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.
Assuming an equal chance of choosing each chip, the chances of pulling #505 is: \[\frac{1}{540}\]
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.)
\[ S = {(A,E,F),(A,E,V),(A,T,F),(A,T,V),(C,E,F),(C,E,V),(C,T,F),(C,T,V)}\]
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.
Since there are 13 hearts, we have a \(\frac{13}{52}\) chance of getting a heart on any draw. However, since we also do not want face cards (J,Q,K), we need to remove 3 from our possible desired outcomes, leaving us with a probability of: \[\frac{10}{52}\]
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.
There are \(6 \times 6 = 36\) possible outcomes. Of those, we can count up the favorable results: \[ P = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}\] So, we have a probability of: \[\frac{10}{36}\]
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:
| Residence | Males | Females |
|---|---|---|
| Apartment | 233 | 208 |
| Dorm | 159 | 138 |
| With Parent(s) | 102 | 280 |
| Sorority/Fraternity House | 220 | 265 |
| 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.
male <- 233 + 159 + 102 + 220 + 250We add the total number of male customers, 964, and divide by the total number of customers, 2001 to get: \[\frac{964}{2001}\]
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.
Since we are replacing the cards after each draw, these are independent events. To find the conjunction of these events, we multiply the probabilities of each.
For clubs, we have 13 possibilities out of 52. Black cards represent half of the deck, or 26 of 52. For face cards, there are 3 of each of 4 suits, so 12 total.
Our probabilities are therefore:
p1 <- 13/52
p2 <- 26/52
p3 <- 12/52
p <- round(p1 * p2 * p3,4)\[\frac{13}{52} \times \frac{26}{52} \times \frac{12}{52} = 0.0288\]
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.
If the first card drawn is a heart, that leaves only 51 cards left in the deck for the second draw, and all of the spades are still present. So, the probability of a spade on the second draw is: \[\frac{13}{51}\]
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.
The probability of choosing a heart in the first draw is \(\frac{13}{52}\). Since we are not replacing, there are now 51 cards left in the deck.
Normally there are 26 red cards, but since we chose a heart the first time, there are only 25 left that we can pick on the second draw. So, the probability of getting a red card is now \(\frac{25}{51}\).
p1 <- 13/52
p2 <- 25/51
p <- round(p1 * p2,4)The combined probability of both events is: \[\frac{13}{52} \times \frac{25}{51} = 0.1225\]
There are 85 students in a basic math class. The instructor must choose two students at random:
| Residence | Males | Females |
|---|---|---|
| Freshmen | 12 | 12 |
| Sophmores | 19 | 15 |
| Juniors | 12 | 4 |
| Seniors | 7 | 4 |
What 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.
total <- 12 + 12 + 19 + 15 + 12 + 4 + 7 + 4
jf <- 4
fm <- 12
p <- (jf/total) + (fm/(total-1)) # first the junior female and then the freshman male\[\frac{4}{85} + \frac{12}{84} = 0.189916\]
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?
\[\frac{52}{141}\] > 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.
\[\frac{52}{102}\]
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?
6 * 5 * 3## [1] 90A doctor visits her patients during morning rounds. In how many ways can the doctor visit 5 patients during the morning rounds?
factorial(5)## [1] 120A 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?
permut <- function(n,r)
{
return(factorial(n)/factorial(n-r))
}
permut(8,5)## [1] 6720A person rolls a standard six-sided die 9 times. In how many ways can he get 3 fours, 5 sixes and 1 two?
factorial(9)/(factorial(5)*factorial(3)*factorial(1))## [1] 504\[\frac{9!}{5!3!1!}=504\]
How many ways can Rudy choose 6 pizza toppings from a menu of 14 toppings if each topping can only be chosen once?
factorial(14)/(factorial(14-6)*factorial(6))## [1] 30033 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?
factorial(52)/(factorial(52-3)*factorial(3))## [1] 22100You 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?
12 * 9 * 5## [1] 540You 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?
ltrs <- factorial(26)/factorial(26-5)
digits <- factorial(5)/factorial(5-3)
ltrs * digits## [1] 473616000Evaluate the following expression. \(_9P_4\)
factorial(9) / factorial(9-4)## [1] 3024Evaluate the following expression. \(_11C_8\)
factorial(11)/(factorial(11-8)*factorial(8))## [1] 165Evaluate the following expression. \(\frac{_{12}P_8}{_{12}C_4}\)
\[\frac{\frac{12!}{4!}}{\frac{12!}{8!4!}} = \frac{12!}{4!} \times \frac{8!4!}{12!} = \frac{8!4!}{4!} = 8!\]
factorial(8)## [1] 40320The 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?
factorial(13)/factorial(13-7)## [1] 8648640In how many ways can the letters in the word ‘Population’ be arranged?
There are 10 letters in the work population, so there would normally be \(10!\) ways to arrange those letters. HOwever, because there are 2 letter “p”s, we have duplicate combinations we need to remove. Thus, we have \(\frac{10!}{2!}\) ways to arrange the letters:
factorial(10)/factorial(2)## [1] 1814400Consider the following data:
| \(x\) | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|
| \(p(x)\) | 0.1 | 0.2 | 0.3 | 0.2 | 0.2 |
Step 1. Find the expected value \(E(X)\)
E <- round((0.1)*(5)+(0.2)*(6)+(0.3)*(7)+(0.2)*(8)+(0.2)*(9),1)\[E(X)=(0.1)(5)+(0.2)(6)+(0.3)(7)+(0.2)(8)+(0.2)(9) = 7.2\]
Step 2. Find the variance.
\[\sigma^2 = E[(x - \mu)^2]\]
varx <- (0.1*(5-E)^2 + 0.2*(6-E)^2 + 0.3*(7-E)^2 + 0.2*(8-E)^2 + 0.2*(9-E)^2)
round(varx,1)## [1] 1.6Step 3. Find the standard deviation.
\[\sigma = \sqrt{\sigma^2}\]
round(varx^.5,1)## [1] 1.2Step 4. Find the value of \(P(X) \lt 9\).
\[ 1 - 0.2 = 0.8 \]
Step 5. Find the value of \[P(x) \ge 7\]
\[ P(x) \ge 7 = P(7) + P(8) + P(9) = 0.3 + 0.2 + 0.2 = 0.7\]
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.
p_freethrow <- 188/376
p3_freethrows <- p_freethrow ^ 3
e <- (p3_freethrows * 23 + (1-p3_freethrows) * -4)
e## [1] -0.625Step 2. If you played this game 994 times how much would you expect to win or lose?
e * 994## [1] -621.25Flip 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.
Getting 8 tails or less on 11 flips can be represented by $P(0 X 8) of the binomial distribution:
p <- sum(dbinom(0:8,11,0.5)) # Probability of getting 8 or fewer tails on 11 flips
q <- 1-p # Probability of getting > 8 tails on 11 flips
E <- round(p*1 + q*-7,2)
E## [1] 0.74So, we’d expect to win 0.74 dollars on such a bet.
If you played this game 615 times how much would you expect to win or lose?
I would expect to win 455.1 dollars
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.
The probability of drawing a club on the first draw is \(\frac{13}{52}\) and the probaility of drawing a second club is \(\frac{12}{51}\). So the total probability would be \(\frac{13}{52} \times \frac{12}{51} = \frac{13}{221}\)
p <- 13/221
q <- 1 - p
E <- round((p*583)+(q*-35),2)
E## [1] 1.35Step 2. If you played this game 632 times how much would you expect to win or lose?
I would expect to win about 853.2 dollars.
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(sum(dbinom(0:2,10,0.30)),2)## [1] 0.38A 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?
For a binomial distribution: \[ E(X) = p\times n = \frac{3}{10}\times 5 = 1.5 \space \mathrm {defective}\space \mathrm{bulbs}\]
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(1 - sum(dpois(0:5,5.5)),4) # 1 minus the probability of getting 0-5 orders## [1] 0.4711At 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(1 - sum(dpois(0:4,5.7)),4) # 1 minus the probability of getting 0-4 cars## [1] 0.6728The 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(sum(dpois(0:1,2.8)),4) # probability of getting 0-1 crashes in a week## [1] 0.2311A 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?
1 - sum(dhyper(0:1,6,19,8)) # 1 minus the probability of selecting 0 or 1 employee over the age of 50## [1] 0.6505929Unknown 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?
sum(dhyper(1:6,10,15,8))## [1] 0.9923445