# 機率期中考題

## 第二章

### 2.29
> In a fuel economy study, each of 3 race cars is tested using 5 different brands of gasoline at 7 test sites located in different regions of the country. If 2 drivers are used in the study, and test runs are made once un- der each distinct set of conditions, how many test runs are needed?
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> 在一項燃油經濟性研究中，在位於該國不同地區的 7 個測試地點，使用 5 種不同品牌的汽油對 3 輛賽車中的每輛進行了測試。 如果在研究中使用 2 個驅動程序，並且在每組不同的條件下進行一次測試運行，那麼需要多少次測試運行？
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:::success
$7\times5\times3\times2=210$
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### 2.34
> (1) How many distinct permutations can be made from the letters of the word COLUMNS?
(2) How many of these permutations start with the let-ter M?
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:::spoiler
> (1) 單詞 COLUMNS 的字母可以做出多少種不同的排列？
(2) 這些排列中有多少以字母 M 開頭？
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(1) $7!=5040$
(2) $6!=720$
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### 2.36
> (1) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 if each digit can be used only once?
(2) How many of these are odd numbers?
(3) How many are greater than 330?
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:::spoiler
> (1) 0、1、2、3、4、5、6，如果每個數字只能出現一次，可以組成多少個三位數？
(2) 其中有多少是奇數？
(3) 有多少大於 330？
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(1) $6\times6\times5=180$
(2) $3\times5\times5=75$
(3)
> $3\times6\times5=90$ (大於400)
> $1\times3\times5=15$ (330至400)
> $90+15=105$
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### 2.38
> Four married couples have bought 8 seats in the same row for a concert. In how many different ways can they be seated.
(1) with no restriction ?
(2) if each couple is sit together ?
(3) if all the men sit together to the right of all the women ?
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> 四對已婚夫婦為一場音樂會購買了同排8個座位。 他們可以以多少種不同的方式就座。
(1) 沒有限制 ?
(2) 如果每對情侶坐在一起？
(3) 如果所有的男人都坐在所有女人的右邊？
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(1) $8!=40320$
(2) $4!\times(2!)^4=2304$
(3) $4!\times4!=576$
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### 2.46
> In how many ways can 3 oaks, 4 pines, and 2 maples be arranged along a property line if one does not distinguish among trees of the same kind?
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> 如果不區分同一種類的樹木，沿著地產界線排列 3 棵橡樹、4 棵松樹和 2 棵楓樹有多少種方式？
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$\dfrac{9!}{3!\times4!\times2!}=1260$
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### 2.56
> An automobile manufacturer is concerned about a possible recall of its best-selling four-door sedan. If there were a recall, there is a probability of 0.25 of a defect in the brake system, 0.18 of a defect in the trans- mission. 0.17 of a defect in the fuel system, and 0.40 of a defect in some other area.
(1) What is the probability that the defect is the brakes or the fueling system if the probability of defects in both systems simultaneously is 0.15?
(2) What is the probability that there are no defects in either the brakes or the fueling system?
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> 一家汽車製造商擔心其最暢銷的四門轎車可能會被召回。 如果召回，則制動系統存在缺陷的概率為 0.25，變速箱存在缺陷的概率為 0.18。 0.17 的燃油系統缺陷，0.40 的其他區域缺陷。
(1) 如果兩個系統同時出現故障的概率為 0.15，則故障是剎車系統或加油系統的概率是多少？
(2) 剎車或加油系統沒有缺陷的概率是多少？
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> P(B)=0.25
> P(T)=0.18
> P(F)=0.17
> P(O)=0.40
>
(1)
> P(B$\cap$F)=0.15
> P(B$\cup$F)=P(B)+P(F)-P(B$\cap$F)
> P(B$\cup$F)=0.25+0.17-0.15=0.27

(2)
> P(B$\cup$F)'=1-P(B$\cup$F)
> P(B$\cup$F)'=1-0.27=0.73
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### 2.65
> Given an electronic component. Let A be the event that the component fails a particular test and B be the event that the component displays strain but does not actually fail. Event A occurs with probability 0.20 and event B occurs with probability 0.35.
(1) What is the probability that the component does not fail the test?
(2) What is the probability that a component works perfectly well (i.e., neither displays strain nor fails the test)?
(3) What is the probability that the component either fails or shows strain in the test?
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> 給定一個電子元件。 假設 A 是組件未通過特定測試的事件，B 是組件顯示應變但實際上並未失敗的事件。 事件 A 的發生概率為 0.20，事件 B 的發生概率為 0.35。
(1) 組件通過測試的概率是多少？
(2) 一個組件完全正常工作的概率是多少？
(3) 部件在測試中失效或出現應變的概率是多少？
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> P(A)=0.20
> P(B)=0.35
>
(1) P(A)'=1-P(A)=1-0.2=0.8
(2) P(A'$\cap$B')=1-P(A$\cup$B)=1-0.20-0.35=0.45
(3) P(A$\cup$B)=0.20+0.35=0.55
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### 2.77
> In the senior year of a high school graduating class of 100 students, 42 studied mathematics, 68 studied psychology, 54 studied history, 22 studied both mathematics and history, 25 studied both mathematics and psychology, 7 studied history but neither mathematics nor psychology, 10 studied all three subjects, and 8 did not take any of the three. Randomly select a student from the class and find the probabilities of the following events.
(1) A person enrolled in psychology takes all three subjects.
(2) A person not taking psychology is taking both history and mathematics.
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> 在一個 100 名學生的高三畢業班中，42 人學習數學，68 人學習心理學，54 人學習歷史，22 人同時學習數學和歷史，25 人同時學習數學和心理學，7 人學習歷史但既不數學也不心理學 , 10 人學習了所有三門科目，8 人沒有修讀這三門科目中的任何一門。 從班級中隨機抽取一名學生，求出下列事件發生的概率。
(1) 心理學的人修了全部三門科目。
(2) 不修心理學但是同時修歷史和數學。
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(1) $\dfrac{10}{68}$
(2) $\dfrac{12}{24}=\dfrac{1}{2}$
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### 2.82
> For married couples living in a certain suburb, the probability that the husband will vote on a bond referendum is 0.21, the probability that the wife will vote on the referendum is 0.28, and the probability that both the husband and the wife will vote is 0.15. What is the probability that
(1) at least one member of a married couple will vote?
(2) a wife will vote, given that her husband will vote?
(3) a husband will vote, given that his wife will not vote?
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> 對於居住在某個郊區的已婚夫婦，丈夫投票支持債券公投的概率是 0.21，妻子投票支持公投的概率是 0.28，夫妻雙方都投票的概率是 0.15。求以下的機率。
(1) 至少一對已婚夫婦中的一個成員會投票？
(2) 鑑於丈夫會投票，妻子會投票？
(3) 鑑於妻子不會投票，丈夫會投票？
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> P(A)=0.21
> P(B)=0.28
> P(A$\cap$B)=0.15
>
(1) P(A$\cup$B)=P(A)+P(B)-P(A$\cap$B)=0.21+0.28-0.15=0.34
(2) P(B|A)=$\dfrac{P(A\cap B)}{P(A)}$=$\dfrac{0.15}{0.21}$=$\dfrac{5}{7}$
(3) P(A|B')=$\dfrac{P(A\cap B')}{P(B')}$=$\dfrac{0.21-0.15}{1-0.28}$=$\dfrac{0.06}{0.72}$=$\dfrac{1}{12}$
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### 2.92
> Suppose the diagram of an electrical system is
an given in Figure 2.10. What is the probability that
the system works? Assume the fail independently.
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> 假設電氣系統圖如圖 2.10 所示。 系統工作的概率是多少？假設組件獨立失效。
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P(A)$\times$[1-P(B')$\times$P(C')]$\times$P(D)=0.95$\times$(1-0.3$\times$ 0.2)$\times$ 0.9=0.95$\times$ 0.94$\times$ 0.9=0.8037
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### 2.93
> A circuit system is given in Figure 2.11. Assume the components fail independently.
(1) What is the probability that the entire system works?
(2) Given that the system works, what is the probability that the component A is not working?
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> 圖 2.11 給出了一個電路系統。 假設組件獨立失效。
(1) 整個系統工作的概率是多少？
(2) 假定系統工作，組件 A 不工作的概率是多少？
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> P($x$)=P(A)$\times$P(B)=0.49
> P($y$)=P(C\)$\times$P(D)$\times$P(E)=0.512
>
(1)
> P($x\cup y$)=1-P($x'\cap y'$)
> P($x'\cap y'$)=P($x'$)$\times$P($y'$)=0.51$\times$ 0.488=0.25
> 1-0.25=0.75
>
(2)
> $\dfrac{P(A')\times P(C)\times P(D)\times P(E)}{0.75}$=$\dfrac{0.3\times 0.8\times 0.8\times 0.8}{0.75}$=$\dfrac{0.1536}{0.75}$=0.2048
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### 2.96
> Police plan to enforce speed limits by using radar traps at four different locations within the city limits. The radar traps at each of the locations L1,L2, L3, and L4 will be operated 40%. 30%, 20%, and 30% of the time. If a person who is speeding on her way to work has probabilities of 0.2, 0.1, 0.5, and 0.2, respectively, of passing through these locations, what is the probability that she will receive a speeding ticket?
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> 警方計劃通過在城市範圍內的四個不同位置使用雷達陷阱來實施限速。 L1、L2、L3 和 L4 位置的雷達陷阱將運行 40%。 30%、20% 和 30% 的時間。 如果一個在上班路上超速的人經過這些地點的概率分別為 0.2、0.1、0.5 和 0.2，那麼她收到超速罰單的概率是多少？
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(0.4$\times$ 0.2)+(0.3$\times$ 0.1)+(0.2$\times$ 0.5)+(0.3$\times$ 0.2)=0.27
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### 2.100
> A regional telephone company operates three identical relay stations at different locations. During a one year period, the number of malfunctions reported by each station and the causes are shown below.Suppose that a malfunction was reported and it was found to be caused by other human errors. What is the probability that it came from station C?
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> 一家地區性電話公司在不同地點經營著三個相同的中繼站。 在一年的時間裡，每個站點報告的故障數量和原因如下所示。假設報告了一個故障，發現是其他人為錯誤造成的。 來自C站的概率是多少？
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$\dfrac{5}{7+7+5}$=$\dfrac{5}{19}$=0.2631
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### 2.101
> A paint-store chain produces and sells latex and semigloss paint. Based on long-range sales, the probability that a customer will purchase latex paint is 0.75. Of those that purchase latex paint, 60% also purchase rollers. But only 30% of semigloss paint buyers purchase rollers. A randomly selected buyer purchases a roller and a can of paint. What is the probability that the paint is latex?
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> 一家油漆連鎖店生產和銷售乳膠漆和半光漆。 基於長期銷售，客戶購買乳膠漆的概率為 0.75。 在購買乳膠漆的人中，60% 還購買了滾筒。 但只有 30% 的半光漆買家購買滾筒。 一個隨機選擇的買家購買了一個滾筒和一罐油漆。 油漆是乳膠的概率是多少？
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> P(A)=0.75$\times$ 0.6=0.45
> P(B)=0.25$\times$ 0.3=0.075
>
P($x$)=P(A)+P(B)=0.525
P(A|$x$)=$\dfrac{P(A)}{P(x)}$=$\dfrac{0.45}{0.525}$=0.857
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### 2.110
> The probability that a patient recovers from a delicate heart operation is 0.8. What is the probability that
(1) exactly 2 of the next 3 patients who have this operation survive?
(2) all of the next 3 patients who have this operation survive?
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> 患者從精細心臟手術中康復的概率為 0.8。求出下列事件發生的概率。
(1) 接下來的 3 名接受此手術的患者中恰好有 2 名存活？
(2) 接下來的 3 位接受此手術的患者全部存活？
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(1) 0.8$\times$ 0.8$\times$ 0.2$\times C_1^3$=0.384
(2) 0.8$\times$ 0.8$\times$ 0.8$\times C_0^3$=0.512
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### 2.124
> A firm is accustomed to training operators who do certain tasks on a production line. Those operators who attend the training course are known to be able to meet their production quotas 90% of the time. New operators who do not take the training course only meet their quotas 65% of the time. Fifty percent of new operators attend the course. Given that a new operator meets her production quota, what is the probability that she attended the program?
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> 公司習慣於培訓在生產線上執行特定任務的操作員。 眾所周知，參加培訓課程的操作員能夠在 90% 的時間內完成生產配額。 沒有參加培訓課程的新操作員只有 65% 的時間達到他們的配額。 50% 的新操作員參加了該課程。 假設新操作員達到她的生產配額，她參加該計劃的概率是多少？
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> P(A)=0.90$\times$ 0.5=0.45
> P(B)=0.65$\times$ 0.5=0.325
>
P($x$)=P(A)+P(B)=0.45+0.325=0.775
P(A|$x$)=$\dfrac{P(A)}{P(x)}$=$\dfrac{0.45}{0.775}$=0.581
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### 2.127
> There is a 50-50 chance that the queen carries the gene of hemophilia. If she is a carrier, then each prince has a 50-50 chance of having hemophilia independently. If the queen is not a carrier, the prince will not have the disease. Suppose the queen has had three princes without the disease. What is the probability the queen is a carrier?
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> 蜂王攜帶血友病基因的機率有50-50。 如果她是攜帶者，那麼每個王子都有 50-50 的機會單獨患上血友病。 如果王后不是攜帶者，王子就不會得病。 假設王后已經有過三個沒有病的王子。 女王是攜帶者的概率是多少？
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> P(A)=$0.5\times 0.5^3$=0.0625
> P(B)=$0.5\times 1^3$=0.5
>
P($x$)=P(A)+P(B)=0.0625+0.5=0.5625
P(A|$x$)=$\dfrac{P(A)}{P(x)}$=$\dfrac{0.0625}{0.5625}$=0.111...
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## 第三章

### 3.5
> Determine the value c so that each of the following functions can serve as a probability distribution of the discrete random variable X:
> 

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> 確定c的值，使得下列各函數都可以作為離散隨機變量X的概率分佈：
> 
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### 3.8
> Find the probability distribution of the random variable W in Exercise 3.3. assuming that the coin is biased so that a head is twice as likely to occur as a tail.
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> 找出練習 3.3 中隨機變量 W 的概率分佈。 假設硬幣有偏差，出現正面的可能性是出現反面的可能性的兩倍。
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### 3.12
> An investment firm offers its customers municipal bonds that mature after varying numbers of years. Given that the cumulative distribution function of T. the number of years to maturity for a randomly selected bond.
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> 一家投資公司向其客戶提供在不同年限後到期的市政債券。 鑑於 T 的累積分佈函數。隨機選擇的債券的到期年數。
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### 3.21
> Consider the density function
(a) Evaluate k.
(b) Find F($x$) and use it to evaluate P(0.3<X<0.6).
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> 考慮密度函數
(a) 評估 k。
(b) 找到 F($x$) 並用它來評估 P(0.3<X<0.6)。
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### 3.25
> From a box containing 4 dimes and 2 nickels, 3 coins are selected at random without replacement. Find the probability distribution for the total T of the 3 coins. Express the probability distribution graphically as a probability histogram.
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> 從裝有 4 個 10 美分硬幣和 2 個 5 美分硬幣的盒子中，隨機抽取 3 枚硬幣，不放回原處。 找出 3 個硬幣的總 T 的概率分佈。 以圖形方式將概率分佈表示為概率直方圖。
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### 3.26
> From a box containing 4 black balls and 2 green balls, 3 balls are drawn in succession, each ball being replaced in the box before the next draw is made. Find the probability distribution for the number of green balls.
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> 從一個裝有 4 個黑球和 2 個綠球的盒子中，連續抽取 3 個球，在下一次抽取之前將每個球放回盒子中。 找出綠球數量的概率分佈。
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0
> $\dfrac{4}{6}\times\dfrac{4}{6}\times\dfrac{4}{6}\times1$=$\dfrac{8}{27}$
>
1
> $\dfrac{4}{6}\times\dfrac{4}{6}\times\dfrac{2}{6}\times3$=$\dfrac{4}{9}$
>
2
> $\dfrac{4}{6}\times\dfrac{2}{6}\times\dfrac{2}{6}\times3$=$\dfrac{2}{9}$
>
3
> $\dfrac{2}{6}\times\dfrac{2}{6}\times\dfrac{2}{6}\times1$=$\dfrac{1}{27}$
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### 3.27
> The time to failure in hours of an important piece of electronic equipment used in a manufactured DVD player has the density function.
(a) Find F($x$).
(b) Determine the probability that the component (and thus the DVD player) lasts more than 1000 hours before the component needs to be replaced.
(c\) Determine the probability that the component fails before 2000 hours.
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> 製造的 DVD 播放器中使用的重要電子設備的故障時間（以小時為單位）具有密度函數。
(a) 求 F($x$)。
(b) 確定組件（以及 DVD 播放器）在需要更換之前持續使用超過 1000 小時的概率。
(c\) 確定部件在 2000 小時之前失效的概率。
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> $\int_{-\infty}^x f(t) \mathrm{d} t =\int_0^x \dfrac{1}{2000}e^{\dfrac{-t}{2000}}\mathrm{d}t=-\dfrac{2000}{2000}\times e^{\dfrac{-t}{2000}}\\=-e^{\dfrac{-x}{2000}}+e^0\\=1-e^{\dfrac{-x}{2000}}$

(a) 

(b)
> P($x\gt 1000$)=1-P($x\le 1000$)
> P($x\le 1000$)=$F(1000)$=$1-e^{\dfrac{-1000}{2000}}$
> P($x\gt 1000$)=1-($1-e^{\dfrac{-1000}{2000}}$)=$e^{-\dfrac{1}{2}}$=0.6065

(c\)
> P($x\lt 2000$)=P($x\le 2000$)=$F(2000)$
> $F(2000)$=$1-e^{\dfrac{-2000}{2000}}$=$1-e^{-1}$=0.6321
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### 3.36
> On a laboratory assignment, if the equipment is working, the density function of the observed outcome, X. is
(a) Calculate P(X ≤ 1/3).
(b) What is the probability that X will exceed 0.5?
(c\) Given that X ≥ 0.5, what is the probability that X will be less than 0.75?
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> 在實驗室任務中，如果設備正在工作，則觀察到的結果的密度函數 X. 是
(a) 計算 P(X ≤ 1/3)。
(b) X 超過 0.5 的概率是多少？
(c\) 假定 X ≥ 0.5，X 小於 0.75 的概率是多少？
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:::success
> $\int_{-\infty}^x f(t) \mathrm{d} t =\int_0^x 2(1-t)\mathrm{d}t=-(1-t)^2\mid_0^x=1-(1-x)^2$
> $F(x)=\{\array{
1-(1-x)^2 ,&0<x<1\\
0,&otherwise
}$

(a)
> P(X ≤ $\dfrac{1}{3}$)=$F(\dfrac{1}{3})$=$\dfrac{5}{9}$

(b)
> P(X > 0.5)=1-P(X $\le$ 0.5)=1-$F(0.5)$=$\dfrac1 4$

(c\)
> $\dfrac{P(0.5\lt X \lt 0.75)}{P(0.5 ≤ X)}$=$\dfrac{F(0.75)-F(0.5)}{1-F(0.5)}$=$\dfrac3 4$
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### 3.39
> From a sack of fruit containing 3 oranges, 2 apples, and 3 bananas, a random sample of 4 pieces of fruit is selected. If X is the number of oranges and Y is the number of apples in the sample, find
(a) the joint probability distribution of X and Y
(b) P(X.Y) $\in$ A], where A is the region that is given by { (x, y) $\mid$ x + y ≤ 2 }.
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> 從一袋包含 3 個橙子、2 個蘋果和 3 個香蕉的水果中，隨機抽取 4 個水果作為樣本。 如果 X 是橙子的數量，Y 是樣本中蘋果的數量，求
(a) X和Y的聯合概率分佈
(b) P(X.Y) $\in$ A]，其中 A 是由 { (x, y) $\mid$ x + y ≤ 2 }。
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> 

(a) 

(b) 
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### 3.40
> A fast-food restaurant operates both a drivethrough facility and a walkin facility. On a randomly selected day, let X and Y, respectively, be the proportions of the time that the drivethrough and walkin facilities are in use, and suppose that the joint density function of these random variables is
(a) Find the marginal density of X.
(b) Find the marginal density of Y.
(c\) Find the probability that the drivethrough facility is busy less than onehalf of the time.
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> 一家快餐店同時經營免下車設施和步行設施。 在隨機選擇的一天，設 X 和 Y 分別為免下車和步行設施使用時間的比例，並假設這些隨機變量的聯合密度函數為
(a) 找出 X 的邊際密度。
(b) 找出 Y 的邊際密度。
(c\) 求得來速設施忙碌時間少於一半的概率。
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(a) 

(b) 

(c\) 
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### 3.47
> The amount of kerosene, in thousands of liters, in a tank at the beginning of any day is a random amount Y from which a random amount X is sold during that day. Suppose that the tank is not resupplied during the day so that x ≤ y, and assume that the joint density function of these variables is
(a) Determine if X and Y are independent.
(b) Find P(1/4 < x < 1/2 | Y=3/4).
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> 任何一天開始時，油箱中的煤油量（以千升為單位）是隨機量 Y，當天從中銷售隨機量 X。 假設坦克在白天沒有補給，使得 x ≤ y，並假設這些變量的聯合密度函數是
(a) 確定 X 和 Y 是否獨立。
(b) 求 P(1/4 < x < 1/2 | Y=3/4)。
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(a) 

(b) 
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### 3.52
> A coin is tossed twice. Let Z denote the number of heads on the first toss and W the total number of heads on the 2 tosses. If the coin is unbalanced and a head has a 40% chance of occurring, find
(a) the joint probability distribution of W and Z
(b) the marginal distribution of W
(c\) the marginal distribution of Z
(d) the probability that at least 1 head occurs.
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> 一枚硬幣被拋兩次。 令 Z 表示第一次拋擲中正面朝上的次數，W 表示兩次拋擲中正面朝上的總數。 如果硬幣不平衡，出現正面朝上的概率為 40%，求
(a) W和Z的聯合概率分佈
(b) W的邊際分佈
(c\) Z 的邊際分佈
(d) 出現至少 1 個正面的概率。
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(a)

(b)

(c\)

(d)

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### 3.57
> Let X, Y, and Z have the joint probability density function
(a) Find k.
(b) Find P( X < 1/4 , Y > 1/2 , 1 < Z < 2 ).\
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> 令 X、Y 和 Z 具有聯合概率密度函數
(a) 找到 k。
(b) 求 P( X < 1/4 , Y > 1/2 , 1 < Z < 2 )。
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(a)

(b)
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### 3.64
> A service facility operates with two service lines. On a randomly selected day, let X be the proportion of time that the first line is in use whereas Y is the proportion of time that the second line is in use. Suppose that the joint probability density function for (X, Y) is
(a) Compute the probability that neither line is busy more than half the time.
(b) Find the probability that the first line is busy more than 75% of the time.
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> 一個服務設施有兩條服務線。 在隨機選擇的一天，令 X 為第一條線路的使用時間比例，而 Y 為第二條線路的使用時間比例。 假設 (X, Y) 的聯合概率密度函數是
(a) 計算兩條線路佔線時間均超過一半的概率。
(b) 找出第一條線路佔線時間超過 75% 的概率。
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(a)(b)
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### 3.80
> Consider a system of components in which there are 5 independent components, each of which possesses an operational probability of 0.92. The system does have a redundancy built in such that it does not fail if 3 out of the 5 components are operational. What is the probability that the total system is operational?
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> 考慮一個組件系統，其中有 5 個獨立組件，每個組件的運行概率為 0.92。 該系統確實有一個內置的冗餘，如果 5 個組件中的 3 個都在運行，它就不會出現故障。 整個系統運行的概率是多少？
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## 第四章

### 4.9
> A private pilot wishes to insure his airplane for 200,000. The insurance company estimates that a total loss will occur with probability 0.002, a 50% loss with probability 0.01, and a 25% loss with probability 0.1. Ignoring all other partial losses, what premium should the insurance company charge each year to realize an average profit of 500?
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> 一位私人飛行員希望為他的飛機投保 200,000 美元。 保險公司估計發生全損的概率為 0.002，50% 的損失概率為 0.01，25% 的損失概率為 0.1。 忽略所有其他部分損失，保險公司每年應收取多少保費才能實現 500 美元的平均利潤？
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### 4.16
> Suppose that you are inspecting a lot of 1000 light bulbs, among which 20 are defectives. You choose two light bulbs randomly from the lot without replacement.
> 
Find the probability that at least one light bulb chosen is defective.
[Hint: Compute P(X₁ + X₂ = 1)]
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> 假設您正在檢查 1000 個燈泡，其中 20 個是次品。 您從該批次中隨機選擇兩個燈泡，無需更換。
> 
找出至少選擇一個燈泡有缺陷的概率。 [提示：計算 P(X₁ + X₂ = 1)]
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### 4.19
> A large industrial firm purchases several new word processors at the end of each year, the exact number depending on the frequency of repairs in the previous year. Suppose that the number of word processors, $x$. purchased each year has the following probability distribution:
> 
If the cost of the desired model is 1200 per unit and at the end of the year a refund of $50x^2$ dollars will be issued, how much can this firm expect to spend on new word processors during this year?
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> 一家大型工業公司在每年年底購買幾台新的文字處理器，具體數量取決於前一年的維修頻率。 假設每年購買的文字處理器數量 $x$. 具有以下概率分佈：
> 
如果所需型號的成本為每台 1200 美元，並且在年底將發放 $50x^2$ 美元的退款，那麼這家公司預計今年在新文字處理器上的支出是多少？
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### 4.23
> Suppose that X and Y have the following joint probability function:
(a) Find the expected value of g(X,Y) = XY².
(b) Find $μx$ and $μy$ .
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> 假設 X 和 Y 具有以下聯合概率函數：
(a) 求 g(X,Y) = XY² 的期望值。
(b) 找出 $μx$ 和 $μy$。
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(a)

(b)
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### 4.24
> Referring to the random variables whose joint probability distribution is given in Exercise 3.39 on page 105.
(a) find $E(x^2y-2xy)$;
(b) find $μx-μy$
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> 參考第 105 頁練習 3.39 中給出的聯合概率分佈的隨機變量。
(a) 求 $E(x^2y-2xy)$
(b) 找到 $μx-μy$
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### 4.36
> Suppose that the probabilities are 0.4, 0.3, 0.2, and 0.1, respectively, that 0, 1, 2, or 3 power failures will strike a certain subdivision in any given year. Find the mean and variance of the random variable X repre senting the number of power failures striking this subdivision
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> 假設概率分別為 0.4、0.3、0.2 和 0.1，即 0、1、2 或 3 次電力故障將在任何給定年份發生在某個分區。 求隨機變量 X 的均值和方差，表示該分區的停電次數
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### 4.43
> The length of time, in minutes, for an airplane to obtain clearance for takeoff at a certain airport is a random variable Y=3X-2, where X has the density function.
Find the mean and variance of the random variable Y.
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> 飛機在某個機場獲得起飛許可的時間長度（以分鐘為單位）是一個隨機變量 Y=3X-2，其中 X 具有密度函數。
找出隨機變量 Y 的均值和方差。
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### 4.47
> For the random variables X and Y whose joint density function is given in Exercise 3.40 on page 105, find the covariance.
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> 對於聯合密度函數在第 105 頁練習 3.40 中給出的隨機變量 X 和 Y，求協方差。
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