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Statistics for Business & Economics 14e Metric Version Chapter4
Statistic
National Taipei University
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Chapter 4 Introduction to Probability
Learning Objectives
- Obtain an appreciation of the role probability information plays in the decision-making process.
- Understand probability as a numerical measure of the likelihood of occurrence.
- Know the three methods commonly used for assigning probabilities and understand when they should be used.
- Know how to use the laws that are available for computing the probabilities of events.
- Understand how new information can be used to revise initial (prior) probability estimates using Bayes’ theorem.
Solutions
Number of experimental outcomes = (3)(2)(4) = 24
63 3!3! (3 2 1)(3 2 1)6! 654321 20
ABC ACE BCD BEF ABD ACF BCE CDE ABE ADE BCF CDF ABF ADF BDE CEF ACD AEF BDF DEF
- P 3
6 () 63 6! !()()()654 120
BDF BFD DBF DFB FBD FDB
4. a.
b. Let: H be head and T be tail: (H,H,H) (T,H,H) (H,H,T) (T,H,T) (H,T,H) (T,T,H) (H,T,T) (T,T,T) c. The outcomes are equally likely, so the probability of each outcome is 1/8. 5. P (Ei) = 1/5 for i = 1, 2, 3, 4, 5
P (Ei) ≥ 0 for i = 1, 2, 3, 4, 5 P (E 1 ) + P (E 2 ) + P (E 3 ) + P (E 4 ) + P (E 5 ) = 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 1 The classical method was used. 6. P (E 1 ) = .40, P (E 2 ) = .26, P (E 3 ) =. The relative frequency method was used.
H T
H T
H T
H T H T H T H T (H,H,H) (H,H,T) (H,T,H) (H,T,T) (T,H,H) (T,H,T) (T,T,H) (T,T,T)
1st Toss 2nd Toss 3rd Toss
10. a.
Programmer Total Lines of Code Written
Number of Lines of Code Requiring Edits
Probability
Liwei 23,789 4,589 0.
Andrew 17,962 2,780 0.
Jaime 31,025 12,080 0.
Sherae 26,050 3,780 0.
Binny 19,586 1,890 0.
Roger 24,786 4,005 0.
Dong-Gil 24,030 5,785 0.
Alex 14,780 1,052 0.
Jay 30,875 3,872 0.
Vivek 21,546 4,125 0. b. Probability = 4589 / 23789 = 0. c. Probability = 1 – 3780 / 26050 = 1 – 0 = 0. d. The lowest probability is Alex at 0; the highest probability is Jaime at 0. 11. a. P (adult in the Tri-State Region smokes) = 118/(118+492) = 118/610 =. b. P(adult in KY smokes) = 47/223 =. P (adult in IN smokes) = 32/166 =. P (adult in OH smokes) = 39/221 =.
For these three states, Ohio has the lowest estimated probability of an adult being a smoker. 12. Initially a probability of .20 would be assigned if selection is equally likely. Data do not appear to confirm the belief of equal consumer preference. For example, using the relative frequency method we would assign a probability of 5/100 = .05 to the design 1 outcome, .15 to design 2, .30 to design 3, .40 to design 4, and .10 to design 5. 13. a. Step 1—Use the counting rule for combinations:
ቀ 696 ቁൌ5!69 െ 5ሻ!69! ൌሺ69ሻሺ68ሻሺ67ሻሺ66ሻሺ65ሻሺ5ሻሺ4ሻሺ3ሻሺ2ሻሺ1ሻ ൌ 11,238, Step 2—There are 26 ways to select the red Powerball from digits 1 to 26. Total number of Powerball lottery outcomes: (11,238,513) x (26) = 292,201, b. Probability of winning the lottery: 1 chance in 292,201, = 1/(292,201,338) =.
14. a. P (E 2 ) = 1/ b. P (any 2 outcomes) = 1/4 + 1/4 = 1/ c. P (any 3 outcomes) = 1/4 + 1/4 + 1/4 = 3/ 15. a. S = {ace of clubs, ace of diamonds, ace of hearts, ace of spades} b. S = {2 of clubs, 3 of clubs,... , 10 of clubs, J of clubs, Q of clubs, K of clubs, A of clubs} c. There are 12; jack, queen, or king in each of the four suits. d. For a: 4/52 = 1/13 =. For b: 13/52 = 1/4 =.
b. Let N = corporate headquarters located in New York T = corporate headquarters located in Texas P ( N ) = 50/500 =. P ( T ) = 52/500 =. Located in California, New York, or Texas:
P ( C ) P ( N ) P ( T ).106.100.104.
c. Let A = corporate headquarters located in one of the eight states Total number of companies with corporate headquarters in the eight states = 283 P ( A ) = 283/500 =. More than half the Fortune 500 companies have corporate headquartered located in these eight states. 19. a. Probability = 131/(134 + 131 + 2) = 131/(267) =. b. Probability = (432)/(293 + 432 + 8) = 432/733 =. c. Probability = (134 + 293) /(267 + 733) = 427/1,000 =. d. Older respondents appear to be less concerned about global warming being a threat in their lifetime than are younger respondents.
20. a.
Age Experimental Outcome Financially Independent
Number of Responses Probability
E 1 16 to 20 191 191/944 = 0. E 2 21 to 24 467 467/944 = 0. E 3 25 to 27 244 244/944 = 0. E 4 28 or older 42 42/944 = 0. 944 b. P (Age <25) P ( E 1 ) P ( E 2 ).2023.4947. c. P (Age >24) P ( E 3 ) P ( E 4 ).2585.0445. d. The probability of being financially independent before age 25, .6970, seems high given the general economic conditions. The teenagers who responded to this survey may have had unrealistic expectations about becoming financially independent at a relatively young age.
21. a. 𝑃ሺ𝑝𝑜𝑙𝑒,𝑝𝑜𝑠𝑡ሻൌଵସଶଵସଵൌ.
b. 𝑃ሺ𝑔𝑢𝑎𝑟𝑑 𝑟𝑎𝑖𝑙ሻൌଵସଶ଼ଽ ൌ. c. The fixed object that is least likely to be involved in a fatal collision is a bridge with probability 𝑃ሺ𝑏𝑟𝑖𝑑𝑔𝑒ሻൌଵସଶଶଷଵ ൌ. d. The fixed object that is most likely to be involved in a fatal collision is a shrubbery or tree with probability 𝑃ሺ𝑠ℎ𝑟𝑢𝑏𝑏𝑒𝑟𝑦, 𝑡𝑟𝑒𝑒ሻൌଵସଶଶହ଼ହ ൌ.
25. Let F = Uses Facebook L = Uses LinkedIn a. 𝑃ሺ𝐹∪𝐿ሻൌ𝑃ሺ𝐹ሻ𝑃ሺ𝐿ሻെ𝑃ሺ𝐹∩𝐿ሻൌ .68 .25 െ .22 ൌ. b. 1െ𝑃ሺ𝐹∪𝐿ሻൌ 1 െ .71 ൌ.
26. a. Let D = Domestic Equity Fund P ( D ) = 16/25 =. b. Let A = 4- or 5-star rating Thirteen funds were rated 3 star or less; thus, 25 – 13 = 12 funds must be 4 star or 5 star. P ( A ) = 12/25 =. c. Seven Domestic Equity funds were rated 4 star, and two were rated 5 star. Thus, nine funds were Domestic Equity funds and were rated 4 star or 5 star: P ( D A ) = 9/25 =. d. P ( D A ) = P ( D ) + P ( A ) - P ( D A ) = .64 + .48 - .36 =. 27. Let A = the event that a randomly selected U. adult uses social media B = the event that a randomly selected U. adult is age 18– a. P(A) = 1 - .35 =. b. P(B) = 1 – .78 =. c. The question asks for the probability of A intersected with B. From the addition law, PA B PA PB PA B ()( )()()
so, PA B PA PB PA B ( ) ( ) ( ) ( ) .65 .22 .672.
- Let: B = rented a car for business reasons P = rented a car for personal reasons a. P (B P) = P (B) + P (P) - P (B P) = .54 + .458 - .30 =. b. P (Neither) = 1 – .698 =. 29. a.
P ( E ) =
P ( R ) = 2851854 . P ( D ) = 2851964 .2851
b. Yes; P ( E D ) = 0 c. Probability = 10332375 . d. Let F denote the event that a student who applies for early admission is deferred and later admitted during the regular admission process. Events E and F are mutually exclusive and the addition law applies. P ( E F ) = P ( E ) + P ( F ) P ( E ) = .3623 from part (a) Of the 964 early applicants who were deferred, we expect 18%, or .18(964) students, to be admitted during the regular admission process. Thus, for the total of 2,851 early
Y = responded yes, N = responded no b. P( M ) = .492, P( W ) =. P( Y ) = .396, P( N ) =. c. P( Y | M ) = .210/ =. d. P( Y | W ) = .186/ =. e. P( Y ) = .396/1 =. f. P( M ) = .492 in the sample. Yes, this seems like a good representative sample based on gender.
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33. a.
Undergraduate Major Business Engineering Other Totals Intended Enrollment Status Full-Time 0 0 0 0. Part-Time 0 0 0 0. Totals 0 0 0 1.
U = US Airways flight Given: P ( O | J ) = .768 P ( O | N ) = .715 P ( O | U ) =. P ( J ) = .30 P ( N ) = .32 P ( U ) =. Joint probabilities using the multiplication law P ( J O ) P ( J ) P ( O | J )(.30)(.768). P ( N O ) P ( N ) P ( O | N )(.32)(. 7 15).22 88 P ( U O ) P ( U ) P ( O | U )(.38)(.822). With the marginal probabilities P ( J ) = .30, P ( N ) = .32, and P ( U ) = .38 given, the joint probability table can then be shown as follows. On Time Late Total Jet Blue .2304 .0696. United .2288 .0912. US Airways .3124 .0676. Total .7716 .2284 1. b. Using the joint probability table, the probability of an on-time flight is the marginal probability P ( O ) = .2304 + .2288 + .3124 =.
c. Because US Airways has the highest percentage of flights into terminal C, US Airways with P ( U ) = .38 is the most likely airline for Flight 1382. d. From the joint probability table, P ( L ) = .2284:
P ( JL ) P ( PJ ( L ) L ).0696.
P ( NL ) P ( PN ( L ) L ).0912.
P ( UL ) P ( UP ( L ) L ).0676.
Most likely airline for Flight 1382 is now United with a probability of .3992. US Airways is now the least likely airline for this flight with a probability of .2961. 35. a. The total sample size is 1,010. Dividing each entry by 1,010 provides the following joint probability table. I Am My Spouse We Are Equal Husband .2752 .1257 .1010. Wife .2871 .1099 .1010. .5624 .2356. Let I = I am S = My spouse E = We are equal H = Husband W = Wife b. Using the marginal probabilities, P(I) = .5642, P(S) = .2356 and P ( E) = .2020. “I am” is the most likely response. It is over twice as likely as either “My spouse” or “We are equal.”
c. ()(). 2752. 5483 (). PI H PI H PH
P (Miss the Shot) P (Miss the Shot) = 1 – .9951 =.
d. For the player who makes 58% of his free throws, we have:
P (Make the Shot) = .58 for each foul shot, so the probability that this player will make two consecutive foul shots is P (Make the Shot) P (Make the Shot) = (.58)(.58) = .3364. Again, there are three unique ways that this player can make at least one shot: He can make the first shot and miss the second shot, miss the first shot and make the second shot, or make both shots. Because the event “Miss the Shot” is the complement of the event “Make the Shot” P (Miss the Shot) = 1 – P (Make the Shot) = 1 – .58 = .42. Thus, P (Make the Shot) P (Miss the Shot) = (.58)(.42) =. P (Miss the Shot) P (Make the Shot) = (.42)(.58) =. P (Make the Shot) P (Make the Shot) = (.58)(.58) =. . We can again find the probability the 58% free-throw shooter will miss both shots in two ways. We can calculate the probability directly: P (Miss the Shot) P (Miss the Shot) = (.42)(.42) =. Or we can recognize that the event “Miss Both Shots” is the complement of the event “Make at Least One of the Two Shots,” so P (Miss the Shot) P (Miss the Shot) = 1 – .9951 =.
Intentionally fouling the 58% free-throw shooter is a better strategy than intentionally fouling the 93% shooter. 37. Let C = person could give up cell phone T = person could give up television a. 𝑃ሺ𝐶ሻൌ. b. 𝑃ሺ𝑇|𝐶ሻൌሺ்∩ሻሺሻ ൌ.ଷଵ.ସ଼ൌ. c. 𝑃ሺ𝑇|𝐶ሻൌሺ்∩ሺሻሻൌ.ଷ଼.ହଶൌ.73 (Note: 𝐶 is the complement of C, or in other words, person could not give up cell phone.) d. The probability a person could give up television if they could not give up a cell phone is higher than the probability a person could give up television if they could up a cell phone. 38. Let Y = has a college degree N = does not have a college degree D = a delinquent student loan a. From the table, P ( Y ). b. From the table, P ( N ). c. P ( D | Y ) P ( PD ( Y ) Y ).16.
d. P ( D | N ) P ( PD ( N ) N ).34. e. Individuals who obtained a college degree have a .3810 probability of a delinquent student loan, whereas individuals who dropped out without obtaining a college degree have a .5862 probability of a delinquent student loan. Not obtaining a college degree
Statistics for Business & Economics 14e Metric Version Chapter4
Course: Statistic
University: National Taipei University
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