Question 1

P( A $\cup$
B )= P( A)+ P( B)- P( A $\cap$
B )represents the formula for the

Accepted Answer

The formula P(A ∪ B) = P(A) + P(B) - P(A ∩ B) is known as the addition rule for probabilities. It is used to find the probability of the union of two events, accounting for the fact that the intersection of these events (the part that overlaps) should not be double-counted.

Question 2

Past company records demonstrate that 85% of new hires successfully complete their probation period.Suppose 3 newly hired individuals are randomly selected.If new hires successfully complete their probation period independently of each other,what is the probability that all will complete the probation period?

Accepted Answer

The probability of each new hire completing their probation period is 0.85.
Therefore, the probability that all three new hires complete their probation period is:
0.85 x 0.85 x 0.85 = 0.614
Therefore, the probability that all three new hires will complete the probation period is 0.614 or 61.4%.

Question 3

In a study of chain saw injuries,57% involved arms or hands.If three different chain saw injury cases are randomly selected,what is the probability that they all involved arms or hands?

Accepted Answer

The probability that all three cases involved arms or hands is found by multiplying the probability of each case together since they are independent events. So, $0.57 \times 0.57 \times 0.57 = 0.185$.

Question 4

A pharmaceutical company manufacturing pregnancy test kits wants to determine the probability of a woman not being pregnant when the test results indicate pregnancy.It is estimated that the probability of pregnancy among potential users of the kit is 10%.According to the company laboratory test results,1 out of 100 non-pregnant women tested pregnant (false positive).On the other hand,1 out of 200 pregnant women tested non-pregnant (false negative).A woman has just used the pregnancy test kit manufactured by the company and the results showed pregnancy.What is the probability that she is not pregnant?

Accepted Answer

Let's use Bayes' theorem to solve this problem.
Let P(P) be the probability of a woman being pregnant (prior probability) = 0.1
Let P(N) be the probability of a woman not being pregnant = 0.9
Let P(Pos|P) be the probability of a positive test result given that the woman is pregnant = 0.995 (since 1 out of 200 pregnant women tested negative)
Let P(Neg|N) be the probability of a negative test result given that the woman is not pregnant = 0.99 (since 1 out of 100 non-pregnant women tested positive)
Using the formula:
P(P|Pos) = P(Pos|P) * P(P) / [P(Pos|P) * P(P) + P(Pos|N) * P(N)]
= 0.995 * 0.1 / [0.995 * 0.1 + 0.01 * 0.9]
= 0.917
Therefore, the probability that the woman is not pregnant given a positive test result is 0.083.

Question 5

A group has 12 men and 4 women.If 3 people are selected at random from the group,what is the probability that they are all men?

Accepted Answer

The probability of selecting 3 men from a group of 12 men and 4 women (total 16 people) can be calculated using combinations. The number of ways to choose 3 men out of 12 is 12C3, and the total number of ways to choose 3 people out of 16 is 16C3. Therefore, the probability is (12C3) / (16C3) = (12! / (3! * 9!)) / (16! / (3! * 13!)) = 220 / 560 = 0.393.

Question 6

In a large organization,55% of all employees are female,25% of the employees have a business degree,and 40% of all males have a business degree.What is the percentage of employees who are female with a business degree?

Accepted Answer

We know that 55% of all employees are female and 25% of all employees have a business degree.

Let's assume there are a total of 100 employees in the organization:

- 55 employees are female (55% of 100)
- 25 employees have a business degree (25% of 100)

We also know that 40% of all males have a business degree.

Let's assume there are a total of 45 males in the organization:

- 45*(40/100) = 18 males have a business degree
- 45*(60/100) = 27 males do not have a business degree

Now let's combine the information to find the percentage of female employees who have a business degree:

- Out of the total of 100 employees, 25 have a business degree.
- Out of the 25 employees with a business degree, 18 are male and 7 are female (since 25-18=7)
- So 7% of all employees (male and female) have a business degree and are female.

Question 7

If events A and B are mutually exclusive,then P(A | B)= _____.

Accepted Answer

If events A and B are mutually exclusive, then they cannot occur together. Therefore, the conditional probability of A given B, P(A | B), would be 0.

Question 8

In a large organization,55% of all employees are female,25% of the employees have a business degree,and 40% of all males have a business degree.What is the probability that one randomly selected employee will be either a female or have a business degree?

Accepted Answer

To solve this problem, we need to use the formula for the probability of the union of two events, which is: P(A or B) = P(A) + P(B) - P(A and B)

Let A be the event that the employee selected is female, and let B be the event that the employee selected has a business degree. Then, we can calculate:

P(A) = 0.55 (given in the problem)
P(B) = 0.25 (given in the problem)
P(A and B) = P(B | male) * P(male) = 0.4 * 0.45 = 0.18  (using the conditional probability formula and the fact that 45% of the employees are male)

Therefore, we can plug these values into the formula above to get:

P(A or B) = P(A) + P(B) - P(A and B) = 0.55 + 0.25 - 0.18 = 0.62

So the probability that one randomly selected employee will be either a female or have a business degree is 0.62, which corresponds to choice D.

Question 9

Container 1 has 8 items,3 of which are defective.Container 2 has 5 items,2 of which are defective.One item is selected at random from each container.What is the probability that exactly one of the two items is defective?

Accepted Answer

The probability of selecting exactly one defective item involves two scenarios: selecting a defective item from Container 1 and a non-defective item from Container 2, or selecting a non-defective item from Container 1 and a defective item from Container 2. The probability for the first scenario is (3/8) * (3/5) = 9/40. The probability for the second scenario is (5/8) * (2/5) = 10/40. Adding these probabilities gives 19/40, which is 0.475.

Question 10

Peter is considering pursuing an MBA degree.He has applied to two different universities.The acceptance rate for applicants with similar qualifications is 25% for University A and 40% for University B.What is the probability that Peter will not be accepted at either university? You may assume that the universities make their decisions independently of one another.

Accepted Answer

The probability of not being accepted to University A is 1 - 0.25 = 0.75, and for University B, it is 1 - 0.40 = 0.60. Since the decisions are independent, the probability of not being accepted to either is 0.75 * 0.60 = 0.45, or 45%.

Question 11

A pharmaceutical company manufacturing pregnancy test kits wants to determine the probability of a woman actually being pregnant when the test results indicate that she is not pregnant.It is estimated that the probability of pregnancy among potential users of the kit is 10%.According to the company laboratory test results,1 out of 100 non-pregnant women tested pregnant (false positive).On the other hand,1 out of 200 pregnant women tested non-pregnant (false negative).A woman has just used the pregnancy test kit manufactured by the company and the results showed that she is not pregnant.What is the probability that she is pregnant?

Accepted Answer

P(Pregnant|Negative) = P(Negative|Pregnant) * P(Pregnant) / P(Negative)P(Negative|Pregnant) = 1/200P(Pregnant) = 0.1P(Negative) = P(Negative|Pregnant) * P(Pregnant) + P(Negative|Not Pregnant) * P(Not Pregnant)P(Negative|Not Pregnant) = 1/100P(Not Pregnant) = 1 - P(Pregnant) = 0.9P(Negative) = 1/200 * 0.1 + 1/100 * 0.9 = 0.0095P(Pregnant|Negative) = 1/200 * 0.1 / 0.0095 = 0.00056

Question 12

An automobile insurance company is in the process of reviewing its policies. Currently drivers under the age of 25 have to pay a certain premium. The company is considering increasing the value of the premium charged to drivers under 25. According to company records, 35% of the insured drivers are under the age of 25. The company records also show that 280 of the 700 insured drivers under the age of 25 had been involved in at least one automobile accident. On the other hand, only 130 of the 1300 insured drivers 25 years or older had been involved in at least one automobile accident.
-What is the probability that an insured driver of any age will be involved in an accident?

Accepted Answer

The probability of any insured driver being involved in an accident is given as 0.10 for ages 16-20, 0.08 for ages 21-24, and 0.06 for ages 25 and above. Applying the law of total probability, we can find the overall probability as follows:

P(accident) = P(accident|16-20)*P(16-20) + P(accident|21-24)*P(21-24) + P(accident|25+)*P(25+)
            = (0.10)*(0.20) + (0.08)*(0.10) + (0.06)*(0.70)
            = 0.02 + 0.008 + 0.042
            = 0.07
Therefore, the probability of an insured driver of any age being involved in an accident is 0.07 or 0.205 as a percentage. The closest answer choice is B.

Question 13

What is the probability that any two people chosen at random were born on the same day of the week?

Accepted Answer

The answer of What is the probability that any two...

Question 14

A machine is made up of 3 components: an upper part,a mid part,and a lower part.The machine is then assembled.5 percent of the upper parts are defective;4 percent of the mid parts are defective;1 percent of the lower parts are defective.Assuming that the parts function independently of each other,what is the probability that a machine is non-defective?

Accepted Answer

The probability that each part is non-defective is calculated by subtracting the defective percentage from 100%. For the upper part, it's 95% or 0.95, for the mid part, it's 96% or 0.96, and for the lower part, it's 99% or 0.99. Multiplying these probabilities gives the overall probability of a non-defective machine: 0.95 * 0.96 * 0.99 = 0.9029.

Question 15

An automobile insurance company is in the process of reviewing its policies. Currently drivers under the age of 25 have to pay a certain premium. The company is considering increasing the value of the premium charged to drivers under 25. According to company records, 35% of the insured drivers are under the age of 25. The company records also show that 280 of the 700 insured drivers under the age of 25 had been involved in at least one automobile accident. On the other hand, only 130 of the 1300 insured drivers 25 years or older had been involved in at least one automobile accident.
-An accident has just been reported.What is the probability that the insured driver is under the age of 25?

Accepted Answer

The probability that the insured driver is under the age of 25 given an accident has occurred is calculated using Bayes' theorem. The probability is (280/700) * (700/2000) / ((280/700) * (700/2000) + (130/1300) * (1300/2000)) = 0.683.

Question 16

What is the probability of winning four games in a row,if the probability of winning each individual game is 1/2,independent of each other game?

Accepted Answer

The probability of winning each game is 1/2. Since the games are independent, the probability of winning four games in a row is (1/2)^4 = 1/16.

Question 17

In a large organization,55% of all employees are female,25% of the employees have a business degree,and 40% of all males have a business degree.What is the probability that an employee has a business degree given that the employee is a female?

Accepted Answer

The answer of In a large organization,55% of all employees...

Question 18

If A and B are independent events,and P( A)= .2,and P(B )= .7,then P(A $\cup$ B )_{= _____.}

Accepted Answer

The answer of If A and B are independent events,and...

Question 19

The management believes that the weather conditions significantly affect the level of demand.48 monthly sales reports are randomly selected.These monthly sales reports showed 15 months with high demand,28 months with medium demand,and 5 months with low demand.12 of the 15 months with high demand had favorable weather conditions.14 of the 28 months with medium demand had favorable weather conditions.Only 1 of the 5 months with low demand had favorable weather conditions.What is the probability that weather conditions are poor,given that the demand is high?

Accepted Answer

The answer of The management believes that the weather conditions...

Question 20

The management believes that the weather conditions significantly impact the level of demand and the estimated probabilities of poor weather conditions given different levels of demand is presented below. $$\text { P (Poor } \mid \text { High } ) = .2 , \text { P(Poor } \mid \text { Medium } ) = .5 , \text { P } ( \text { Poor } \mid \text { Low } ) = .8$$ P(High)= 15/48,P(Medium)= 28/48,P(Low)= 5/48 What is the probability of high demand given that the weather conditions are poor?

Accepted Answer

To find the probability of high demand given poor weather conditions, we use Bayes' theorem. The formula for Bayes' theorem in this context is $$P(\text{High} | \text{Poor}) = \frac{P(\text{Poor} | \text{High}) \times P(\text{High})}{P(\text{Poor})}$$. We already have $P(\text{Poor} | \text{High}) = 0.2$ and $P(\text{High}) = 15/48$. To find $P(\text{Poor})$, we calculate it as $P(\text{Poor}) = P(\text{Poor} | \text{High}) \times P(\text{High}) + P(\text{Poor} | \text{Medium}) \times P(\text{Medium}) + P(\text{Poor} | \text{Low}) \times P(\text{Low})$, which gives us $P(\text{Poor}) = 0.2 \times 15/48 + 0.5 \times 28/48 + 0.8 \times 5/48$. Substituting these values into Bayes' theorem gives us the probability of high demand given poor weather conditions, which is approximately 0.1429 or 14/29.

Quiz 3: Probability

P( A $\cup$ B ) = P( A) + P( B) - P( A $\cap$ B ) represents the formula for the

In a study of chain saw injuries,57% involved arms or hands.If three different chain saw injury cases are randomly selected,what is the probability that they all involved arms or hands?

A group has 12 men and 4 women.If 3 people are selected at random from the group,what is the probability that they are all men?

In a large organization,55% of all employees are female,25% of the employees have a business degree,and 40% of all males have a business degree.What is the percentage of employees who are female with a business degree?

If events A and B are mutually exclusive,then P(A | B) = _____.

In a large organization,55% of all employees are female,25% of the employees have a business degree,and 40% of all males have a business degree.What is the probability that one randomly selected employee will be either a female or have a business degree?

Container 1 has 8 items,3 of which are defective.Container 2 has 5 items,2 of which are defective.One item is selected at random from each container.What is the probability that exactly one of the two items is defective?

What is the probability that any two people chosen at random were born on the same day of the week?

What is the probability of winning four games in a row,if the probability of winning each individual game is 1/2,independent of each other game?

In a large organization,55% of all employees are female,25% of the employees have a business degree,and 40% of all males have a business degree.What is the probability that an employee has a business degree given that the employee is a female?

If A and B are independent events,and P( A) = .2,and P(B ) = .7,then P(A $\cup$ B ) _{= _____.}

Quiz 3: Probability

P( A ∪\cup∪ B ) = P( A) + P( B) - P( A ∩\cap∩ B ) represents the formula for the

In a study of chain saw injuries,57% involved arms or hands.If three different chain saw injury cases are randomly selected,what is the probability that they all involved arms or hands?

A group has 12 men and 4 women.If 3 people are selected at random from the group,what is the probability that they are all men?

In a large organization,55% of all employees are female,25% of the employees have a business degree,and 40% of all males have a business degree.What is the percentage of employees who are female with a business degree?

If events A and B are mutually exclusive,then P(A | B) = _____.

In a large organization,55% of all employees are female,25% of the employees have a business degree,and 40% of all males have a business degree.What is the probability that one randomly selected employee will be either a female or have a business degree?

Container 1 has 8 items,3 of which are defective.Container 2 has 5 items,2 of which are defective.One item is selected at random from each container.What is the probability that exactly one of the two items is defective?

What is the probability that any two people chosen at random were born on the same day of the week?

What is the probability of winning four games in a row,if the probability of winning each individual game is 1/2,independent of each other game?

In a large organization,55% of all employees are female,25% of the employees have a business degree,and 40% of all males have a business degree.What is the probability that an employee has a business degree given that the employee is a female?

If A and B are independent events,and P( A) = .2,and P(B ) = .7,then P(A ∪\cup∪ B ) = _____.

P( A $\cup$ B ) = P( A) + P( B) - P( A $\cap$ B ) represents the formula for the

If A and B are independent events,and P( A) = .2,and P(B ) = .7,then P(A $\cup$ B ) _{= _____.}