1 Problem Set 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.

1.1 Answer

As getting one red or one blue marble randomly is mutually exclusive, \(P(red \cup blue) = P(red) + P(blue)\)

\[P(getting\;red\;or\;blue)\] \[=P(red \cup blue) =P(red) + P(blue)\] \[=\frac{54}{54+9+75} + \frac{75}{54+9+75}\] \[=\frac{129}{138}\] \[=\frac{43}{46}\]

2 Problem Set 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.

2.1 Answer

\[P(getting\;red\;golf\;ball)\] \[=\frac{\#of\;red\;balls}{Total\;\#\;of\;balls}\] \[=\frac{20}{19+20+24+17}\] \[=\frac{20}{80}\] \[=\frac{1}{4}\]

3 Problem Set 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 1399 customers. The data is summarized in the table below.

Gender and Residence of Customers
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.

3.1 Answer

P(not male or does not live with parents) = P(not(male and live with parents)) = 1 - P(male and live with parents) \[=1-\frac{215}{1399}\] \[\frac{1184}{1399}\]

4 Problem Set 4

Determine if the following events are independent: 1) Going to the gym, 2) Losing weight.

Choices: A) Dependent B) Independent

4.1 Answer

\(\because\) Going to the gym and losing weight has a causal relationship.

\(\therefore\) My answer is A, dependent.

5 Problem Set 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?

5.1 Answer

Using the combination rule, the number of ways \(=8C3 \times 7C3 \times 3C1 =56 \times 35 \times 3 =5880\)

6 Problem Set 6

Determine if the following events are independent: 1) Jeff runs out of gas on the way to work, 2) Liz watches the evening news.

Choices: A) Dependent B) Independent

6.1 Answer

\(\because\) The two events do not have a causal relationship.

\(\therefore\) My answer is B, independent.

7 Problem Set 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?

7.1 Answer

\(\because\) Rank of the 8 spots matters,

\(\therefore\) Using permutation rule, the number of ways \(=14P8 =121,080,960\)

8 Problem Set 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.

8.1 Answer

Total number of jellybeans \(=9+4+9=22\)

Using combination rule,

\[P(given\;event) = \frac{\#\;of\;ways\;of\;getting\;the\;given\;event}{\#\;of\;ways\;getting\;any\;4\;jellybeans}\]

\[=\frac{9C0 \times 4C1 \times 9C3}{22C4}\]

\[=\frac{1\times4\times84}{7315}\]

\[=\frac{48}{1045}\]

9 Problem Set 9

Evaluate the following expression: \(\frac{11!}{7!}\)

9.1 Answer

By definition, \(n!=n\times(n-1)\times(n-2)\times\cdots\times3\times2\times1\) and \(0!=1\).

\[\therefore \frac{11!}{7!} = \frac{11\times10\times9\times\cdots\times3\times2\times1}{7\times6\times5\times4\times3\times2\times1}\] \[=11\times10\times9\times8\] \[=7920\]

10 Problem Set 10

Describe the complement of the given event: 67% of subscribers to a fitness magazine are over the age of 34.

10.1 Answer

The complement of the given event is 33% of subscribers to a fitness magazine are below or at the age of 34.

11 Problem Set 11

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.

Step 2. If you played this game 559 times how much would you expect to win or lose? (Losses must be entered as negative.)

11.1 Step 1 Answer

Note: Exactly three heads in four tosses = Exactly one tail in four tosses

The expected value of the proposition \(=(\frac{1}{2})^{4}\times C^{4}_{1}\times97+(1-\frac{1}{2^{4}}\times C^{4}_{1})\times(-30)\) \[=\frac{1}{16}\times4\times97+(1-\frac{1}{16}\times4)\times(-30)\] \[=\frac{97}{4}-\frac{90}{4}\] \[=1.75\]

11.2 Step 2 Answer

The expected value of paying this game 559 times would be \(\$1.75 \times 559 =\$978.25\)

12 Problem Set 12

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.

Step 2. If you played this game 994 times how much would you expect to win or lose? (Losses must be entered as negative.)

12.1 Step 1 Answer

Getting 4 tails or less is the same as choosing 4 tails or less from the 9 coins.

Thus, the probability of getting 4 tails or less \(=\frac{9C4+9C3+9C2+9C1+9C0}{2^{9}}=\frac{126+84+36+9+1}{512}=\frac{1}{2}\)

\(\therefore\) The expected value of the proposition \(=\frac{1}{2}\times23+(1-\frac{1}{2})\times(-26)=11.5-13=-\$1.5\)

12.2 Step 2 Answer

The expected value of paying this game 994 times would be \(-\$1.5 \times 994 =-\$1491\)

\(\therefore\) I would expect to lose $1,491.

13 Problem Set 13

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.

  1. 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.)

  2. 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.)

  3. 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.

13.1 Answer

By definition and given information,

  1. sensitivity = True Positive / (True Positive + False Negative) = 0.59

  2. Specificity = True Negative / (True Negative + False Positive) = 0.90

  3. Total percentage of people would truly lie = 0.20

Liar_Actual Truth_Teller_Actual Total
Liar_Predict True Positive False Positive TP+FP
Truth_Teller_Predict False Negative True Negative FN+TN
Total TP+FN = 0.2 FP+TN = 1-0.2 = 0.8 ALL = 1

By using the 3 information and table listed above, I have

  • True Positive \(=0.59\times0.2=0.118\)

  • True Negative \(=0.9\times(1-0.2)=0.72\)

  • False Negative \(=0.2-0.118=0.082\)

  • False Positive \(=0.8-0.72=0.08\)

  • TP+FP \(=0.118+0.08=0.198\)

  • FN+TN \(=0.082+0.72=0.802\)

Part a.

P(an individual is actually a liar given that the polygraph detected him/her as such)

\(=\)P(actual liar | Predicted as liar)

\(=\)P(actual liar and predicted as liar) / P(Predicted as liar)

\(=\)TP/(TP+FP)

\(=\frac{0.118}{0.118+0.08}\)

\(=\frac{59}{99}\)

Part b.

P(an individual is actually a truth-teller given that the polygraph detected him/her as such)

\(=\)P(actual truth-teller | Predicted as truth-teller)

\(=\)P(actual truth-teller and predicted as truth-teller) / P(Predicted as truth-teller)

\(=\)TN/(FN+TN)

\(=\frac{0.72}{0.802}\)

\(=\frac{360}{401}\)

Part c.

P(an individual is either a liar or was identified as a liar by the polygraph)

\(=\)P(actual liar or predicted as liar)

\(=TP+FN+FP\)

\(=0.118+0.082+0.08\)

\(=0.28\)