1. 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.
A. On the first choice, the individual probabilities are:
P(Red) = 54/138 = 0.3841
P(White) = 9/38 = 0.0652
P(Blue) = 75/138 = 0.5435
Note that the probabilities sum to 1. The probability of red or blue is:
P(Red or Blue) = P(Red) + P(Blue) = 0.9348
Which is the complement of P(White) =
P(WhiteC) = 1 - P(Red or Blue) = 0.9348.
2. 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.
A. On the first choice, the probabilities are:
P(Green) = 19/80 = 0.2375
P(Red) = = 20/80 = 0.25
P(Blue) = 24/80 = 0.30
P(Yellow) = 17/80 = 0.2125
Note that probabilities sum to 1.
3. A pizza delivery company classifies its customers by gender and location of residence. The research department has gathered data from a random sample of 1,399 customers. The data is summarized in the table below.
pander(customers)| Males | Females | |
|---|---|---|
| Apartment | 81 | 228 |
| Dorm | 116 | 79 |
| With Parent(s) | 215 | 252 |
| Sorority/Fraternity | 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.
A. We use the rule of addition:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
In ths case:
P(not male U not at home) = P(not male) + P(not at home) - P(not male ∩ not at home)
= 1-P(male) + 1-P(at home) - 1-P(male) x 1-P(at home)
= 0.8399
4. Determine if the following events are independent. Going to the gym. Losing weight.
Answer: A) Dependent B) Independent
Assuming there is excercise.
5. 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?
A. We use the basic counting technique described in Grimstead p. where
Possible veggie combinations = 56
Possible condiment combos = 35
Possible tortilla combos = 3
Total combinations = veggies x condiments x tortillas = 5880
6. Determine if the following events are independent. Jeff runs out of gas on the way to work. Liz watches the evening news.
Answer: A) Dependent B) Independent
P(Jeff runs out | Liz watches) = P(Jeff runs out)
and
P(Liz watches | Jeff runs out) = P(Liz watches)
Although, maybe not, if Jeff was supposed to take Liz out to dinner and never arrived.
7. 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?
A. This is a permutation problem, where n is the number of candidates to choose from and r is the number of jobs to fill.
\(\frac{n!}{(n - r)!}\)
= 121080960
8. A 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.
A. We’re working with combinations. Multiply all combinations for the individual choices and divide by all possible combinations.
Combinations:
No reds = 1
One orange = 4
Three green = 84
All combos = 7315
Prob = 0.0459
9. Evaluate the following expression. 11!/7!
A. Factorials. A couple ways to do this in R:
cat(factorial(11)/factorial(7), prod(seq(1,11,1))/prod(seq(1,7,1)))## 7920 7920
10. Describe the complement of the given event.
67% of subscribers to a fitness magazine are over the age of 34.
A. 33 percent of subscribers are under the age of 34.
Rationale:
Take the iconic dice roll, where the sample space of a single role is the set S = {1, 2, 3, 4, 5, 6}. If we assume that outcome R is the probability of rolling a 1 or 2, then the complement RC is the probability of all outcomes that are not in R, or {3,4,5,6}. An event and its outcome are disjoint.
Some properties:
P(A or AC) = 1 P(A or AC)) = P(A) + P(AC) P(A) + P(AC) = 1 and P(A) = 1 - P(AC)
- If you throw exactly three heads in four tosses of a coin you win $97. If not, you pay me $30.
Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
A. The probability of 3 heads in 4 tosses is:
\(\left( \begin{matrix} n \\ k \end{matrix} \right) ⋅{ p }^{ k }{ ⋅(1-p) }^{ n-k }=\left( \begin{matrix} 4 \\ 3 \end{matrix} \right) ⋅{ \left( 1/2 \right) }^{ 3 }⋅{ \left( 1/2 \right) }^{ 1 }= 0.25\)
The expected value is:
\(E(X)=\sum _{ i }^{ \infty }{ { x }_{ i } } ⋅{ p }_{ i }=97⋅0.25-30⋅0.75=1.75\)
Step 2. If you played this game 559 times how much would you expect to win or lose? (Losses must be entered as negative.)
Total winnings = E(X) * N of games = $1.75 * 559 = $976.5
- Flip a coin 9 times. If you get 4 tails or less, I will pay you $23. Otherwise you pay me $26. Step 1. Find the expected value of the proposition. Round your answer to two decimal places.
A. Again, the probability of 4 successes in 9 binomial trials is:
Using the pbinom() function in R (to make this a whole lot easier) : 0.5
The expected value is:
\(E(X)=\sum _{ i }^{ \infty }{ { x }_{ i } } ⋅{ p }_{ i }=23⋅0.5-26⋅0.5=-1.5\)
Step 2. If you played this game 994 times how much would you expect to win or lose? (Losses must be entered as negative.)
Total winnings = E(X) * N of games = $-1.50 * 994 = $-1491
OUCH!!!
- 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.
A. Given the sensitivity, specificity and rate of deception, we have enough information to create a confusion matrix.
pander(poly)| Liar | Saint | Total | |
|---|---|---|---|
| Positive Test | 11.8 | 8 | 19.8 |
| Negative Test | 8.2 | 72 | 80.2 |
| Total | 20 | 80 | 100 |
- What is the probability that an individual is actually a liar given that the polygraph detected him/her as such? (Show me the table or the formulaic solution or both.)
A. Given that 20 percent of those screened are liars and the probability of detecting a liar is 0.59:
P(Liar) = 0.20 x 0.59 = 0.118
- What is the probability that an individual is actually a truth-teller given that the polygraph detected him/her as such? (Show me the table or the formulaic solution or both.)
A. Given that 80 percent of those screened are truthful and the probability of detecting one is 0.90:
P(Truth) = 0.80 x 0.90 = 0.72
- What is the probability that a randomly selected individual is either a liar or was identified as a liar by the polygraph? Be sure to write the probability statement.
P(Liar ∪ Tested Liar) = P(Liar) + P(Tested liar) - P(Liary) x P(Tested Liar)
Based on our confusion matrix:
=(0.2+0.198)-(0.2x0.198)
= 0.3584