the probability of any event can be obtained by the sum disjoint events that constitute the event.
Page 6 Read page 6 as text
5 (b) The event ∪ can be obtained as the disjoint union of the three sets . Hence by applying the third axiom of probability, we obtain [ ∪ ] = [ ∪ ( ∪ )] = [ ] + [ ∪ ] = [ ] + [ ] + [ ] Now the event can be written as the disjoint union of and (Axiom 3). Therefore [ ] = [ ] + [ ] = ⇒ [ ] = [ ] − [ ] Similarly [ ] = [ ] + [ ] = ⇒ [ ] = [ ] − [ ] Therefore [ ∪ ] = [ ] + ( [ ] − [ ]) + ( [ ] − [ ]) = [ ] + [ ] − [ ] . 13. We fi rst form our mathematical model by setting outcomes ς = ( 1 2 ) where 1 corresponds to the label on the fi rst ball drawn, and 2 corresponds to the label on the second ball drawn. We can also write the outcomes as strings ς = 1 2 The sample space Ω can then be identi fi ed with the 2-D array 11 12 13 14 15 21 22 23 24 25 31 32 33 34 35 41 42 43 44 45 51 52 53 54 55 There are thus 25 outcomes in the sample space. Now the word problem statement uses the phrase ’at random’ to describe the drawing. This is a technical term that can be read ’equally likely.’ Thus all the elementary events { 1 2 } in our mathematical model must have equal probability, i.e. [ { 1 2 } ] = 1 25 Armed thusly we can attack the given problem as follows. De fi ne the event = {’sum of labels equals fi ve’}, or precisely = { 41 32 23 14 } Then we decompose this event into four singleton events as = { 41 } ∪ { 32 } ∪ { 23 } ∪ { 14 } Since di ff erent singleton events are disjoint, probability adds, and we have [ ] = 1 25 + 1 25 + 1 25 + 1 25 = 4 25 "Dim" ignored that outcome is di ff erent (distinguishable) from outcome . "Dense" talked about the sums and correctly noted that there were nine of them. However, he incorrectly assumed that each sum was equally likely. Looking at our sample space above, we can see that the sum 2 has only one favorable outcome 11, while the sum 6 has fi ve favorable outcomes, just looking at the anti-diagonals of this matrix. 14. First we show ∩ ( ∪ ) ⊂ ( ∩ ) ∪ ( ∩ ) . Let ∈ ∩ ( ∪ ) . Then ∈ and ∈ ( ∪ ) . ∈ and ∈ or ∈ .
Page 7 Read page 7 as text
6 Say if ∈ . Then ∈ and ∈ (Step k) Thus ∈ ( ∩ ) . And therefore ∈ ( ∩ ) ∪ ( ∩ ) . Similar arguments can be made if we consider ∈ in step k, in which case we will show that ∈ ( ∩ ) and hence ∈ ( ∩ ) ∪ ( ∩ ) . Thus we have shown that ∩ ( ∪ ) ⊂ ( ∩ ) ∪ ( ∩ ) . Now we show that ( ∩ ) ∪ ( ∩ ) ⊂ ∩ ( ∪ ) . Suppose ∈ ( ∩ ) ∪ ( ∩ ) . Then ∈ ( ∩ ) or ∈ ( ∩ ) . Say ∈ ( ∩ ) Then ∈ and ∈ . Or ∈ and ∈ ( ∪ ) . Or in other words, ∈ ∩ ( ∪ ) . Similar arguments can be used to show that if ∈ ( ∩ ) , then ∈ ∩ ( ∪ ) . Thus ( ∩ ) ∪ ( ∩ ) ⊂ ∩ ( ∪ ) . Thus we have shown that both sets are contained in each other. Hence ∩ ( ∪ ) = ( ∩ ) ∪ ( ∩ ) . 15. We use the set identity Ω = ∪ Since this union is disjoint, by the additivity of probability (i.e. axiom 3), we get 1 = [ Ω ] = [ ] + [ ] which with rearranging becomes the desired result. 16. (a) ∩ = { 1 2 } ∩ { 4 5 6 } = . Therefore, [ ∩ ] = [ ] = 1 − [ Ω ] ( ∵ Ω ∩ Φ = 1 = [ Ω ∪ ] = [ Ω ] + [ ]) = 1 − 1 ( because [ Ω ] = 1) = 0 (b) [ ∪ ∪ ] = [ { 1 2 } ∪ { 2 3 } ∪ { 4 5 6 } = [ { 1 2 3 4 5 6 } ] = [ Ω ] = 1 . (c) We see that ∩ = and so [ ] = 0 . For and to be independent, [ ] = [ ] [ ] . Therefore, if either [ ] = 0 or [ ] = 0 or both are zeros, and will be independent. 17. This problem uses only set theory and just two axioms of probability to get these general results. (a) We need to show [ ] = 0 We write the disjoint decomposition Ω = Ω ∪ and then use the additivity of probability (axiom 3) to get [ Ω ] = [ Ω ∪ ] = [ Ω ] + [ ] So we must have [ ] = 0 (b) Using set theory, we can write the disjoint decomposition = ∪ Then by axiom 3, the additivity of probability, we have [ ] = [ ∪ ] = [ ] + [ ]
Page 8 Read page 8 as text
7 or what is the same [ ] = [ ] − [ ] (c) Here we simply note ∪ = Ω is a disjoint decomposition, so that again by axiom 3, [ Ω ] = [ ] + [ ] = 1 by axiom 2, which is the same as [ ] = 1 − [ ] 18. The outcome is the result of a probabilistic experiment. An event is a collection (set) of outcomes. The fi eld of events is the complete collection of events that are relevant for the given probability problem. 19. We start with the mutually exclusive decomposition ∪ = ∪ ∪ yielding [ ∪ ] = [ ] + [ ] + [ ] Then consider the two simple disjoint decompositions = ∪ and = ∪ which yield [ ] = [ ] + [ ] and [ ] = [ ] + [ ] Putting them all together, we have [ ∪ ] = [ ] + [ ] + [ ] = ( [ ] − [ ]) + [ ] + ( [ ] − [ ]) = [ ] + [ ] − [ ] 20. From Eq. 1.4-3, we see that ⊕ = ( − ) ∪ ( − ) = ∪ . We see that and are disjoint, i.e., ( ) ∩ ( ) = . Therefore, the probability of the union of and are the sum of the probabilities of the two events. In other words, ( ⊕ ) = ( ∪ ) = ( ) + ( ) 21. We have already (Problem 17) seen that we can write [ ] = [ ] − [ ] and [ ] = [ ] − [ ] . Therefore, ( ⊕ ) = ( ) + ( ) = [ ] + [ ] − 2 [ ] . 22. (a) For simplicity associate as follows: cat=1, dog=2, goat=3, and pig=4. The outcomes then become the integers 1,2,3, and 4. The sample space Ω = { 1 2 3 4 } For probability information we are given: [ { 1 2 } ] = 0 9 [ { 3 4 } ] = 0 1 [ { 4 } ] = 0 05 and [ { 2 } ] = 0 5 Now for every event in our fi eld of events, we must be able to specify the probability. This is equivalent to being able to supply the probability for all the singleton events. To see if we can do this, we note that singleton events {1} and {3} are missing probabilities, so we fi rst write { 1 } = { 1 2 } − { 2 } so that [ { 1 } ] = [ { 1 2 } ] − [ { 2 } ] = 0 9 − 0 5 = 0 4
Page 9 Read page 9 as text
8 Doing the same for the other missing singleton probability [ { 3 } ] , we write { 3 } = { 3 4 } − { 4 } so that [ { 3 } ] = [ { 3 4 } ] − [ { 4 } ] = 0 1 − 0 05 = 0 05 Thus we have enough probability information for all the singleton events, and hence all 16 = 2 4 subsets of Ω = { 1 2 3 4 } The appropriate fi eld F of events then consists of the following events along with their probabilities: { 1 } [ { 1 } ] = 0 4 { 2 } [ { 2 } ] = 0 5 { 3 } [ { 3 } ] = 0 05 { 4 } [ { 4 } ] = 0 05 { 1 2 } [ { 1 2 } ] = 0 9 { 1 3 } [ { 1 3 } ] = 0 45 { 1 4 } [ { 1 4 } ] = 0 45 { 2 3 } [ { 2 3 } ] = 0 55 { 2 4 } [ { 2 4 } ] = 0 55 { 3 4 } [ { 3 4 } ] = 0 1 { 1 2 3 } [ { 1 2 3 } ] = 0 95 { 1 2 4 } { 1 2 4 } ] = 0 95 { 1 3 4 } [ { 1 3 4 } ] = 0 5 { 2 3 4 } [ { 2 3 4 } ] = 0 6 { 1 2 3 4 } (= Ω ) [ { 1 2 3 4 } ] = 1 = [ Ω ] [ ] = 0 (b) Now the above is not an appropriate fi eld of events if some of the events do not have known probabilities. So if [ ’pig’ = { 4 } ] = 0 05 is removed, then we cannot determine the probabilities of some of the above events. In particular we cannot fi nd [ { 3 } ] The alternative then is to treat { 3 4 } , whose probability is still given, as a singleton and form a smaller fi eld with just the 8 events formed by unions of {1}, {2}, and {3,4}. The resulting fi eld, along with its probabilities is as follows: { 1 } [ { 1 } ] = 0 4 { 2 } [ { 2 } ] = 0 5 { 1 2 } [ { 1 2 } ] = 0 9 { 3 4 } [ { 3 4 } ] = 0 1 { 1 3 4 } [ { 1 3 4 } ] = 0 5 { 2 3 4 } [ { 2 3 4 } ] = 0 6 { 1 2 3 4 } (= Ω ) [ { 1 2 3 4 } ] = 1 = [ Ω ] [ ] = 0 23. First we show that ∪ ( ∩ ) ⊂ ( ∪ ) ∩ ( ∪ ) . Suppose ∈ ∩ ( ∪ ) Then ∈ Therefore ∈ ( ∪ ) , and ∈ ( ∪ ) Hence, ∈ ( ∪ ) ∩ ( ∪ ) . Now we show that ( ∪ ) ∩ ( ∪ ) ⊂ ∪ ( ∩ ) . Suppose ∈ ( ∪ ) ∩ ( ∪ )
Page 10 Read page 10 as text
9 Then ∈ ( ∪ ) and ∈ ( ∪ ) ∈ and ∈ If ∈ , then ∈ ∪ ( ∩ ) (because ⊂ ( ∪ ( ∩ )) ) If ∈ , then ∈ and ∈ . Or in other words, ∈ ( ∩ ) ∈ ∪ ( ∩ ) . Thus we have shown that both the sets are contained in each other. Therefore, ∪ ( ∩ ) = ( ∪ ) ∩ ( ∪ ) . 24. The probability of is [ ] = [ { 1 2 } ] = [ { 1 } ] + [ { 2 } ] = 1 4 + 1 4 = 1 2 The event (set) in terms of the outcomes is = { 3 4 } The probabilty of is [ ] = [ { 3 4 } ] = [ { 3 } ] + [ { 4 } ] = 1 4 + 1 4 = 1 2 Note that we are told that the four outcomes are equally likely. This means that the four singleton (atomic) events have equal probability. We verify [ ] = 1 2 = 1 − [ ] = 1 − 1 2 25. The composition of the urn is: ( ) , ( ) , ( ) , ( ) , ( ) , ( ) , ( ) , ( ) . [ ] = 6 8 , [ ] = 6 8 , [ ] = = 4 8 is not equal to [ ] [ ] = 9 16 . Therefore and are not independent. 26. Let = 1 2 represent the outcome of the th toss. Since the tosses are independent: [ 1 2 ] = [ 1 ] [ 2 ] = 1 6 · 1 6 [ 1 + 2 = 7 | 1 = 3] = [ 2 = 4 | 1 = 3] = [ 1 = 3 2 = 4] [ 1 = 3] = [ 1 = 3] [ 2 = 4] [ 1 = 3] (because tosses are independent) = 1 6 27. Clearly [ ] = 4 52 and [ ] = 26 52 = 1 2 Then [ ] = [ { pick one of two red aces in 52 cards } ] = 2 52 Is [ ] = [ ] [ ]? Now [ ] = 2 52 = 4 52 1 2 = [ ] [ ] so, yes and are independent events. 28. Since it is a fair die, the successive tosses are independent with probability = 1 6 for each face. From the provided information, we equivalently want the probability of getting a total of 5 on the two remaining tosses. This can happen in just 4 equally likely outcomes, i.e. (4,1), (3,2), (2,3), and (1,4). The desired probability this then 4 36 = 1 9 .
Page 11 Read page 11 as text
10 29. We can look at the compound outcomes ς = ( 1 2 ) as corresponding to the locations in the 9 × 9 array 11 21 31 41 51 61 · · · 91 12 22 32 42 52 · · · . . . 13 23 33 43 · · · 14 24 34 · · · 15 25 · · · . . . 16 · · · . . . . . . . . . 19 · · · · · · 99 with 81 equally likely outcomes. We agree to call the sample space for the fi rst experiment Ω 1 , the sample space for the second experiment Ω 2 , and the compound sample space simply Ω To get the sum Σ , 1 + 2 = 7 we need one of the following outcomes 16 25 34 43 52 61 located on a 45 ◦ diagonal in the above table. So there are 6 favorable outcomes for the event { Σ = 7 } The event { Σ = odd } contains 40 outcomes and the event { Σ = even } contains the remaining 81 − 40 = 41 even-sum outcomes. Now the joint event { Σ = 7 } ∩ { Σ = odd } = { Σ = 7 } since the sum 7 is an odd number. We can now calculate the needed probabilities [ { Σ = odd } ] = 40 81 and [ { Σ = 7 } ] = 6 81 The answer for the fi rst question is then [ { Σ = 7 }|{ Σ = odd } ] = [ { Σ = 7 } ∩ { Σ = odd } ] [ { Σ = odd } ] (by de fi nition) = [ { Σ = 7 } ] [ { Σ = odd } ] (by above result) = 6 40 The next question is to fi nd [( { 1 7 } × Ω 2 ) ∪ ( Ω 1 × { 2 7 } ) |{ Σ 10 } ] For simplicity of notation, let’s agree to write the compound events { 1 7 } × Ω 2 and Ω 1 × { 2 7 } as simply { 1 7 } and { 2 7 } respectively, for the rest of this calculation. So we must count the relevant number of outcomes from the above 9 × 9 array, where the various sums are found on 45 ◦ diagonals. For the event { Σ 10 } we count 36 outcomes. For the joint event ( { 1 7 } ∪ { 2 7 } ) ∩ { Σ 10 } , we fi nd it easier to consider the set of outcomes that make up the remainder of the event { Σ 10 } i.e. the event { 1 ≤ 7 } ∩ { 2 ≤ 7 } ∩ { Σ 10 } which is equal, in words, to the event ’ 1 ≤ 7 and 2 ≤ 7 and Σ 10 ’. We could call this the complement with respect to { Σ 10 } of the event ( { 1 7 } ∪ { 2 7 } ) ∩ { Σ 10 } Anyway, we fi nd from the 9 × 9 array that the number of outcomes in { 1 ≤ 7 } ∩ { 2 ≤ 7 } ) ∩ { Σ 10 } is composed of the following 10 cases: P = 11 = 6 + 5 = 5 + 6 = 7 + 4 = 4 + 7 and P = 12 = 5 + 7 = 7 + 5 = 6 + 6 and P = 13 = 6 + 7 = 7 + 6 and P = 14 = 7 + 7 So we subtract these 10 outcomes from the 36 outcomes in the event { Σ 10 } to obtain 26 outcomes in the compound event ( { 1 7 } ∪ { 2 7 } ) ∩ { Σ 10 } The relevant probabilities are then [ { Σ 10 } ] = 36 81 and [( { 1 7 } ∪ { 2 7 } ) ∩ { Σ 10 } ] = 26 81
Page 12 Read page 12 as text
11 The desired conditional probability is then [( { 1 7 } ∪ { 2 7 } ) |{ Σ 10 } ] = 26 81 36 81 = 26 36 ≈ 0 72 Finally to compute [ { Σ = odd }|{ 1 8 } ] we proceed as follows. For the combined experiment, we know there is only one possibility for 1 8 and that is 1 = 9 , along with any value for 2 Thus there are 9 outcomes in the compound event { 1 8 } 1 , so that it’s probability is 9 81 Now the joint event { Σ = odd } ∩ { 1 8 } = { 1 = 9 } ∩ { Σ = odd } = { (9 2) (9 4) (9 6) (9 8) } with four outcomes. Thus since all outcomes are equally likely, we have [ { Σ = odd } ∩ { 1 8 } ] = 4 81 The desired conditional probability is then [ { Σ = odd }|{ 1 8 } ] = [ { Σ = odd } ∩ { 1 8 } ] [ { 1 8 } ] = 4 81 9 81 = 4 9 ≈ 0 44 30. We are given that [ ] = 0 001 , where is the event ’disease is present.’ Let denote the event ’test is positive,’ so that is the event ’test is negative.’ We are additionally given [ | ] = 1 and [ | ] = 0 005 We are asked to compute [ | ] , i.e. the probability that ’disease is present given the test is positive.’ We use Bayes’ rule and Theorem as follows [ | ] = [ ] [ ] = [ | ] [ ] [ | ] [ ] + [ | ] [ ] = 1 × 0 001 1 × 0 001 + 0 005 × 0 999 = 1 1 + 4 995 ≈ 0 167 Thus in only about 17% of the cases will a positive test result actually con fi rm that you su ff er from the disease. The other 83% of the time you will be needlessly worried! 31. Let 1 denote the set of occupations and let 2 denote the set of interests and/or hobbies. Then 1 = { ’o ffi ce manager’, ’engineer’, ’doctor’, ’teacher’, ... } 2 = { ’nat. defense’, ’books’, ’music’, ’cooking’,... } Let denote Henrietta’s occupation and her interests. Then [ = ’o ffi ce manager’ = ’nat. defense’ ] = [ = ’o ffi ce manager’ ] [ = ’nat. defense’ | = ’o ffi ce m ≤ [ = ’o ffi ce manager’ ] since 0 ≤ [ = ’nat. defense’ | = ’o ffi ce manager’ ] ≤ 1 . 1 Remember, we decided above to write simply { 1 } for the compound event { 1 } × Ω 2 This since, in this problem, we only compute probabilities for events in the compound experiment.
Page 13 Read page 13 as text
12 32. Directly from the problem statement [ = 3] = 3 · [ = 1] [ = 2] = 2 · [ = 1] But we also know [ = 3] + [ = 2] + [ = 1] = 1 which is always true by axiom 2 [ Ω ] = 1 . Therefore [ = 1] = 1 6 , [ = 2] = 1 3 , and [ = 3] = 1 2 . Using Bayes’ Theorem, we then compute [ = 1 | = 1] = [ = 1 | = 1] [ = 1] P 3 =1 [ = 1 | = ] [ = ] = (1 − )1 6 (1 − ) 1 6 + 2 1 3 + 2 1 2 = 1 − 1 − + + 3 2 33. Let , {examinee knows}, , {examinee guesses}, and , {getting right answer} . Then [ ] = [ ] = 1 − [ | ] = 1 and [ | ] = 1 So [ | ] = [ | ] [ ] [ ] = 1 · [ | ] [ ] + [ | ] [ ] = + 1 (1 − ) = + (1 − ) 34. There are contestants and only one most beautiful. Hence [ { pick most beautiful} ] = 1 35. Let e , { random drawn chip ∈ } , e , { random drawn chip ∈ } , and e , { random drawn chip ∈ } . Also, let , { random drawn chip is defective } Then [ ] = [ | e ] [ e ] + [ | e ] [ e ] + [ | e ] [ e ] = 0 05 × 0 25 + 0 04 × 0 35 + 0 02 × 0 40 = 0 0345
Page 14 Read page 14 as text
13 Hence [ e | ] = [ | e ] [ e ] [ ] = 0 05 × 0 25 0 0345 = 0 363 [ e | ] = [ | e ] [ e ] [ ] = 0 04 × 0 35 0 0345 = 0 406 [ e | ] = [ | e ] [ e ] [ ] = 0 02 × 0 40 0 0345 = 0 232 36. From the example [ ] ' log We set , and construct the following table. [ ] [ ] 0.0 0.0 0.5 0.346 0.1 0.23 0.6 0.31 0.2 0.32 0.7 0.25 0.3 0.361 0.8 0.18 0.4 0.367 0.9 0.10 The peak is quite shallow, therefore the choice of is not critical near the peak. 37. (a) If we associate the 103 villagers with = 103 balls and the = 30 tents with 30 cells, this becomes a classical occupancy problem. (b) The result is given by Eq.1.8-6, which is repeated here as μ + − 1 ¶ = μ 30 + 103 − 1 103 ¶ = 132! 103!29! (c) The result is μ − 1 − ¶ = μ 103 − 1 103 − 30 ¶ = 102! 73!29! To obtain numerical evaluations of these factorial expressions, one might want to use Stirling’s formulas: ! ≈ (2 ) 1 2 +1 2 − 38. The most natural set of outcomes here are the strings (or vectors) of length , indicating where each ball has landed. There are such strings. They are all equally likely. The number of favorable outcomes would be ! since there are choices for the fi rst preselected location, − 1 choices for the second location, etc. The desired probability is then = ! Now, since the balls are indistinguishable, we could have considered the so-called distinguish- able outcomes, μ + − 1 ¶ in number, however from the description of the experiment in the problem statement, they would not all be equally likely. So we could not rely on classical theory then to give us the probabilities of these outcomes.
Page 15 Read page 15 as text
14 39. As in problem 1.38, the number of favorable ways is ! . However, the total number of ways is not since cells can at most hold one ball. For the fi rst ball, there are cells; for the second ball, − 1 cells, etc. Thus = ( − 1) · · · ( − + 1) = ! ( − )! Thus = ! ³ ! ( − )! ´ = !( − )! ! = μ ¶ − 1 40. (a) Let the tribal leaders be the cells and the ri fl es be the balls. Then the three tribal leaders collecting the fi ve ri fl es is the analog of putting fi ve balls into three cells. (b) These are the distributions shown in non-bold. There are fi fteen such distributions. (c) Careful here! If we count only the outcomes in bold we shall get the wrong answer i.e., 6/21=0.286. The reason this answer is wrong is that the outcomes in the columns are not equally likely. The correct answer is computed using Eq.(1.8-9) i.e., 41. (a) The probability that a speci fi ed number appears on the face of a dice is 1/6. Hence the probability of getting three speci fi ed numbers is or 1 in 216. Hence if you win you should get $216 for every dollar bet. But the casino payout is only $180:1. (b) The face value of the fi rst dice is irrelevant. The probability that the second dice matches the fi rst is 1/6. The probability that the third dice matches the fi rst is 1/6. Hence the probability of getting three unspeci fi ed matches is or 1 in 36. (c) Let denote that dice = 1 2 3 shows a speci fi ed number. Then the probability that (at least) two speci fi ed numbers appear is [ 1 2 3 ] + [ 1 3 2 ] + [ 3 2 1 ] + [ 1 2 3 ] = 3 × 5 6 × 1 6 × 1 6 + 1 6 × 1 6 × 1 6 = 0 0741 or about 1 in 14. So per dollar bet you should get $14 but the casino payout is only $10. d.-i. The next six parts can be solved by enumeration i.e., counting. However there is a systematic procedure based on the mathematical operation of convolution that can yield all of the answers from reading a graph. The details are given in Example 3.3.-5.
Page 16 Read page 16 as text
15 d. Refer to the table below: We note that there are only three ways of getting a 4: 1+1+2; 1+2+1,2+1+1. Hence the probability that the sum equals 4 is 3 (6 × 6 × 6) = 1 72 . Thus the fair payout should be 1:72 instead of 1:60. e. The number of ways of getting a 5 is 6: 3+1+1; 1+3+1; 1+1+3; 2+2+1; 2+1+2; 1+2+2. Hence the probability that the sum equals 5 is 6 (6 × 6 × 6) = 1 36 . A fair payout would be 1:36 instead of 1:30. f.-i. follow the same enumeration. j. Let’s think of this a series of throws. The probability that the fi rst throw matches one of the two speci fi ed numbers is 2/6. The probability that the next throw matches a speci fi ed number is 1/6. The last throw should not match either of the numbers. Its probability is 4/6. In a throw of three dice this can happen in three ways. Hence the probability is 3 × 2 6 × 1 6 × 4 6 = 1 9 or 1:9. But the payout is only 1:5. 42. For packets there are ! ways of arranging themselves, but only one way of doing it correctly. All the packet arrangements are equally likely. Hence [ { correct reassembly } ] = 1 ! = (3628800) − 1 for = 10 , ≈ 2 76 × 10 − 7 43. For three packets, there are six di ff erent arrangements ( 3 · 2 · 1 = 6) , but only one correct one. Hence on any try , [ { success } ] and , [ { failure } ] = 1 6 = 1 − = 5 6 For a fi rst correct reassembly on the th try, there must be − 1 failures followed by on success on the th try, thus ( ) = 1 6 μ 5 6 ¶ − 1 ≥ 1 We note in passing that this is a valid PMF, i.e. it sums to one over its support [1 ∞ ) . To fi nd the smallest such that X =1 1 6 μ 5 6 ¶ − 1 ≥ 0 95 we note that the complementary event is no successes in trials, with probability 1 − ¡ 5 6 ¢ Thus we seek instead the smallest such that 1 − ¡ 5 6 ¢ ≤ 1 − 0 95 = 0 05 Thus ' ln(0 05) ln(5 6) = 16 5 So the answer is = 17 .
327 more pages in the full document
Unlock it once and it stays in your library, ready to chat with or turn into flashcards.
Unlock the full documentMore Statistics study materials
See all
Exam GuidesTest Bank for Biostatistics for the Biological and Health Sciences , 2nd Edition
Statistics
182 pages161downloadsFeb 2025
Exam GuidesElementary Statistics, 13th Edition Test Bank
Statistics
239 pages160downloadsFeb 2025
Textbook GuidesSolution Manual For Statistics for Managers Using Microsoft Excel, 7th Edition
Statistics
1418 pages158downloadsFeb 2025
Textbook GuidesStatistics for Managers Using Microsoft Excel 8th Edition Solution Manual
Statistics
1343 pages156downloadsFeb 2025
Textbook GuidesEssentials of Social Statistics for a Diverse Society Second Edition Solution Manual
Statistics
137 pages135downloadsFeb 2025
Exam GuidesTest Bank For Elementary Statistics, 3rd Edition
Statistics
969 pages126downloadsFeb 2025
Textbook GuidesSolution Manual for Business Statistics, 4th Edition
Statistics
582 pages125downloadsFeb 2025
Exam GuidesTest Bank for Business Statistics: A Decision-Making Approach, 10th Edition
Statistics
881 pages110downloadsFeb 2025
Study this document with CramX
Once it is in your library, every tool can work from it.