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.

Answer:

Total number of color marbles = 54 + 9 + 75
= 138

P(red or blue) = (red marbles + blue marbles) / total number of color marbles
= (54 + 75)/138
= 129/138
= 0.9348

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.

Answer:

Total number of golf balls = 19 + 20 + 24 +17
= 80

P(red) = red balls / total number of golf balls
= 20/80
= 1/4
= 0.25

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.

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.

Answer:

Total number of random sample of customers = 1399

P(not male or does not live with parents) = (P(is female) + P(does not live with parents)
- P(female and does not live with parents)) / total number of sample customers
= (228+79+252+97+72) + (1399-215-252) - (228+79+97+72) / 1399
= (728 + 932 - 476) / 1399
= 1184/1399
= 0.8463

Problem Set 4

Determine if the following events are independent.

Going to the gym. Losing weight.

  1. Dependent B) Independent

Answer:

The events are A) Dependent because one event affects the other where going to gymn can result in losing weight. There is a causal connection between these two events.

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?

Answer:

Number of different veggie wraps can be made is

= \({ _{ 8 }{ C }_{ 3 } }\bullet { _{ 7 }{ C }_{ 3 } }\bullet { _{ 3 }{ C }_{ 1 } }\)
= \(\frac { 8! }{ (8-3)!3! } \bullet \frac { 7! }{ (7-3)!3! } \bullet \frac { 3! }{ (3-1)!1! }\)
= (40320/720)(5040/144)(6/2)
= (56)(35)(3)
= 5880

Problem Set 6

Determine if the following events are independent.

Jeff runs out of gas on the way to work. Liz watches the evening news.

  1. Dependent B) Independent

Answer:

The events are A) Independent because one event does not affect the other where Jeff runs out of gas does not affect the probability of Liz watches the evening news. There is no causal connection between these events.

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?

Answer:

As the rank (order) is matter here, we need to apply permutation formula, \(\frac { n! }{ (n-r)! }\)

Number of different ways = \(\frac { 14! }{ (14-8)! }\)
= 87178291200/720
= 121,080,960

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.

Answer:

Total number of jellybeans = 9 + 4 + 9
= 22

P(0 red, 1 orange, and 3 green) = (Number of ways to select 0 red, 1 orange, and 3 green) / (Number of ways to select 22 jellybeans)
= \({ _{ 9 }{ C }_{ 0 } }\bullet { _{ 4 }{ C }_{ 1 } }\bullet { _{ 9 }{ C }_{ 3 } }\)
/ \({ _{ 22 }{ C }_{ 4 } }\) = \(\frac { 9! }{ (9-0)!0! } \bullet \frac { 4! }{ (4-1)!1! } \bullet \frac { 9! }{ (9-3)!3! }\)
/ \(\frac { 22! }{ (22-4)!4! }\) = (1)(24/6)(362880/4320) / 7315
= (1)(4)(84) / 7315
= 336 / 7315
= 0.04593

Problem Set 9

Evaluate the following expression.

\[\frac { 11! }{ 7! }\]

Answer:

\(\frac { 11! }{ 7! }\) = (11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1) / (7)(6)(5)(4)(3)(2)(1)
= (11)(10)(9)(8)
= 7920

Problem Set 10

Describe the complement of the given event.

67% of subscribers to a fitness magazine are over the age of 34.

Answer:

The complement is 31% of subscribers to a fitness magazine are 34 years old or younger.

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.)

Answer:

1.) Possible combinations of four tosses of a coin is 2^4 = 16. The sample space is below:

4H: {HHHH}
3H and 1T: {HHHT, HHTH, HTHH, THHH}
2H and 2T: {HHTT, HTTH, TTHH, HTHT, THTH, THHT}
1H and 3T: {HTTT, THTT, TTHT, TTTH}
4T: {TTTT}

P(exactly 3 heads) = 4/16
= 1/4

P(other than exactly 3 heads) = 1 - 1/4
= 3/4

So, the expected value of the proposition is:

Expected value = (1/4) * 97 - (3/4) * 30
= 97/4 - 90/4
= $7/4
= $1.75

2.) If I played this game 559 times, I would expect to win $978.25

Expected Win = $1.75 * 559
= $978.25

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.)

Answer:

1.) As the number of flip increases, this time we use the binomial distribution formula, \(P(X=x)={ _{ n }{ C }_{ x } }\bullet { P }^{ x }\bullet { (1-P) }^{ n-x }=\frac { n! }{ (n-x)!x! } \bullet { P }^{ x }\bullet { (1-P) }^{ n-x }\).

where in this case,

n (number of flip) = 9 p (probability of getting tail) = 0.5

So, probability of getting 4 tails or less is

\(P(X\le4)\) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)
= \(\frac { 9! }{ (9-0)!0! } \bullet { 0.5 }^{ 0 }\bullet { (1-0.5) }^{ 9-0 }\)
+ \(\frac { 9! }{ (9-1)!1! } \bullet { 0.5 }^{ 1 }\bullet { (1-0.5) }^{ 9-1 }\)
+ \(\frac { 9! }{ (9-2)!2! } \bullet { 0.5 }^{ 2 }\bullet { (1-0.5) }^{ 9-2 }\)
+ \(\frac { 9! }{ (9-3)!3! } \bullet { 0.5 }^{ 3 }\bullet { (1-0.5) }^{ 9-3 }\)
+ \(\frac { 9! }{ (9-4)!4! } \bullet { 0.5 }^{ 4 }\bullet { (1-0.5) }^{ 9-4 }\)
= 0.5

And probability of getting 5 tails or more is

\(P(X\ge5)\) = 1 - 0.5
= 0.5

So, the expected value of the proposition is:

Expected value = (0.5) * 23 - (0.5) * 26
= 11.5 - 13
= -$1.5

2.) If I played this game 994 times, I would expect to lose -$1,491

Expected Loss = -$1.5 * 994
= - $1,491

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.

Answer:
Liar Truth Teller
Liar True Positive False Positive
Truth Teller False Negative True Negative


a.) Let X = liar, + = detected as a liar. X’ = truth teller, and +’ = detected as a truth teller.

The probability of an individual is actually a liar given that detected him/her as a liar is

P(X|+) = P(+ and X) / P(+)

Using the Bayes’s formula, P(+) = P(+|X)P(X) + P(+|X’)P(X’), which simply states that the probability of event + is the sum of the conditional probabilities of event + given that event X has or has not occurred. So,

P(+ and X) / P(+) = P(+|X)P(X) / (P(+|X)P(X) + P(+|X’)P(X’))
= (0.59)(0.2) / ((0.59)(0.2) + (1-0.9)(1-0.2))
= 0.118 / (0.118 + 0.08)
= 0.118 / 0.198
= 0.596

So, P(X|+) = 0.596

This is simply equivalent to the True Positive / (True Positive + False Positive) as shown in table above, where TP = 0.118, FP = 0.08.

b.) Let X = liar, + = detected as a liar. X’ = truth teller, and +’ = detected as a truth teller.

The probability of an individual is actually a truth-teller given that detected him/her as a truth-teller is

P(X’|+‘) = P(+’ and X’) / P(+’)

Again, using the Bayes’s formula, P(+‘) = P(+’|X’)P(X’) + P(+‘|X)P(X), which simply states that the probability of event +’ is the sum of the conditional probabilities of event +’ given that event X’ has or has not occurred. So,

P(+’ and X’) / P(+‘) = P(+’|X’)P(X’) / (P(+‘|X’)P(X’) + P(+’|X)P(X))
= (0.9)(0.8) / ((0.9)(0.8) + (1-0.59)(0.2))
= 0.72 / (0.72 + 0.082)
= 0.72 / 0.802
= 0.898

So, P(X’|+’) = 0.898

This is simply equivalent to the True Negative / (True Negative + False Negative) as shown in table above, where TN = 0.72, FN = 0.082.

c.) Let X = liar, + = detected as a liar. X’ = truth teller, and +’ = detected as a truth teller.

The probability that a randomly selected individual is either a liar or was identified as a liar is equivalent to saying that either a liar or those detected incorrectly as liars (False Positive).

P(X or +) = P(X) + P(+) - P(X and +)

We know that

P(X) = 0.2

P(+) = P(+|X)P(X) + P(+|X’)P(X’)
= (0.59)(0.2) + (1-0.9)(1-0.2)
= 0.198

P(X and +) = P(+|X)P(X)
= (0.59)(0.2)
= 0.118

So, P(X or +) = 0.2 + 0.198 - 0.118
= 0.28