Question 1

The department in which you work in your company has 10 employees: 10 women and 14 men.A team of anywhere from one employee up to 9 employees must be chose to attend a meeting at the company's headquarters.How many different teams can be selected?

Accepted Answer

The answer of The department in which you work in...

Question 2

Suppose that the probability that LIBOR (London inter-bank offered rate)will increase next trimester is 0.025.Assume also that the probability that your company will open a new branch overseas is 0.20.If the probability that at least one of the two events occur, i.e., LIBOR increases and/or your company opens a new overseas branch, is 0.21, what is the probability that both events occur?

Accepted Answer

The probability that both events occur can be found using the formula: P(A and B) = P(A) + P(B) - P(A or B). Substituting the given probabilities, we get 0.025 + 0.20 - 0.21 = 0.015.

Question 3

The department in which you work in your company has 24 employees.A team of 3 employees must be selected to represent the department at a companywide meeting in the headquarters of the company.How many different teams of 3 can be selected?

Accepted Answer

The number of ways to select a team of 3 employees from 24 can be calculated using the combination formula, which is $C(n, k) = \frac{n!}{k!(n-k)!}$, where $n$ is the total number of items (in this case, employees), and $k$ is the number of items to choose (in this case, team members). So, $C(24, 3) = \frac{24!}{3!(24-3)!} = \frac{24!}{3!21!} = \frac{24 \times 23 \times 22}{3 \times 2 \times 1} = 2024$.

Question 4

The table below provides summary information about the students in a class.The sex of each individual and their age is given. $$\begin{array} { | l | c | c | c | } 
\hline & \text { Male } & \text { Fernale } & \text { Total } \
\hline \text { Under 20 yrs old } & 10 & 8 & 18 \
\hline \begin{array} { l } 
\text { Between } 20 \text { and } 25 \
\text { yrs old }
\end{array} & 12 & 18 & 30 \
\hline \text { Older tharn 25 yrs. } & 26 & 26 & 52 \
\hline \text { Total } & 48 & 52 & 100 \
\hline
\end{array}$$ A student is randomly selected from this group, and it is found that the student is older than 25 years.What is the probability that the student is a male?

Accepted Answer

Given that the student is older than 25 years, we have a total of 52 students in this category. Out of these, 26 are male. Therefore, the probability that the student is a male is 26/52 = 0.50.

Question 5

A market research firm is investigating the appeal of three package designs.The table below gives information obtained through a sample of 200 consumers.The three package designs are labeled A, B, and C.The consumers are classified according to age and package design preference. $$\begin{array} { | c | c | c | c | c | } 
\hline & \text { A } & \text { B } & \text { C } & \text { Total } \
\hline \text { Under 25 years } & 22 & 34 & 40 & 96 \
\hline 25 \text { or older } & 54 & 28 & 22 & 104 \
\hline \text { Total } & 76 & 62 & 62 & 200 \
\hline
\end{array}$$ If one of these consumers is randomly selected and is under 25, what is the probability that the person prefers design A?

Accepted Answer

The answer of A market research firm is investigating the...

Question 6

A market research firm is investigating the appeal of three package designs.The table below gives information obtained through a sample of 200 consumers.The three package designs are labeled A, B, and C.The consumers are classified according to age and package design preference. $$\begin{array} { | c | c | c | c | c | } 
\hline & \text { A } & \text { B } & \text { C } & \text { Total } \
\hline \text { Under 25 years } & 22 & 34 & 40 & 96 \
\hline 25 \text { or older } & 54 & 28 & 22 & 104 \
\hline \text { Total } & 76 & 62 & 62 & 200 \
\hline
\end{array}$$ Are "B" and "25 or older" independent and why or why not?

Accepted Answer

Two events are independent if the probability of one event occurring does not affect the probability of the other event. To check if "B" and "25 or older" are independent, we calculate P(25 or older | B) and compare it to P(25 or older). P(25 or older | B) = 28/62, since 28 out of the 62 people who preferred B are 25 or older. P(25 or older) = 104/200, since 104 out of the 200 total people are 25 or older. These probabilities are not equal, indicating that the events are not independent.

Question 7

A market research firms conducts studies regarding the success of new products.The company is not always perfect in predicting the success.Suppose that there is a 50% chance that any new product would be successful (and a 50% chance that it would fail).In the past, for all new products that ultimately were successful, 80% were predicted to be successful (and the other 20% were inaccurately predicted to be failures).Also, for all new products that were ultimately failures, 70% were predicted to be failures (and the other 30% were inaccurately predicted to be successes).For any randomly selected new product, what is the probability that the market research firm would predict that it would be a success?

Accepted Answer

The probability that a product is predicted to be a success is the sum of the probabilities of two events: (1) the product is actually successful and predicted as such, and (2) the product is actually a failure but is incorrectly predicted to be a success. This can be calculated as (0.5 * 0.8) + (0.5 * 0.3) = 0.4 + 0.15 = 0.55.

Question 8

The department in which you work in your company has 24 employees: 10 women and 14 men.A team of 4 employees must be selected to represent the department at a companywide meeting in the headquarters of the company.The team must have two women and two men.How many different teams of 4 can be selected?

Accepted Answer

The answer of The department in which you work in...

Question 9

Your company provides services to three different industries: automotive, construction, and financial services.These services can be either on a yearly or on a monthly contract.The cross-tabulation below summarizes this information: $$\begin{array} { | l | l | l | l | } 
\hline & \text { Morthly } & \text { Yearly } & \text { Total } \
\hline \text { Automotive } & 125 & 85 & 210 \
\hline \text { Construction } & 212 & 351 & 563 \
\hline \text { Finaricial } & 357 & 210 & 567 \
\hline \text { Total } & 694 & 646 & 1340 \
\hline
\end{array}$$ If a client is randomly selected, what is the likelihood that it will be from the financial industry or on a monthly contract?

Accepted Answer

The probability of selecting a client from the financial industry is 567/1340. The probability of selecting a client on a monthly contract is 694/1340. To find the probability of selecting a client from the financial industry or on a monthly contract, we add these probabilities but subtract the overlap (financial industry and monthly contract) to avoid double counting: (567 + 694 - 357) / 1340 = 904 / 1340 = 67.5%.

Question 10

Suppose 5% of the population have a certain disease.A laboratory blood test gives a positive reading for 95% of people who have the disease and 10% positive reading of people who do not have the disease.What is the probability of testing positive?

Accepted Answer

The probability of testing positive can be calculated using the law of total probability. Let's denote:- $P(D) = 0.05$ as the probability of having the disease,- $P(\neg D) = 0.95$ as the probability of not having the disease,- $P(+|D) = 0.95$ as the probability of testing positive given that the person has the disease,- $P(+|\neg D) = 0.10$ as the probability of testing positive given that the person does not have the disease.The total probability of testing positive, $P(+)$, is given by:$$P(+) = P(+|D)P(D) + P(+|\neg D)P(\neg D)$$$$P(+) = (0.95)(0.05) + (0.10)(0.95)$$$$P(+) = 0.0475 + 0.095 = 0.1425$$

Question 11

It is known that 20% of all students in some large university are overweight, 20% exercise regularly and 2% are overweight and exercise regularly.What is the probability that a randomly selected student is overweight given that this student exercises regularly?

Accepted Answer

The answer of It is known that 20% of all...

Question 12

A market research firms conducts studies regarding the success of new products.The company is not always perfect in predicting the success.Suppose that there is a 50% chance that any new product would be successful (and a 50% chance that it would fail).In the past, for all new products that ultimately were successful, 80% were predicted to be successful (and the other 20% were inaccurately predicted to be failures).Also, for all new products that were ultimately failures, 70% were predicted to be failures (and the other 30% were inaccurately predicted to be successes).What is the a priori probability that a new product would be a success?

Accepted Answer

The a priori probability that a new product would be a success is given as 50%, which means before considering any specific information about the product's predicted success or failure, there is a 50% chance of success.

Question 13

A market research firm is investigating the appeal of three package designs.The table below gives information obtained through a sample of 200 consumers.The three package designs are labeled A, B, and C.The consumers are classified according to age and package design preference. $$\begin{array} { | r | r | r | r | r | } 
\hline & \text { A } & \text { B } & \text { C } & \text { Total } \
\hline \text { Under 25 years } & 22 & 34 & 40 & 96 \
\hline \text { 25 or older } & 54 & 28 & 22 & 104 \
\hline \text { Total } & 76 & 62 & 62 & 200 \
\hline
\end{array}$$ If one of these consumers is randomly selected and prefers design B, what is the probability that the person is 25 or older?

Accepted Answer

The probability that a person is 25 or older given they prefer design B is calculated by dividing the number of people 25 or older who prefer design B (28) by the total number of people who prefer design B (62). Thus, the probability is 28/62 = 0.45.

Question 14

Suppose you toss a fair coin three times in a row and obtain tails, tails, and tails.What it the probability that the fourth toss will give heads?

Accepted Answer

The probability of getting heads on a fair coin toss is always 0.50, regardless of previous outcomes, due to the independence of each toss.

Question 15

An analysis of personal loans at a local bank revealed the following facts: 10% of all personal loans are in default (D), 90% of all personal loans are not in default (D΄), 20% of those in default are homeowners (H | D), and 70% of those not in default are homeowners (H | D΄).If a personal loan is selected at random P (H $\cap$ D')= ___________.

Accepted Answer

P(H ∩ D') = P(H | D') * P(D') = 0.70 * 0.90 = 0.63. This is because 70% of those not in default are homeowners and 90% of all loans are not in default.

Question 16

Suppose 5% of the population have a certain disease.A laboratory blood test gives a positive reading for 95% of people who have the disease and 10% positive reading of people who do not have the disease.What is the probability that a randomly selected person has the disease given that this person is testing positive?

Accepted Answer

To solve this, we use Bayes' theorem. Let D be the event that a person has the disease, and T+ be the event that a person tests positive. We want to find P(D|T+), the probability that a person has the disease given that they test positive.P(D|T+) = [P(T+|D) * P(D)] / [P(T+|D) * P(D) + P(T+|D') * P(D')]Where:- P(T+|D) = 0.95 (probability of testing positive given the person has the disease)- P(D) = 0.05 (probability of having the disease)- P(T+|D') = 0.10 (probability of testing positive given the person does not have the disease)- P(D') = 0.95 (probability of not having the disease)Plugging in the numbers:P(D|T+) = (0.95 * 0.05) / (0.95 * 0.05 + 0.10 * 0.95) = 0.0475 / (0.0475 + 0.095) = 0.0475 / 0.1425 = 1/3 ≈ 0.3333Therefore, the probability that a person has the disease given they test positive is approximately 0.3333 or 33.33%.

Question 17

Suppose that your company is sending invitations to its 50 most important clients for an end-of-the-year event where a new product will be exhibited.These are printed invitation cards mailed through USPS, and each card is personalized with the name of the corresponding client.You realize that the cards have been accidentally shuffled before placing them in the envelopes.What is the probability that only one card is in the wrong envelope?

Accepted Answer

It's impossible for only one card to be in the wrong envelope because if one card is in the wrong envelope, at least one other card must also be in the wrong envelope to accommodate the displacement of the first card.

Question 18

An analysis of personal loans at a local bank revealed the following facts: 10% of all personal loans are in default (D), 90% of all personal loans are not in default (D΄), 20% of those in default are homeowners (H | D), and 70% of those not in default are homeowners (H | D΄).If a personal loan is selected at random, P (D | H)= ___________.

Accepted Answer

P (D | H) = P (D ∩ H) / P (H)= P (H | D) * P (D) / P (H)= 0.2 * 0.1 / (0.2 * 0.1 + 0.7 * 0.9)= 0.02 / 0.63= 0.03

Question 19

A market research firm conducts studies regarding the success of new products.The company is not always perfect in predicting the success.Suppose that there is a 50% chance that any new product would be successful (and a 50% chance that it would fail).In the past, for all new products that ultimately were successful, 80% were predicted to be successful (and the other 20% were inaccurately predicted to be failures).Also, for all new products that were ultimately failures, 70% were predicted to be failures (and the other 30% were inaccurately predicted to be successes).If the market research predicted that the product would be a success, what is the probability that it would actually be a success?

Accepted Answer

To find the probability that a product is actually successful given that it was predicted to be a success, we use Bayes' theorem. Let's denote S as the event that a product is successful and F as the event that a product fails. Let PS be the event that a product is predicted to be successful and PF as the event that a product is predicted to fail.We are given:- P(S) = 0.5 (probability of success)- P(F) = 0.5 (probability of failure)- P(PS|S) = 0.8 (probability of being predicted successful given it's successful)- P(PF|F) = 0.7 (probability of being predicted a failure given it's a failure)- P(PS|F) = 0.3 (probability of being predicted successful given it's a failure)We want to find P(S|PS), the probability that a product is actually successful given that it was predicted to be successful.Using Bayes' theorem:$$P(S|PS) = \frac{P(PS|S) \cdot P(S)}{P(PS|S) \cdot P(S) + P(PS|F) \cdot P(F)}$$Substituting the given values:$$P(S|PS) = \frac{0.8 \cdot 0.5}{0.8 \cdot 0.5 + 0.3 \cdot 0.5} = \frac{0.4}{0.4 + 0.15} = \frac{0.4}{0.55} = \frac{4}{5.5} = \frac{8}{11}$$$$P(S|PS) = \frac{8}{11} \approx 0.727$$Therefore, the probability that a product is actually successful given that it was predicted to be a success is approximately 0.73.

Question 20

Your company provides services to three different industries: automotive, construction, and financial services.These services can be either on a yearly or on a monthly contract.The cross-tabulation below summarizes this information: $$\begin{array} { | l | l | l | l | } 
\hline & \text { Morthly } & \text { Yearly } & \text { Total } \
\hline \text { Automotive } & 125 & 85 & 210 \
\hline \text { Construction } & 212 & 351 & 563 \
\hline \text { Finaricial } & 357 & 210 & 567 \
\hline \text { Total } & 694 & 646 & 1340 \
\hline
\end{array}$$ Suppose you know that a given client is a construction company and you are interested in the likelihood that this client is on a yearly contract.This is an example of _______.

Accepted Answer

This is an example of conditional probability because it involves finding the probability of an event (a construction company being on a yearly contract) given that another event has occurred (the company is in the construction industry).

Quiz 4: Probability

The department in which you work in your company has 10 employees: 10 women and 14 men.A team of anywhere from one employee up to 9 employees must be chose to attend a meeting at the company's headquarters.How many different teams can be selected?

The department in which you work in your company has 24 employees.A team of 3 employees must be selected to represent the department at a companywide meeting in the headquarters of the company.How many different teams of 3 can be selected?

Suppose 5% of the population have a certain disease.A laboratory blood test gives a positive reading for 95% of people who have the disease and 10% positive reading of people who do not have the disease.What is the probability of testing positive?

It is known that 20% of all students in some large university are overweight, 20% exercise regularly and 2% are overweight and exercise regularly.What is the probability that a randomly selected student is overweight given that this student exercises regularly?

Suppose you toss a fair coin three times in a row and obtain tails, tails, and tails.What it the probability that the fourth toss will give heads?