Test Name: Data Science Math (Second week) Submitted by: Sudhan Maharjan 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. Here, Total identical chips = 540 Random number drawn (C505) = 1 P(C505) = 1/540 = 0.0019 > round(1/540, 4) [1] 0.0019

Therefore, the probability of getting chip numbered 505 is 0.019%.

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.)

Here, A = asparagus C = cheese E = eggs T = turkey F = French V = vinaigrette We have total choice of Toppings = 2 Meats = 2 Dressings = 2 Total combinations = 222 = 8. Therefore, in total we can get 8 different kinds of combination from the sample. > Toppings <- c(“A”,“A”,“A”,“A”,“C”,“C”,“C”,“C”) > Meats <- c(“E”,“E”,“T”,“T”,“E”,“E”,“T”,“T”) > Dressings <- c(“F”,“V”,“F”,“V”,“F”,“V”,“F”,“V”) > Salad <- data.frame(Toppings, Meats, Dressings) > Salad Toppings Meats Dressings 1 A E F 2 A E V 3 A T F 4 A T V 5 C E F 6 C E V 7 C T F 8 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.

Here, Total cards in a deck SD = 52 Total hearts in a deck H = 13 Total hearts which are face card HFC= 3 Total hearts without face card HWFC = 13 - 3 = 10 Probability of getting total hearts without face cards from a standard deck P(HWFC) = 10/52 = 0.192307 Therefore, there is a 19.2307% chance of getting hearts without face cards.

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.

Here, x = number of dots on the dice. P(x) = probability of obtaining x dots in a dice In Dice 1 P1(x = 1) = 1 P1(x = 2) = 2 P1(x = 3) = 3 P1(x = 4) = 4 P1(x = 5) = 5 P1(x = 6) = 6

In Dice 1 P2(x = 1) = 1 P2(x = 2) = 2 P2(x = 3) = 3 P2(x = 4) = 4 P2(x = 5) = 5 P2(x = 6) = 6

Total possible combinations = 6 * 6 = 36. So, the possible outcomes whose sum will be less than 36 are: P1(x = 1) + P2(x = 1) = 1 + 1 = 2 P1(x = 1) + P2(x = 2) = 1 + 2 = 3 P1(x = 1) + P2(x = 3) = 1 + 3 = 4 P1(x = 1) + P2(x = 4) = 1 + 4 = 5 P1(x = 2) + P2(x = 1) = 2 + 1 = 3 P1(x = 2) + P2(x = 2) = 2 + 2 = 4 P1(x = 2) + P2(x = 3) = 2 + 3 = 5 P1(x = 3) + P2(x = 1) = 3 + 1 = 4 P1(x = 3) + P2(x = 2) = 3 + 2 = 5 P1(x = 4) + P2(x = 1) = 4 + 1 = 5

Therefore, we have that 10 different combinations to get the results that is less than 6. P(P1(x) +P2(x)) < 6 P(P1(x) +P2(x)) = 10/36 = 0.2777778 The probability of getting the combination less than 6 is 27.78%

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.

Gender and Residence of Customers Males Females Apartment 233 208 Dorm 159 138 With Parents(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. Here, Total Males M = 964 Total Females F = 1037 Total Customers C = 2001 P(M) = 964/2001 = 0.481759 Therefore, the probability of a customer being male is 48.18%

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.

Here, 3 cards are drawn from a standard deck with replacement Total cards in a Standard Card deck SC = 52 Number of clubs = 13 Number of black cards = 26 Number of faces in a deck = 12

P(C) = 13/52 = 0.25

P(B) = 26/52 = 0.5 P(F) = 12/52 = 0.230769 Now, P(C) * P(B) * P(F) = 0.25 * 0.5 * 0.2308 = 0.02885 Therefore, the probability is 2.885%.

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.

Here, Since, 2 cards are drawn from the deck without replacement Total cards in Standard Deck SC1 = 52 Total cards in Standard Deck SC2 = 51 Number of spades in first deck S1 = 13 Number of spades in second deck S2 = 13 Now, P(S2) = 13/51 = 0.2549 Therefore, the probability of getting spade for the second card drawn without replacement is 25.49%.

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.

Here, Since, 2 cards are drawn from the deck without replacement Total cards in Standard Deck SC1 = 52 Total cards in Standard Deck SC2 = 51 Number of hearts in first deck H1 = 13 P(H1) = 13/52 = 0.25 Since, in a deck there are 26 red cards and 1 red is not replaced in the deck we have now total Red (R2) = 25 Now, P(R2) = 25/51 = 0.4901 Again, P(H1) * P(R2) = 0.25 * 0.4901 = 0.1225 Therefore, the probability of getting heart and then a red card without replacements is 12.25%.

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

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. Here, Total students S1 = 85 Total students in second selection S2= 84 Total Males M = 50 Total Females F = 35 P(JF) * P(FM) = 4/85 * 12/84 = 0.047 * 0.1428 = 0.00671 Therefore, the probability for a junior female and then a freshmen male chosen at random is 0.671%.

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.

Here, Probability of graduate and male P(G and M) = 52/300 = 0.1733 Probability of male P(M) = 141/300 = 0.47 Now, P(G|M) = 0.1733/0.47 = 0.3687 Therefore, the probability of randomly chosen applicant who has a graduate degree, given that they are male is 36.87%

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.
Here, Probability of graduate and male P(G and M) = 52/300 = 0.1733 Probability of male P(G) = 102/300 = 0.34 Now, P(G|M) = 0.1733/0.34 = 0.5097 Therefore, the probability of randomly chosen applicant who has a graduate degree, given that they are male is 50.97%.

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?

Here, Different types of drink D = 6 Different types of sandwiches S = 5 Different types of chips C = 3 Different combinations = D * S * C = 6 * 5 * 3 = 90 Therefore, there are 90 different value meal packages possible.

A doctor visits her patients during morning rounds. In how many ways can the doctor visit 5 patients during the morning rounds?

Here, The doctor makes morning rounds and visits 5 different patients. Now, 5! = 54321 = 120 Therefore, there are 120 different ways to visit 5 patients during the morning round.

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?

Here, We have songs n = 8, and we can select x = 5 in random order to lineup P(n, x) = n! / (n-k)! = 8! / (8 - 5)! = (8 * 7 * 6 * 5 * 4 * 3!) / 3! = 6720 Therefore, there will be 6720 possible lineups for the songs.

A person rolls a standard six-sided die 9 times. In how many ways can he get 3 fours, 5 sixes and 1 two?

Here, We have n = 9, and we can get fours = 3, sixes = 5, two = 1 Different ways = 9! / (3! * 5! * 1!) = 504 Therefore, there are 540 different ways to get 3 fours, 5 sixes and 1 two. How many ways can Rudy choose 6 pizza toppings from a menu of 14 toppings if each topping can only be chosen once? Here, We have Different pizza n = 6 Different toppings t = 14 P(k, n) = k! / (k - n)! = 14! / (14 - 6)! = 2162160 Therefore, Rudy can choose 6 pizza toppings in 2162160 different ways from a menu of 14 toppings.

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?

Here, Total cards in standard deck SD = 52 Total cards drawn = 3 without replacement C(52, 3) = P(52, 3) / 3! = 52! / ((52 - 3)! * 3!) = 22100 Therefore, there are 22100 different ways we can draw the cards without replacement.

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?

Here, Different ways to choose TV = 12 Different ways to choose surround sound systems = 9 Different ways to choose DVD players = 5 Now, Different ways home theater system can be built = TV * Sound System * DVD = 12 * 9 * 5 = 540 Therefore, there are 540 different ways to build the home theater systems.

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?

Here, Since, we cannot repeat the characters and digits Alphabets = 26 Total odd numbers from 0-9 are 1, 3, 5, 7, 9 but we can choose only 3 numbers. Length of a password = 8 Different ways to create password = P(26, 5) * P(5, 3) = 26!/(26-5)! * 5!/(5-3)! = 473616000 Therefore, there are 473616000 possible choices to create the password.

Evaluate the following expression 9P4.

Here, By using nPr formula, = n! / (n - r)! = 9! / (9 - 4)! = 3024 Therefore, 9P4 = 3024.

Evaluate the following expression 11C8

Here, 11C8 = nPr(n - r)! / r! = 11! / ((11 - 8)! * 8!) = 165 Therefore, 11C8 = 165.

Evaluate the following expression.

12P8 12C4 = 40320 Therefore, the result is 40320.

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?

Here, Since, 1 person can hold only one position, we have 13P7 = 13! / (13 - 7)! = 8648640 Therefore, there are 8648640 different ways a president can elect the candidate.

In how many ways can the letters in the word 'Population' be arranged?

Here, Total letters in Population = 10 Repeat letter P = 2 Repeat letter o = 2 Now, Different ways we can arrange = 10! / (2! * 2!) = 907200 Therefore, the word Population can be arranged in 907200 different ways.

Consider 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 ????????( ???????? ). Round your answer to one decimal place. E(X) = ???_(k=0)^n??????(x_k.p(x_k ))??? E(X) = 7.2

Step 2. Find the variance. Round your answer to one decimal place. Var(x) = ???_in??????[(???x_i)???^2*p(x_i )]-E???(x)]???2??? = 1.6 The variance is V(x) = 1.6.

Step 3. Find the standard deviation. Round your answer to one decimal place. Here, Standard Deviation SD(x) = ???V(x) = 1.3

Step 4. Find the value of ????(???? ê 9). Round your answer to one decimal place. Here, E(x >= 9) = ?????????(x???9)*p(x???9) ??? = 0.2 P(x >= 9) = 0.2

Step 5. Find the value of ????(???? Ç7). Round your answer to one decimal place. Here, P(x <=7) = 0.6

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. Here, Probability of event E is P(E) = number of ways an event can occur Total number of possible outcomes For P1 = 188 / 376 = 0.5 Therefore, the probability for an event E is 0.5. Probability of making 3 shots = (P1)3 = 0.125 Now, the probability of not making the shots = 1 - 0.125 = 0.875 Again, Expected value of the proposition E = (23 * 0.125) + (-4 * 0.875) = -0.62 Therefore, the expected value will be -$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.) Here, E994 = 994 * E = 994 * -0.62 = -616.28 If ever I played this game 994 times there is a chance for me to lose $616.28.

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. Step 2. If you played this game 615 times how much would you expect to win or lose? (Losses must be entered as negative.)

Here, In a coin there are two sides n = 2 and there is 50 % chance so, ?? = 0.5

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. Here, Number of cards in a deck = 52 Number of clubs in a deck = 13 P(E) = (13/52) * (12/51) = 0.25 * 0.23529 = 0.058822 Probability of failure P(F) = 1 - 0.05822 = 0.941178 Expected value of the proposition = 583 * (0.05822) + (-35) * 0.941178 = 33.94 - 32.94 = 1 The expected value of the proposition is 1.

Step 2. If you played this game 632 times how much would you expect to win or lose? (Losses must be entered as negative.) Here, If I played this game 632 times, I will lose 632 * $1 = $632.00

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)

Here, Sample of bulbs = 10 If the number of defective bulb is less than or equal to 2 the inspection passed. 30% of the bulbs are defective = 0.3 defective Using formula p < - round(spw(10, 2, 0.3), 3) p = 0.355 The probability of passing the lot inspection is 35.5%.

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.

Here, Number of bulbs n = 5 Defective bulbs is d = 30% EV = n* d = 0.3* 5 = 1.5 Therefore, the expected value of defective bulbs is 1.5.

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)

Here, Mean = 5.5 Now, we have p(k) = (??2* e(-??))/k! = 0.5289 Now, Upper Tail = 1 - 0.5289 = 0.4711 Therefore, the number of special orders sent out that will be more than 5 is 47.11%.

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)

Here, Mean = 5.7 Now, we have p(k) = (??2* e(-??))/k! = 0.3272 Now, Upper tail = 1 - 0.3272 = 0.6728 Therefore, the probability of more than 4 customers in any hour is 67.28%.

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)

Here, We have a daily mean of 0.4 per day Daily lambda = 0.4 Weekly lambda = 0.4 * 7 With this transformation we need to take our k = 1 week as well. We are asked for the lower Tail, by using the above function; we obtain

Lt <-round(p(1, 0.4 * 7),4) = 0.2311 Therefore, the probability in any 7-day week, computer will crash no more than 1 time is 23.11%

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. 


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.