stat_sol.json

[
 {
  "problem_text": "An insurance company sells several types of insurance policies, including auto policies and homeowner policies. Let $A_1$ be those people with an auto policy only, $A_2$ those people with a homeowner policy only, and $A_3$ those people with both an auto and homeowner policy (but no other policies). For a person randomly selected from the company's policy holders, suppose that $P\\left(A_1\\right)=0.3, P\\left(A_2\\right)=0.2$, and $P\\left(A_3\\right)=0.2$. Further, let $B$ be the event that the person will renew at least one of these policies. Say from past experience that we assign the conditional probabilities $P\\left(B \\mid A_1\\right)=0.6, P\\left(B \\mid A_2\\right)=0.7$, and $P\\left(B \\mid A_3\\right)=0.8$. Given that the person selected at random has an auto or homeowner policy, what is the conditional probability that the person will renew at least one of those policies?",
  "answer_latex": " 0.686",
  "answer_number": "0.686",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.3.11 ",
  "comment": " ",
  "solution": "The desired probability is\r\n$$\r\n\\begin{aligned}\r\nP\\left(B \\mid A_1 \\cup A_2 \\cup A_3\\right) & =\\frac{P\\left(A_1 \\cap B\\right)+P\\left(A_2 \\cap B\\right)+P\\left(A_3 \\cap B\\right)}{P\\left(A_1\\right)+P\\left(A_2\\right)+P\\left(A_3\\right)} \\\\\r\n& =\\frac{(0.3)(0.6)+(0.2)(0.7)+(0.2)(0.8)}{0.3+0.2+0.2} \\\\\r\n& =\\frac{0.48}{0.70}=0.686 .\r\n\\end{aligned}\r\n$$"
 },
 {
  "problem_text": "What is the number of possible 13-card hands (in bridge) that can be selected from a deck of 52 playing cards?",
  "answer_latex": " 635013559600",
  "answer_number": "635013559600",
  "unit": " ",
  "source": "stat",
  "problemid": " Example 1.2.10",
  "comment": " ",
  "solution": "The number of possible 13-card hands (in bridge) that can be selected from a deck of 52 playing cards is\r\n$$\r\n{ }_{52} C_{13}=\\left(\\begin{array}{l}\r\n52 \\\\\r\n13\r\n\\end{array}\\right)=\\frac{52 !}{13 ! 39 !}=635,013,559,600 .\r\n$$\r\n"
 },
 {
  "problem_text": "What is the number of ways of selecting a president, a vice president, a secretary, and a treasurer in a club consisting of 10 persons?",
  "answer_latex": " 5040",
  "answer_number": "5040",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.2.5 ",
  "comment": " ",
  "solution": "The number of ways of selecting a president, a vice president, a secretary, and a treasurer in a club consisting of 10 persons is\r\n$$\r\n{ }_{10} P_4=10 \\cdot 9 \\cdot 8 \\cdot 7=\\frac{10 !}{6 !}=5040 .\r\n$$"
 },
 {
  "problem_text": "At a county fair carnival game there are 25 balloons on a board, of which 10 balloons 1.3-5 are yellow, 8 are red, and 7 are green. A player throws darts at the balloons to win a prize and randomly hits one of them. Given that the first balloon hit is yellow, what is the probability that the next balloon hit is also yellow?",
  "answer_latex": "$\\frac{9}{24}$",
  "answer_number": "0.375",
  "unit": " ",
  "source": "stat",
  "problemid": " Example 1.3.5",
  "comment": " ",
  "solution": "Of the 24 remaining balloons, 9 are yellow, so a natural value to assign to this conditional probability is $9 / 24$."
 },
 {
  "problem_text": "What is the number of ordered samples of 5 cards that can be drawn without replacement from a standard deck of 52 playing cards?",
  "answer_latex": " 311875200",
  "answer_number": "311875200",
  "unit": " ",
  "source": "stat",
  "problemid": " Example 1.2.8",
  "comment": " ",
  "solution": "\r\nThe number of ordered samples of 5 cards that can be drawn without replacement from a standard deck of 52 playing cards is\r\n$$\r\n(52)(51)(50)(49)(48)=\\frac{52 !}{47 !}=311,875,200 .\r\n$$\r\n"
 },
 {
  "problem_text": "A bowl contains seven blue chips and three red chips. Two chips are to be drawn successively at random and without replacement. We want to compute the probability that the first draw results in a red chip $(A)$ and the second draw results in a blue chip $(B)$. ",
  "answer_latex": "$\\frac{7}{30}$",
  "answer_number": "0.23333333333",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.3.6 ",
  "comment": " ",
  "solution": "It is reasonable to assign the following probabilities:\r\n$$\r\nP(A)=\\frac{3}{10} \\text { and } P(B \\mid A)=\\frac{7}{9} \\text {. }\r\n$$\r\nThe probability of obtaining red on the first draw and blue on the second draw is\r\n$$\r\nP(A \\cap B)=\\frac{3}{10} \\cdot \\frac{7}{9}=\\frac{7}{30}\r\n$$"
 },
 {
  "problem_text": "From an ordinary deck of playing cards, cards are to be drawn successively at random and without replacement. What is the probability that the third spade appears on the sixth draw?",
  "answer_latex": " 0.064",
  "answer_number": "0.064",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.3.7 ",
  "comment": " ",
  "solution": "Let $A$ be the event of two spades in the first five cards drawn, and let $B$ be the event of a spade on the sixth draw. Thus, the probability that we wish to compute is $P(A \\cap B)$. It is reasonable to take\r\n$$\r\nP(A)=\\frac{\\left(\\begin{array}{c}\r\n13 \\\\\r\n2\r\n\\end{array}\\right)\\left(\\begin{array}{c}\r\n39 \\\\\r\n3\r\n\\end{array}\\right)}{\\left(\\begin{array}{c}\r\n52 \\\\\r\n5\r\n\\end{array}\\right)}=0.274 \\quad \\text { and } \\quad P(B \\mid A)=\\frac{11}{47}=0.234\r\n$$\r\nThe desired probability, $P(A \\cap B)$, is the product of those numbers:\r\n$$\r\nP(A \\cap B)=(0.274)(0.234)=0.064\r\n$$\r\n"
 },
 {
  "problem_text": "What is the probability of drawing three kings and two queens when drawing a five-card hand from a deck of 52 playing cards?",
  "answer_latex": " 0.0000092",
  "answer_number": "0.0000092",
  "unit": " ",
  "source": "stat",
  "problemid": " Example 1.2.11",
  "comment": " ",
  "solution": "Assume that each of the $\\left(\\begin{array}{c}52 \\\\ 5\\end{array}\\right)=2,598,960$ five-card hands drawn from a deck of 52 playing cards has the same probability of being selected. \r\nSuppose now that the event $B$ is the set of outcomes in which exactly three cards are kings and exactly two cards are queens. We can select the three kings in any one of $\\left(\\begin{array}{l}4 \\\\ 3\\end{array}\\right)$ ways and the two queens in any one of $\\left(\\begin{array}{l}4 \\\\ 2\\end{array}\\right)$ ways. By the multiplication principle, the number of outcomes in $B$ is\r\n$$\r\nN(B)=\\left(\\begin{array}{l}\r\n4 \\\\\r\n3\r\n\\end{array}\\right)\\left(\\begin{array}{l}\r\n4 \\\\\r\n2\r\n\\end{array}\\right)\\left(\\begin{array}{c}\r\n44 \\\\\r\n0\r\n\\end{array}\\right)\r\n$$\r\nwhere $\\left(\\begin{array}{c}44 \\\\ 0\\end{array}\\right)$ gives the number of ways in which 0 cards are selected out of the nonkings and nonqueens and of course is equal to 1 . Thus,\r\n$$\r\nP(B)=\\frac{N(B)}{N(S)}=\\frac{\\left(\\begin{array}{l}\r\n4 \\\\\r\n3\r\n\\end{array}\\right)\\left(\\begin{array}{c}\r\n4 \\\\\r\n2\r\n\\end{array}\\right)\\left(\\begin{array}{c}\r\n44 \\\\\r\n0\r\n\\end{array}\\right)}{\\left(\\begin{array}{c}\r\n52 \\\\\r\n5\r\n\\end{array}\\right)}=\\frac{24}{2,598,960}=0.0000092 .\r\n$$"
 },
 {
  "problem_text": "In an orchid show, seven orchids are to be placed along one side of the greenhouse. There are four lavender orchids and three white orchids. How many ways are there to lineup these orchids?",
  "answer_latex": " 35",
  "answer_number": "35",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.2.13",
  "comment": " ",
  "solution": "Considering only the color of the orchids, we see that the number of lineups of the orchids is\r\n$$\r\n\\left(\\begin{array}{l}\r\n7 \\\\\r\n4\r\n\\end{array}\\right)=\\frac{7 !}{4 ! 3 !}=35 \\text {. }\r\n$$"
 },
 {
  "problem_text": "If $P(A)=0.4, P(B)=0.5$, and $P(A \\cap B)=0.3$, find $P(B \\mid A)$.",
  "answer_latex": " 0.75",
  "answer_number": "0.75",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.3.2 ",
  "comment": " ",
  "solution": "$P(B \\mid A)=P(A \\cap B) / P(A)=0.3 / 0.4=0.75$."
 },
 {
  "problem_text": "What is the number of possible 5-card hands (in 5-card poker) drawn from a deck of 52 playing cards?",
  "answer_latex": " 2598960",
  "answer_number": "2598960",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.2.9 ",
  "comment": " ",
  "solution": "The number of possible 5-card hands (in 5-card poker) drawn from a deck of 52 playing cards is\r\n$$\r\n{ }_{52} C_5=\\left(\\begin{array}{c}\r\n52 \\\\\r\n5\r\n\\end{array}\\right)=\\frac{52 !}{5 ! 47 !}=2,598,960\r\n$$\r\n"
 },
 {
  "problem_text": "A certain food service gives the following choices for dinner: $E_1$, soup or tomato 1.2-2 juice; $E_2$, steak or shrimp; $E_3$, French fried potatoes, mashed potatoes, or a baked potato; $E_4$, corn or peas; $E_5$, jello, tossed salad, cottage cheese, or coleslaw; $E_6$, cake, cookies, pudding, brownie, vanilla ice cream, chocolate ice cream, or orange sherbet; $E_7$, coffee, tea, milk, or punch. How many different dinner selections are possible if one of the listed choices is made for each of $E_1, E_2, \\ldots$, and $E_7$ ?",
  "answer_latex": " 2688",
  "answer_number": "2688",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.2.2",
  "comment": " ",
  "solution": " By the multiplication principle, there are\r\n$(2)(2)(3)(2)(4)(7)(4)=2688$\r\ndifferent combinations.\r\n"
 },
 {
  "problem_text": "A rocket has a built-in redundant system. In this system, if component $K_1$ fails, it is bypassed and component $K_2$ is used. If component $K_2$ fails, it is bypassed and component $K_3$ is used. (An example of a system with these kinds of components is three computer systems.) Suppose that the probability of failure of any one component is 0.15 , and assume that the failures of these components are mutually independent events. Let $A_i$ denote the event that component $K_i$ fails for $i=1,2,3$. What is the probability that the system fails?",
  "answer_latex": " 0.9966",
  "answer_number": "0.9966",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.4.5 ",
  "comment": " ",
  "solution": "\r\nBecause the system fails if $K_1$ fails and $K_2$ fails and $K_3$ fails, the probability that the system does not fail is given by\r\n$$\r\n\\begin{aligned}\r\nP\\left[\\left(A_1 \\cap A_2 \\cap A_3\\right)^{\\prime}\\right] & =1-P\\left(A_1 \\cap A_2 \\cap A_3\\right) \\\\\r\n& =1-P\\left(A_1\\right) P\\left(A_2\\right) P\\left(A_3\\right) \\\\\r\n& =1-(0.15)^3 \\\\\r\n& =0.9966 .\r\n\\end{aligned}\r\n$$"
 },
 {
  "problem_text": "Suppose that $P(A)=0.7, P(B)=0.3$, and $P(A \\cap B)=0.2$. Given that the outcome of the experiment belongs to $B$, what then is the probability of $A$ ?",
  "answer_latex": " \\frac{2}{3}",
  "answer_number": "0.66666666666",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.3.3 ",
  "comment": " ",
  "solution": "We are effectively restricting the sample space to $B$; of the probability $P(B)=0.3,0.2$ corresponds to $P(A \\cap B)$ and hence to $A$. That is, $0.2 / 0.3=2 / 3$ of the probability of $B$ corresponds to $A$. Of course, by the formal definition, we also obtain\r\n$$\r\nP(A \\mid B)=\\frac{P(A \\cap B)}{P(B)}=\\frac{0.2}{0.3}=\\frac{2}{3}\r\n$$"
 },
 {
  "problem_text": "A coin is flipped 10 times and the sequence of heads and tails is observed. What is the number of possible 10-tuplets that result in four heads and six tails?",
  "answer_latex": " 210",
  "answer_number": "210",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.2.12 ",
  "comment": " ",
  "solution": "A coin is flipped 10 times and the sequence of heads and tails is observed. The number of possible 10-tuplets that result in four heads and six tails is\r\n$$\r\n\\left(\\begin{array}{c}\r\n10 \\\\\r\n4\r\n\\end{array}\\right)=\\frac{10 !}{4 ! 6 !}=\\frac{10 !}{6 ! 4 !}=\\left(\\begin{array}{c}\r\n10 \\\\\r\n6\r\n\\end{array}\\right)=210 .\r\n$$\r\n"
 },
 {
  "problem_text": "Among nine orchids for a line of orchids along one wall, three are white, four lavender, and two yellow. How many color displays are there?",
  "answer_latex": " 1260",
  "answer_number": "1260",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.2.14 ",
  "comment": " ",
  "solution": "The number of different color displays is\r\n$$\r\n\\left(\\begin{array}{c}\r\n9 \\\\\r\n3,4,2\r\n\\end{array}\\right)=\\frac{9 !}{3 ! 4 ! 2 !}=1260\r\n$$\r\n"
 },
 {
  "problem_text": "A survey was taken of a group's viewing habits of sporting events on TV during I.I-5 the last year. Let $A=\\{$ watched football $\\}, B=\\{$ watched basketball $\\}, C=\\{$ watched baseball $\\}$. The results indicate that if a person is selected at random from the surveyed group, then $P(A)=0.43, P(B)=0.40, P(C)=0.32, P(A \\cap B)=0.29$, $P(A \\cap C)=0.22, P(B \\cap C)=0.20$, and $P(A \\cap B \\cap C)=0.15$. Find $P(A \\cup B \\cup C)$.",
  "answer_latex": " 0.59",
  "answer_number": "0.59",
  "unit": " ",
  "source": "stat",
  "problemid": " Example 1.1.5",
  "comment": " ",
  "solution": "$$\r\n\\begin{aligned}\r\nP(A \\cup B \\cup C)= & P(A)+P(B)+P(C)-P(A \\cap B)-P(A \\cap C) \\\\\r\n& -P(B \\cap C)+P(A \\cap B \\cap C) \\\\\r\n= & 0.43+0.40+0.32-0.29-0.22-0.20+0.15 \\\\\r\n= & 0.59\r\n\\end{aligned}\r\n$$"
 },
 {
  "problem_text": "A grade school boy has five blue and four white marbles in his left pocket and four blue and five white marbles in his right pocket. If he transfers one marble at random from his left to his right pocket, what is the probability of his then drawing a blue marble from his right pocket?",
  "answer_latex": " $\\frac{41}{90}$",
  "answer_number": "0.444444444444444 ",
  "unit": "",
  "source": "stat",
  "problemid": " Example 1.3.10",
  "comment": " ",
  "solution": "For notation, let $B L, B R$, and $W L$ denote drawing blue from left pocket, blue from right pocket, and white from left pocket, respectively. Then\r\n$$\r\n\\begin{aligned}\r\nP(B R) & =P(B L \\cap B R)+P(W L \\cap B R) \\\\\r\n& =P(B L) P(B R \\mid B L)+P(W L) P(B R \\mid W L) \\\\\r\n& =\\frac{5}{9} \\cdot \\frac{5}{10}+\\frac{4}{9} \\cdot \\frac{4}{10}=\\frac{41}{90}\r\n\\end{aligned}\r\n$$\r\nis the desired probability."
 },
 {
  "problem_text": "A faculty leader was meeting two students in Paris, one arriving by train from Amsterdam and the other arriving by train from Brussels at approximately the same time. Let $A$ and $B$ be the events that the respective trains are on time. Suppose we know from past experience that $P(A)=0.93, P(B)=0.89$, and $P(A \\cap B)=0.87$. Find $P(A \\cup B)$.",
  "answer_latex": " 0.95",
  "answer_number": "0.95",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.1.4 ",
  "comment": " ",
  "solution": "$$P(A \\cup B) =P(A)+P(B)-P(A \\cap B)=0.93+0.89-0.87=0.95$$"
 },
 {
  "problem_text": "What is the number of possible four-letter code words, selecting from the 26 letters in the alphabet?",
  "answer_latex": " 358800",
  "answer_number": "358800",
  "unit": " ",
  "source": "stat",
  "problemid": "Example 1.2.4 ",
  "comment": " ",
  "solution": "The number of possible four-letter code words, selecting from the 26 letters in the  alphabet, in which all four letters are different is\r\n$$\r\n{ }_{26} P_4=(26)(25)(24)(23)=\\frac{26 !}{22 !}=358,800 .\r\n$$"
 }
]