MSDS Summer Bridge 2018 Data Science Math - HW 2
Jeff Littlejohn, 7/21/2018
Note: Had issues using Latex with combinatorics.
1. There are 540 identical plastic chips numbered 1 through 540 in a box. What is the probability of reaching into the box and randomly drawing the chip numbered 505? Express your answer as a fraction or a decimal number rounded to four decimal places. \[P(505) = \frac{1}{540}\]
2. Write out the sample space for the given experiment. Separate your answers using commas.
When deciding what you want to put into a salad for dinner at a restaurant, you will choose one of the following extra toppings: asparagus, cheese. Also, you will add one of following meats: eggs, turkey. Lastly, you will decide on one of the following dressings: French, vinaigrette. (Note: Use the following letters to indicate each choice: A for asparagus, C for cheese, E for eggs, T for turkey, F for French, and V for vinaigrette.)
\[S = {(A,E,F),(A,E,V),(A,T,F),(A,T,V),(C,E,F),(C,E,V),(C,T,F),(C,T,V)}\]:
3. A card is drawn from a standard deck of 52 playing cards. What is the probability that the card will be a heart and not a face card? Write your answer as a fraction or a decimal number rounded to four decimal places. \[Event A = {heart}\] \[Event B = {heart face cards}\] \[P(A \cap B) = P(A) * P(B|A)\] \[P(A \cap B) = \frac{13}{52} * \frac{10}{13} = \frac{130}{676} = \frac{5}{26} \]
4. A standard pair of six-sided dice is rolled. What is the probability of rolling a sum less than 6? Write your answer as a fraction or a decimal number rounded to four decimal places.
library(prob)## Loading required package: combinat##
## Attaching package: 'combinat'## The following object is masked from 'package:utils':
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## combn## Loading required package: fAsianOptions## Loading required package: timeDate## Loading required package: timeSeries## Loading required package: fBasics## ## Rmetrics Package fBasics## Analysing Markets and calculating Basic Statistics## Copyright (C) 2005-2014 Rmetrics Association Zurich## Educational Software for Financial Engineering and Computational Science## Rmetrics is free software and comes with ABSOLUTELY NO WARRANTY.## https://www.rmetrics.org --- Mail to: info@rmetrics.org## Loading required package: fOptions## ## Rmetrics Package fOptions## Pricing and Evaluating Basic Options## Copyright (C) 2005-2014 Rmetrics Association Zurich## Educational Software for Financial Engineering and Computational Science## Rmetrics is free software and comes with ABSOLUTELY NO WARRANTY.## https://www.rmetrics.org --- Mail to: info@rmetrics.org## Loading required package: hypergeo## Loading required package: VGAM## Loading required package: stats4## Loading required package: splines##
## Attaching package: 'VGAM'## The following object is masked from 'package:hypergeo':
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## is.zero## The following object is masked from 'package:fAsianOptions':
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## erf##
## Attaching package: 'prob'## The following objects are masked from 'package:base':
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## intersect, setdiff, unionf4 <- rolldie(2)
f4## X1 X2
## 1 1 1
## 2 2 1
## 3 3 1
## 4 4 1
## 5 5 1
## 6 6 1
## 7 1 2
## 8 2 2
## 9 3 2
## 10 4 2
## 11 5 2
## 12 6 2
## 13 1 3
## 14 2 3
## 15 3 3
## 16 4 3
## 17 5 3
## 18 6 3
## 19 1 4
## 20 2 4
## 21 3 4
## 22 4 4
## 23 5 4
## 24 6 4
## 25 1 5
## 26 2 5
## 27 3 5
## 28 4 5
## 29 5 5
## 30 6 5
## 31 1 6
## 32 2 6
## 33 3 6
## 34 4 6
## 35 5 6
## 36 6 6
\(6 x 6 = 36\) possible outcomes Spaces that have sum less than 6: \({(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}\) \[P(sum < 6) = \frac{10}{36} = \frac{5}{18}\]
5.A pizza delivery company classifies its customers by gender and location of residence. The research department has gathered data from a random sample of 2001 customers. The data is summarized in the table below.
rn5 = c("Apartments","Dorm","With Parent(s)","Sorority/Fraternity House","Other")
a5 = c(233,159,102,220,250)
b5 = c(208,138,280,265,146)
df5 = data.frame(rn5,a5,b5)
colnames(df5) = c("Residence Type","Male","Female")
df5## Residence Type Male Female
## 1 Apartments 233 208
## 2 Dorm 159 138
## 3 With Parent(s) 102 280
## 4 Sorority/Fraternity House 220 265
## 5 Other 250 146
What is the probability that a customer is male? Write your answer as a fraction or a decimal number rounded to four decimal places.
m5 <- sum(df5$Male)
m5## [1] 964t5 <- sum(df5$Male) + sum(df5$Female)
t5## [1] 2001\[P(male) = \frac{964}{2001}\]
6. Three cards are drawn with replacement from a standard deck. What is the probability that the first card will be a club, the second card will be a black card, and the third card will be a face card? Write your answer as a fraction or a decimal number rounded to four decimal places.
\[P(club) = \frac{1}{4}\] \[P(black) = \frac{1}{2}\] \[P(face) = \frac{3}{13}\] With replacement, order matters
(1/4) * (1/2) * (3/13)## [1] 0.028846153/104## [1] 0.02884615\[P = \frac{3}{104}\]
7.Two cards are drawn without replacement from a standard deck of 52 playing cards. What is the probability of choosing a spade for the second card drawn, if the first card, drawn without replacement, was a heart? Write your answer as a fraction or a decimal number rounded to four decimal places. \[P(spade|heart) = \frac{13}{51}\]
8. Two cards are drawn without replacement from a standard deck of 52 playing cards. What is the probability of choosing a heart and then, without replacement, a red card? Write your answer as a fraction or a decimal number rounded to four decimal places. \[P(heart then red card) = \frac{13}{52} * \frac{25}{51}\]
(13/52) * (25/51)## [1] 0.12254913 * 25## [1] 32552 * 51## [1] 265225/204## [1] 0.122549\[P(heart then red card) = \frac{25}{204}\]
9. There are 85 students in a basic math class. The instructor must choose two students at random.
Students in a Basic Math Class Males Females Freshmen 12 12 Sophomores 19 15 Juniors 12 4 Seniors 7 4
rn9 = c("Freshmen","Sophmores","Juniors","Seniors")
a9 = c(12,19,12,7)
b9 = c(12,15,4,4)
df9 = data.frame(rn9,a9,b9)
colnames(df9) = c("Year","Male","Female")
df9## Year Male Female
## 1 Freshmen 12 12
## 2 Sophmores 19 15
## 3 Juniors 12 4
## 4 Seniors 7 4What is the probability that a junior female and then a freshmen male are chosen at random? Write your answer as a fraction or a decimal number rounded to four decimal places.
tot9 <- sum(df9$Male) + sum(df9$Female)
tot9## [1] 85\[P(junior female then freshman male) = \frac{4}{85} * \frac{12}{84}\]
(4/85) * (12/84)## [1] 0.00672268984 * 85## [1] 714048/7140## [1] 0.0067226894/595## [1] 0.006722689\[P(junior female then freshman male) = \frac{4}{595}\]
10.Out of 300 applicants for a job, 141 are male and 52 are male and have a graduate degree.
Step 1. What is the probability that a randomly chosen applicant has a graduate degree, given that they are male? Enter your answer as a fraction or a decimal rounded to four decimal places.
\[P(grad degree|male) = \frac{52}{141}\]
Step 2. If 102 of the applicants have graduate degrees, what is the probability that a randomly chosen applicant is male, given that the applicant has a graduate degree? Enter your answer as a fraction or a decimal rounded to four decimal places. \[P(male|grad degree) = \frac{52}{102} = \frac{26}{51}\]
11. A value meal package at Ron’s Subs consists of a drink, a sandwich, and a bag of chips. There are 6 types of drinks to choose from, 5 types of sandwiches, and 3 types of chips. How many different value meal packages are possible?
Multiplication rule of counting
6 * 5 * 3## [1] 9090
12. A doctor visits her patients during morning rounds. In how many ways can the doctor visit 5 patients during the morning rounds?
Factorial rule of counting
factorial(5)## [1] 120\[5! = 120\]
13. A coordinator will select 5 songs from a list of 8 songs to compose an event’s musical entertainment lineup. How many different lineups are possible?
Assuming order does not matter - Combination: \[C(n,r) = \frac{n!}{(n-r)!r!}\]
factorial(8)/(factorial(8-5) * factorial(5))## [1] 5656
Assuming order matters - Permutation: \[P(n,r) = \frac{n!}{(n-r)!}\]
factorial(8)/factorial(8-5)## [1] 67206720
14. A person rolls a standard six-sided die 9 times. In how many ways can he get 3 fours, 5 sixes and 1 two?
This is wrong, but I will keep looking into it.
Total outcomes\(= 6^9 = 10077696\) Getting 3 fours: \(9_C_3 = \frac{1}{6}^3 * (1-\frac{1}{6})^{9-3}\) Getting 5 sixes: \(9_C_5 = \frac{1}{6}^5 * (1-\frac{1}{6})^{9-5}\) Getting 1 two: \(9_C_1 = \frac{1}{6}^1 * (1-\frac{1}{6})^{9-1}\)
\[(\frac{1}{6})^3 * (\frac{5}{6})^6 * (\frac{1}{6})^5 * (\frac{5}{6})^4 * (\frac{1}{6}) * (\frac{5}{6})^8}\]
(1/6)^3 * (5/6)^6 * (1/6)^5 * (5/6)^4 * (1/6) * (5/6)^8## [1] 3.727145e-09
15. How many ways can Rudy choose 6 pizza toppings from a menu of 14 toppings if each topping can only be chosen once?
Combination: \[C(n,r) = \frac{n!}{(n-r)!r!}\]
factorial(14)/(factorial(14-6) * factorial(6))## [1] 3003
16. 3 cards are drawn from a standard deck of 52 playing cards. How many different 3-card hands are possible if the drawing is done without replacement?
\[C(n,r) = \frac{n!}{(n-r)!r!}\]
factorial(52)/(factorial(52-3) * factorial(3))## [1] 22100
17. You are ordering a new home theater system that consists of a TV, surround sound system, and DVD player. You can choose from 12 different TVs, 9 types of surround sound systems, and 5 types of DVD players. How many different home theater systems can you build?
Multiplication rule of counting
12 * 9 * 5## [1] 540540
18. You need to have a password with 5 letters followed by 3 odd digits between 0 - 9 inclusively. If the characters and digits cannot be used more than once, how many choices do you have for your password?
26 * 25 * 24 * 23 * 22 * 10 * 9 * 8## [1] 56833920005683392000
19. Evaluate the following expression.
\[ 9P_4\]??? Permutation: \[P(n,r) = \frac{n!}{(n-r)!}\]
factorial(9)/factorial(9-4)## [1] 30243024
20. Evaluate the following expression.
\[11C_8\] \[C(n,r) = \frac{n!}{(n-r)!r!}\]
factorial(11)/(factorial(11-8) * factorial(8))## [1] 165165
21. Evaluate the following expression.
\[\frac{12P_8}{12C_4}\]
p21 <- factorial(12)/factorial(12-8)
c21 <- factorial(12)/(factorial(12-4) * factorial(4))
p21/c21## [1] 4032040320
???4
22. The newly elected president needs to decide the remaining 7 spots available in the cabinet he/she is appointing. If there are 13 eligible candidates for these positions (where rank matters), how many different ways can the members of the cabinet be appointed? \[13P_7\]
factorial(13)/(factorial(13-7)*factorial(7))## [1] 17161716
23. In how many ways can the letters in the word ‘Population’ be arranged? 10 letters - 2 letters included twice
factorial(10)/(factorial(2) * factorial(2))## [1] 907200907200
24. Consider the following data:
x24 = c(5,6,7,8,9)
px24 = c(.1,.2,.3,.2,.2)
df24 = data.frame(x24,px24)
colnames(df24) = c("x","p(x)")
df24## x p(x)
## 1 5 0.1
## 2 6 0.2
## 3 7 0.3
## 4 8 0.2
## 5 9 0.2Step 1. Find the expected value (\(Ex\)). Round your answer to one decimal place.
exp24 <- sum(x24*px24)
exp24## [1] 7.2
Step 2. Find the variance. Round your answer to one decimal place.
var24 <- sum((x24 - exp24)^2 * px24)
var24## [1] 1.56
Step 3. Find the standard deviation. Round your answer to one decimal place.
sd24 <- var24^(1/2)
sd24## [1] 1.249
Step 4. Find the value of (X>=9). Round your answer to one decimal place.
.2
Step 5. Find the value of (X <= 7). Round your answer to one decimal place. .6
25. Suppose a basketball player has made 188 out of 376 free throws. If the player makes the next 3 free throws, I will pay you $23. Otherwise you pay me $4.
Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
#Odds of making next 3 free throws
(188/376)^3## [1] 0.125x25 = c(23,-4)
px25 = c(.125,.875)
df25 = data.frame(x25,px25)
colnames(df25) = c("x","p(x)")
df25## x p(x)
## 1 23 0.125
## 2 -4 0.875exp25 <- sum(x25*px25)
round(exp25,2)## [1] -0.62
Step 2. If you played this game 994 times how much would you expect to win or lose? (Losses must be entered as negative.)
exp25 * 994## [1] -621.25
26. Flip a coin 11 times. If you get 8 tails or less, I will pay you $1. Otherwise you pay me $7.
Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
Reworded - Get tails 9, 10, or 11 times - I pay $7.
#prob of tails 8 or fewer times
px26lt9 <- pbinom(8,11,.5)
#prob of tails 9-11 times
px26gte9 <- 1 - px26lt9
#expected value
exp26 <- px26lt9 * 1 + px26gte9 * -7
exp26## [1] 0.7382812Expected value = $.74
Step 2. If you played this game 615 times how much would you expect to win or lose? (Losses must be entered as negative.)
round(615 * exp26,2)## [1] 454.04Win $454.04
27. If you draw two clubs on two consecutive draws from a standard deck of cards you win $583. Otherwise you pay me $35. (Cards drawn without replacement.)
Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
#prob of consecutive clubs drawn w/o replacement
px27c <- (13/52) * (12/51)
#prob of complement
px27nc <- 1 - px27c
#expected value
exp27 <- px27c * 583 + px27nc * -35
round(exp27,2)## [1] 1.35Expected value is $1.35 per game.
Step 2. If you played this game 632 times how much would you expect to win or lose? (Losses must be entered as negative.)
round(exp27 * 632,2)## [1] 855.06Win $855.06
28. A quality control inspector has drawn a sample of 10 light bulbs from a recent production lot. If the number of defective bulbs is 2 or less, the lot passes inspection. Suppose 30% of the bulbs in the lot are defective. What is the probability that the lot will pass inspection? (Round your answer to 3 decimal places)
round(pbinom(2,10,.3),3)## [1] 0.383
29. A quality control inspector has drawn a sample of 5 light bulbs from a recent production lot. Suppose that 30% of the bulbs in the lot are defective. What is the expected value of the number of defective bulbs in the sample? Do not round your answer.
5 * .3## [1] 1.5
30. The auto parts department of an automotive dealership sends out a mean of 5.5 special orders daily. What is the probability that, for any day, the number of special orders sent out will be more than 5? (Round your answer to 4 decimal places)
# P(X>5)
round(ppois(5, 5.5,lower.tail = FALSE), 4)## [1] 0.4711
31. At the Fidelity Credit Union, a mean of 5.7 customers arrive hourly at the drive-through window. What is the probability that, in any hour, more than 4 customers will arrive? (Round your answer to 4 decimal places)
# P(X>4)
round(ppois(4, 5.7,lower.tail = FALSE), 4)## [1] 0.6728
32. The computer that controls a bank’s automatic teller machine crashes a mean of 0.4 times per day. What is the probability that, in any 7-day week, the computer will crash no more than 1 time? (Round your answer to 4 decimal places)
# P(X<=1)
round(ppois(1, 2.8), 4)## [1] 0.2311
33. A town recently dismissed 8 employees in order to meet their new budget reductions. The town had 6 employees over 50 years of age and 19 under 50. If the dismissed employees were selected at random, what is the probability that more than 1 employee was over 50? Write your answer as a fraction or a decimal number rounded to three decimal places.
#P(q>=1)
#more than 1 employee
q <- 1
#6 employees over 50
m <- 6
#19 under 50
n <- 19
#8 fired
k <- 8
#lower.tail because P[X > x]
P33 <- phyper(q, m, n, k, lower.tail = F)
round(P33, 3)## [1] 0.651
34. Unknown to a medical researcher, 10 out of 25 patients have a heart problem that will result in death if they receive the test drug. Eight patients are randomly selected to receive the drug and the rest receive a placebo. What is the probability that less than 7 patients will die? Write your answer as a fraction or a decimal number rounded to three decimal places.
#fewer than 7 will die
q34 <- 6
#10 will die with test drug
m34 <- 10
#15 won't die with test drug
n34 <- 15
#8 receive test drug in trial
k34 <- 8
P34 <- phyper(q34, m34, n34, k34)
round(P34, 3)## [1] 0.998