A box contains 54 red marbles, 9 white marbles, and 75 blue marbles. If a marble is randomly selected from the box, what is the probability that it is red or blue? Express your answer as a fraction or a decimal number rounded to four decimal places.
\[ P(R\cup B) = \frac{N(R)+N(B)}{N(T)} = \frac{54+75}{54+75+9} = \frac{43}{46} \]
You are going to play mini golf. A ball machine that contains 19 green golf balls, 20 red golf balls, 24 blue golf balls, and 17 yellow golf balls, randomly gives you your ball. What is the probability that you end up with a red golf ball? Express your answer as a simplified fraction or a decimal rounded to four decimal places.
\[ P(R) = \frac{N(R)}{N(T)} = \frac{20}{19+20+24+17} = \frac{1}{4} \]
A pizza delivery company classifies its customers by gender and location of residence. The research department has gathered data from a random sample of 1399 customers. The data is summarized in the table below.
\(\textbf{Gender and Residence of Customers}\)
| Residence | Males | Females |
|---|---|---|
| Apartment | 81 | 228 |
| Dorm | 116 | 79 |
| With Parent(s) | 215 | 252 |
| Sorority/Fraternity House | 130 | 97 |
| Other | 129 | 72 |
What is the probability that a customer is not male or does not live with parents? Write your answer as a fraction or a decimal number rounded to four decimal places.
| Conditions | Male | Female | Totals |
|---|---|---|---|
| Parents | 215 | 252 | 467 |
| No Parents | 456 | 476 | 932 |
| Totals | 671 | 728 | 1399 |
| Conditions | Male | Female | Totals |
|---|---|---|---|
| Parents | 0.1537 | 0.1801 | 0.3338 |
| No Parents | 0.3259 | 0.3403 | 0.6662 |
| Totals | 0.4796 | 0.5204 | 1.000 |
\[ P(!M\cup!P) = P(!M) + P(!P) - P(!M\cap!P) = 0.5204+0.6662-0.3403 = \boxed{0.8463} \]
Determine if the following events are independent. Going to the gym. Losing weight.
Answer: \(\boxed{A) Dependent}\) B) Independent
There is a causal connection, burning calories, that would make these dependent.
A veggie wrap at City Subs is composed of 3 different vegetables and 3 different condiments wrapped up in a tortilla. If there are 8 vegetables, 7 condiments, and 3 types of tortilla available, how many different veggie wraps can be made?
We need to use combinations, because order does not matter. \[ C(n,k) = \frac{n!}{(n-k)!k!} \] To get the total number of combinations, we need to multiply all the different numbers of combinations for veggies, condiments and tortillas together:
\[ C_{total} = \frac{8!}{(8-3)!3!}*\frac{7!}{(7-3)!3!}*\frac{3!}{(3-1)!1!} \]
wraps <- (factorial(8)/(factorial(5)*factorial(3)))*(factorial(7)/(factorial(4)*factorial(3)))*(factorial(3)/(factorial(2)*factorial(1)))
wraps## [1] 5880Determine if the following events are independent. Jeff runs out of gas on the way to work. Liz watches the evening news. Answer: A) Dependent \(\boxed{B) Independent}\)
There is no causal connection between Jeff running out of gas or Liz watching the evening news.
The newly elected president needs to decide the remaining 8 spots available in the cabinet he/she is appointing. If there are 14 eligible candidates for these positions (where rank matters), how many different ways can the members of the cabinet be appointed?
We need to use permutations, because order (rank) does matter. \[ P(n,k) = \frac{n!}{(n-k)!} \]
\[ P(14,8) = \frac{14!}{(14-8)!} \]
cabnt <- factorial(14)/factorial(6)
cabnt## [1] 121080960A bag contains 9 red, 4 orange, and 9 green jellybeans. What is the probability of reaching into the bag and randomly withdrawing 4 jellybeans such that the number of red ones is 0, the number of orange ones is 1, and the number of green ones is 3? Write your answer as a fraction or a decimal number rounded to four decimal places.
This would be the product of the individual combinations for each color divided by the total combinations for the set \[ P = \frac{9!}{(9-0)!0!}*\frac{4!}{(4-3)!3!}*\frac{9!}{(9-3)!3!}*\frac{(22-4)!4!}{22!} = \frac{4*4!*9!*18!}{3!*6!*22!} \]
choose(9,0)*choose(4,1)*choose(9,3)/choose(22,4)## [1] 0.04593301Evaluate the following expression. \[ \frac{11!}{7!} \] = 11*10*9*8 = 7920
To check:
factorial(11)/factorial(7)## [1] 7920Describe the complement of the given event. 67% of subscribers to a fitness magazine are over the age of 34.
33% of suscribers to a fitness magazine are 34 or younger.
If you throw exactly three heads in four tosses of a coin you win $97. If not, you pay me $30.
Probability of 3 heads in 4 coin tosses is the combinations divided by number of states:
\[ P = \frac{4!}{(4-3)!*3!*4^2} = \frac{4}{16} = 0.25 \]
Round your answer to two decimal places.
| Payout | +97$ | -30$ |
|---|---|---|
| Prob | 0.25 | 0.75 |
\[ E(X) = \sum_{i=1}^{\infty}x_i*p_i = 97\$*0.25+-30\$*0.75 \]
97*0.25-30*.75## [1] 1.75It appears that the payout is large enough to overcome the lower odd of winning.
If you played this game 559 times how much would you expect to win or lose? (Losses must be entered as negative.)
This is simply the product of the number of games times the expected payout.
\[ N*E(X) = 559* \$1.75 = \$978.25 \]
In this case the gambler would slowly win more than they lose. This is not a profitable game for the house.
Flip a coin 9 times. If you get 4 tails or less, I will pay you $23. Otherwise you pay me $26.
To calculate the probability we will use the binomial distribution, 0-4 tails, at 50% and 9 trials:
pbinom(4,9,0.5)## [1] 0.5Note that this is 50% because the range is [0-9], so the subsets of [0-4] and [5-9] are of equal length.
Round your answer to two decimal places.
| Payout | +23$ | -26$ |
|---|---|---|
| Prob | 0.5 | 0.5 |
\[ E(X) = \sum_{i=1}^{\infty}x_i*p_i = 23\$*0.5+-26\$*0.5 \]
23*0.5-26*.5## [1] -1.5The gambler would slowly lose more money than they win. This would be a profitable game for the house.
If you played this game 994 times how much would you expect to win or lose? (Losses must be entered as negative.)
This is simply the product of the number of games times the expected payout.
\[ N*E(X) = 994* \$-1.50 = \$-1,491 \]
The house has a big advantage here in that it is not very obvious that this is a losing proposition for the gambler.
The sensitivity and specificity of the polygraph has been a subject of study and debate for years. A 2001 study of the use of polygraph for screening purposes suggested that the probability of detecting a liar was .59 (sensitivity) and that the probability of detecting a “truth teller” was .90 (specificity). We estimate that about 20% of individuals selected for the screening polygraph will lie.
| Conditions | Lie | Truth | Totals |
|---|---|---|---|
| Positive | 0.118 | 0.080 | 0.198 |
| Negative | 0.082 | 0.720 | 0.802 |
| Totals | 0.200 | 0.800 | 1.000 |
\[ P(L|+) = \frac{P(L\cap+)}{P(+)} = \frac{0.118}{0.198} = 0.5959 \]
\[ P(T|-) = \frac{P(T\cap-)}{P(-)} = \frac{0.720}{0.802} = 0.8977 \]
\[ P(L\cup+) = P(L) + P(+) - P(L\cap+) = 0.2+ 0.198 - 0.118 = 0.28 \]