Question 1

Numbered disks are placed in a box and one disk is selected at random. If there are 6 red disks numbered 1through 6, and 4 yellow disks numbered 7 through 10, find the probability of selecting a yellow disk, given thatthe number selected is less than or equal to 3 or greater than or equal to 8. 
A) $\frac { 1 } { 2 }$
B) $\frac { 3 } { 4 }$
C) $\frac { 3 } { 5 }$
D) $\frac { 3 } { 10 }$

Accepted Answer

The disks that are either less than or equal to 3 or greater than or equal to 8 are 1, 2, 3, 8, 9, 10. Out of these, 3 are yellow (8, 9, 10), and 3 are red (1, 2, 3), making the probability of selecting a yellow disk 3 out of 6, or $\frac{1}{2}$.

Question 2

Find the probability that of 25 randomly selected students, no two share the same birthday.
A) 0.431
B) 0.995
C) 0.569
D) 0.068

Accepted Answer

The probability that no two of the 25 randomly selected students share the same birthday can be calculated using the formula for the birthday problem. Assuming each year has 365 days, the probability for the first student, there's no restriction, so it's 365/365. For the second student, to not share a birthday with the first, it's 364/365, and so on until the 25th student, which would be 341/365. The calculation is: (365/365) * (364/365) * ... * (341/365). This product gives approximately 0.431, making choice A correct.

Question 3

The manager of a used car lot took inventory of the automobiles on his lot and constructed the following tablebased on the age of his car and its make (foreign or domestic). $$\begin{array}{l}
\text {\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad Age of Car (in years) }\
\begin{array} { r | c c c c | c } 
{ \text { Make } } & 0 - 2 & 3 - 5 & 6 - 10 & \text { over } 10 & \text { Total } \
\hline \text { Foreign } & 42 & 24 & 13 & 21 & 100 \
\text { Domestic } & 36 & 23 & 11 & 30 & 100 \
\hline \text { Total } & 78 & 47 & 24 & 51 & 200
\end{array}
\end{array}$$ A car was randomly selected from the lot. Given that the car selected is older than two years old, find theprobability that it is not a foreign car. 
A) $\frac { 32 } { 61 }$
B) $\frac { 29 } { 61 }$
C) $\frac { 16 } { 25 }$
D) $\frac { 29 } { 50 }$ 
Find the indicated probability. Give your answer as a decimal rounded to the nearest thousandth.

Accepted Answer

The probability that the car selected is older than two years old is:$$P(Age > 2) = \frac{47 + 24 + 51}{200} = \frac{122}{200} = 0.61$$The probability that the car selected is not a foreign car is:$$P(Not\ Foreign) = \frac{100}{200} = 0.5$$Therefore, the probability that the car selected is older than two years old and not a foreign car is:$$P(Age > 2 \cap Not\ Foreign) = P(Age > 2) \times P(Not\ Foreign) = 0.61 \times 0.5 = 0.305$$Therefore, the probability that the car selected is older than two years old and not a foreign car is 0.305, which is closest to choice A, $\frac { 32 } { 61 }$.

Question 4

The overnight shipping business has skyrocketed in the last ten years. The single greatest predictor of acompany?s success has been proven time and again to be customer service. A study was conducted to study thecustomer satisfaction levels for one overnight shipping business. In addition to the customer?s satisfaction level,the customers were asked how often they used overnight shipping. The results are shown below in thefollowing table. A customer is chosen at random. Given that the customer uses the company more than fivetimes per month, what is the probability that they expressed low satisfaction with the company? $\begin{array} { c | c c c | c } 
& { \text { Satisfaction level } } & \
\text { Frequency of Use } & \text { High } & \text { Medium } & \text { Low } & \text { TOTAL } \
\hline < 2 \text { per month } & 250 & 140 & 10 & 400 \
2 - 5 \text { per month } & 140 & 55 & 5 & 200 \
> 5 \text { per month } & 70 & 25 & 5 & 100 \
\hline \text { TOTAL } & 460 & 220 & 20 & 700
\end{array}$

A) $\frac { 1 } { 20 }$
B) $\frac { 1 } { 4 }$
C) $\frac { 1 } { 140 }$
D) $\frac { 23 } { 140 }$

Accepted Answer

The answer of The overnight shipping business has skyrocketed in...

Question 5

You are playing roulette at a casino in the United States. The wheel has 18 red slots, 18 black slots, and twogreen slots. In 4 spins of the wheel what is the probability of at least one red? Round to the nearestten-thousandth.
A) 0.9048
B) 0.9375
C) 0.0625
D) 0.0953

Accepted Answer

The probability of not hitting a red in one spin is the probability of hitting either a black or a green, which is 20/38 (since there are 18 black and 2 green slots out of a total of 38 slots). The probability of not hitting a red in 4 consecutive spins is (20/38)^4. To find the probability of hitting at least one red in 4 spins, we subtract this value from 1. So, 1 - (20/38)^4 ≈ 0.9048.

Question 6

The managers of a corporation were surveyed to determine the background that leads to a successful manager.Each manager was rated as being either a good, fair, or poor manager by his/her boss. The manager?seducational background was also noted. The data appear below. Given that a manager is only a fair manager,what is the probability that this manager has no college background? $\begin{array}{l}
\text {\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad Educational Background }\
\begin{array} { r | c c c c | c } 
\begin{array} { c } 
\text { Manager } \
\text { Rating }
\end{array} & \text { H. S. Degree } & \text { Some College } & \text { College Degree } & \text { Master's or Ph.D. } & \text { Totals } \
\hline \text { Good } & 8 & 3 & 22 & 6 & 39 \
\text { Fair } & 7 & 11 & 44 & 25 & 87 \
\text { Poor } & 2 & 4 & 1 & 27 & 34 \
\hline \text { Totals } & 17 & 18 & 67 & 58 & 160
\end{array}
\end{array}$

A) $\frac { 7 } { 87 }$
B) $\frac { 7 } { 17 }$
C) $\frac { 7 } { 160 }$
D) $\frac { 97 } { 160 }$

Accepted Answer

The probability that a fair manager has no college background (H.S. Degree) is the number of fair managers with a H.S. Degree divided by the total number of fair managers: $ \frac{7}{87} $.

Question 7

Two companies, A and B, package and market a chemical substance and claim 0.15 of the total weight of thesubstance is sodium. However, a careful survey of 4,000 packages (half from each company) indicates that theproportion varies around 0.15, with the results shown below. Find the percentage of all chemical B packagesthat contain a sodium total weight proportion above 0.150. $\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\text { Proportion of Sodium }$
$\begin{array}{cccccc}
\hline & <0.100 & 0.100-0.149 & 0.150-0.199 & >0.200 \
\hline & \text { A } & 25 \% & 10 \% & 10 \% & 5 \% \
{\text { Chemcal Brand }}\
& \text { B } & 5 \% & 5 \% & 10 \% & 30 \%
\end{array}$

A) 80%
B) 40%
C) 50%
D) 55%

Accepted Answer

The answer of Two companies, A and B, package and...

Question 8

You toss a fair coin 5 times. What is the probability of at least one head? Round to the nearest ten-thousandth.
A) 0.9688
B) 0.7500
C) 0.5000
D) 0.0313

Accepted Answer

The probability of getting at least one head in 5 tosses of a fair coin is easier to calculate by finding the probability of the opposite event (getting no heads at all, i.e., all tails) and subtracting it from 1. The probability of getting tails in a single toss is 0.5, so for 5 tosses, it's (0.5)^5 = 0.03125. Therefore, the probability of getting at least one head is 1 - 0.03125 = 0.96875, which rounds to 0.9688.

Question 9

The probability that a region prone to hurricanes will be hit by a hurricane in any single year is 101 . What is theprobability of a hurricane at least once in the next 5 years? 
A) $0.40951$
B) $\frac { 1 } { 2 }$
C) $0.99999$
D) $0.00001$

Accepted Answer

The answer of The probability that a region prone to...

Question 10

After completing an inventory of three warehouses, a golf club shaft manufacturer described its stock of 12,246shafts with the percentages given in the table. Suppose a shaft is selected at random from the 12,246 currentlyin stock, and the warehouse number and type of shaft are observed. Given that the shaft is produced inwarehouse 2, find the probability it has an extra stiff shaft. $\begin{array}{l}
\text {\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad Type of Shaft }\
\begin{array} { c c c c c } 
\hline & & \text { Regular } & \text { Stiff } & \text { Extra Stiff } \
\hline& 1 & 19 \% & 8 \% & 17 \% \
 \text { Warehouse }  & 2 & 14 \% & 8 \% & 14 \% \
& 3 & 2 \% & 18 \% & 0 \% \
\hline
\end{array}
\end{array}$

A) $0.389$
B) $0.452$
C) $0.264$
D) $0.53$

Accepted Answer

Given that the shaft is produced in warehouse 2, the probability it has an extra stiff shaft is 14%.

Question 11

The managers of a corporation were surveyed to determine the background that leads to a successful manager.Each manager was rated as being either a good, fair, or poor manager by his/her boss. The manager?seducational background was also noted. The data appear below. Given that a manager is a good manager,what is the probability that this manager has some college background? $\begin{array}{l}
\text {\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad Educational Background }\
\begin{array} { r | c c c c | c } 
\begin{array} { c } 
\text { Manager } \
\text { Rating }
\end{array} &\text { H. S. Degree } & \text { Some College } & \text { College Degree } & \text { Master's or Ph.D. } & \text { Totals } \
\hline \text { Good } & 5 & 1 & 28 & 5 & 39 \
\text { Fair } & 8 & 16 & 45 & 18 & 87 \
\text { Poor } & 9 & 4 & 6 & 15 & 34 \
\hline \text { Totals } & 22 & 21 & 79 & 38 & 160
\end{array}
\end{array}$
A) $\frac { 1 } { 39 }$
B) $\frac { 28 } { 39 }$
C) $\frac { 1 } { 160 }$
D) $\frac { 1 } { 21 }$

Accepted Answer

Given that a manager is a good manager, the probability that this manager has some college background is calculated by dividing the number of good managers with some college background by the total number of good managers. There is 1 good manager with some college background out of 39 good managers, so the probability is $\frac{1}{39}$.

Question 12

The managers of a corporation were surveyed to determine the background that leads to a successful manager.Each manager was rated as being either a good, fair, or poor manager by his/her boss. The manager?seducational background was also noted. The data appear below. Given that a manager is only a fair manager,what is the probability that this manager has a college degree? $\begin{array}{l}
\text {\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad Educational Background }\
\begin{array} { r | c c c c | c } 
\begin{array} { c } 
\text { Manager } \
\text { Rating }
\end{array} & \text { H. S. Degree } & \text { Some College } & \text { College Degree } & \text { Master's or Ph.D. } & \text { Totals } \
\hline \text { Good } & 7 & 2 & 22 & 8 & 39 \
\text { Fair } & 8 & 19 & 48 & 12 & 87 \
\text { Poor } & 3 & 6 & 4 & 21 & 34 \
\hline \text { Totals } & 18 & 27 & 74 & 41 & 160
\end{array}
\end{array}$
A) $\frac { 16 } { 29 }$
B) $\frac { 3 } { 10 }$
C) $\frac { 37 } { 80 }$
D) $\frac { 74 } { 87 }$

Accepted Answer

The answer of The managers of a corporation were surveyed...

Question 13

Investing is a game of chance. Suppose there is a 36% chance that a risky stock investment will end up in a totalloss of your investment. Because the rewards are so high, you decide to invest in five independent risky stocks.Find the probability that at least one of your five investments becomes a total loss. Round to the nearestten-thousandth when necessary.
A) 0.8926
B) 0.302
C) 0.0604
D) 0.006

Accepted Answer

The probability that at least one investment becomes a total loss is 1 minus the probability that none of the investments becomes a total loss. The probability of not losing on a single investment is 1 - 0.36 = 0.64. For five independent investments, the probability of not losing on any is $0.64^5$. Thus, the probability of at least one loss is $1 - 0.64^5 = 1 - 0.1074 = 0.8926$.

Question 14

The manager of a used car lot took inventory of the automobiles on his lot and constructed the following tablebased on the age of his car and its make (foreign or domestic). A car was randomly selected from the lot. Giventhat the car selected was a domestic car, what is the probability that it was older than 2 years? $\begin{array}{l}
\text {\quad\quad\quad\quad\quad\quad\quad\quad\quad Age of Car (in years) }\
\begin{array} { r | c c c c | c } 
\text { Make } & 0 - 2 & 3 - 5 & 6 - 10 & \text { over 10 } & \text { Total } \
\hline \text { Foreign } & 40 & 30 & 10 & 20 & 100 \
\text { Domestic } & 45 & 27 & 11 & 17 & 100 \
\hline \text { Total } & 85 & 57 & 21 & 37 & 200
\end{array}
\end{array}$
A) $\frac { 11 } { 20 }$
B) $\frac { 9 } { 17 }$
C) $\frac { 9 } { 40 }$
D) $\frac { 17 } { 40 }$

Accepted Answer

The probability that the car selected was a domestic car is 100/200 = 1/2. The probability that the car selected was older than 2 years is (27 + 11 + 17)/100 = 55/100 = 11/20. Therefore, the probability that the car selected was a domestic car and was older than 2 years is (1/2) * (11/20) = 11/40.

Question 15

Investment in new issues (the stock of newly formed companies) can be both suicidal and rewarding. Supposethat of 500 newly formed companies in 2010, only 18 appeared to have outstanding prospects. Suppose thatyou had selected two of these 500 companies back in 2010. Find the probability that at least one of yourcompanies had outstanding prospects.
A) 0.0707735
B) 0.3581242
C) 0.9292265
D) 0.0347735

Accepted Answer

The answer of Investment in new issues (the stock of...

Question 16

Find the probability that of 25 randomly selected students, at least two share the same birthday. Round to thenearest thousandth.
A) 0.569
B) 0.068
C) 0.432
D) 0.995

Accepted Answer

The probability that no two students share the same birthday is 365/365 * 364/365 * 363/365 * ... * 341/365 = 0.432. Therefore, the probability that at least two students share the same birthday is 1 - 0.432 = 0.568.

Question 17

The manager of a used car lot took inventory of the automobiles on his lot and constructed the following tablebased on the age of his car and its make (foreign or domestic). A car was randomly selected from the lot. Giventhat the car selected was a foreign car, what is the probability that it was older than 2 years? Age of Car (inyears) $\begin{array}{r|cccc|c}
{\text { Make }} & 0-2 & 3-5 & 6-10 & \text { over 10 } & \text { Total } \
\hline \text { Foreign } & 42 & 22 & 15 & 21 & 100 \
\text { Domestic } & 41 & 20 & 13 & 26 & 100 \
\hline \text { Total } & 83 & 42 & 28 & 47 & 200
\end{array}$
A) $\frac { 29 } { 50 }$
B) $\frac { 21 } { 50 }$
C) $\frac { 58 } { 117 }$
D) $\frac { 14 } { 39 }$

Accepted Answer

There are 100 foreign cars in total. The number of foreign cars older than 2 years is 22 (3-5 years) + 15 (6-10 years) + 21 (over 10 years) = 58. Thus, the probability is 58/100 = 29/50.

Question 18

A researcher at a large university wanted to investigate if a studentʹs seat preference was related in any way tothe gender of the student. The researcher divided the lecture room into three sections (1-front, middle of theroom, 2-front, sides of the classroom, and 3-back of the classroom, both middle and sides) and noted where hisstudents sat on a particular day of the class. The researcherʹs summary table is provided below. Suppose aperson sitting in the front, middle portion of the class is randomly selected to answer a question. Find theprobability the person selected is a female. $\begin{array}{r|ccc|c} 
& \text { Area (1) } & \text { Area (2) } & \text { Area (3) } & \text { Total } \
\hline \text { Males } & 17 & 5 & 11 & 33 \
\text { Females } & 10 & 14 & 15 & 39 \
\hline \text { Total } & 27 & 19 & 26 & 72
\end{array}$
A) $\frac { 10 } { 27 }$
B) $\frac { 10 } { 39 }$
C) $\frac { 9 } { 13 }$
D) $\frac { 5 } { 36 }$

Accepted Answer

The probability that a randomly selected person from the front, middle portion of the class (Area 1) is female is calculated by dividing the number of females in that area (10) by the total number of students in that area (27).

Question 19

A fast-food restaurant chain with 700 outlets in the United States describes the geographic location of itsrestaurants with the accompanying table of percentages. A restaurant is to be chosen at random from the 700 totest market a new style of chicken. Given that the restaurant is located in the eastern United States, what is theprobability it is located in a city with a population of at least 10,000? $\begin{array}{l}
\text {\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad Region }\
\begin{array} { r r r r r c } 
\hline & & \text { NE } & \text { SE } & \text { SW } & \text { NW } \
\hline & < 10,000 & 10 \% & 6 \% & 3 \% & 0 \% \
\text { Population of City } 10,000 &- 100,000 & 15 \% & 5 \% & 12 \% & 5 \% \
& > 100,000 & 20 \% & 4 \% & 9 \% & 11 \%
\end{array}
\end{array}$

A) 0.733
B) 0.543
C) 0.44
D) 0.267

Accepted Answer

The probability that a randomly chosen restaurant is located in a city with a population of at least 10,000, given it is in the eastern United States (NE or SE), is calculated by adding the percentages of restaurants in cities with populations of 10,000-100,000 and those in cities with populations of more than 100,000, then dividing by the total percentage of restaurants in the eastern United States. For NE and SE, the percentages are 15% + 20% for NE and 5% + 4% for SE. The total for the eastern United States is 10% + 15% + 20% + 6% + 5% + 4% = 60%. The probability is then (15% + 20% + 5% + 4%) / 60% = 44% / 60% = 0.733.

Question 20

A study was recently done that emphasized the problem we all face with drinking and driving. Four hundredaccidents that occurred on a Saturday night were analyzed. Two items noted were the number of vehiclesinvolved and whether alcohol played a role in the accident. The numbers are shown below. Given that anaccident involved multiple vehicles, what is the probability that it involved alcohol? $\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\quad $$\text { Number of Vehicles Involved } $
$\begin{array} { r | c c c | c } 
\text { Did Alcohol Play a Role? } & 1 & 2 & 3 \text { or more } & \text { Totals } \
\hline \text { Yes } & 55 & 94 & 21 & 170 \
\text { No } & 25 & 170 & 35 & 230 \
\hline \text { Totals } & 80 & 264 & 56 & 400
\end{array}$
A) $\frac { 23 } { 64 }$
B) $\frac { 23 } { 80 }$
C) $\frac { 3 } { 8 }$
D) $\frac { 21 } { 400 }$

Accepted Answer

The probability that an accident involved alcohol given it involved multiple vehicles is calculated by dividing the number of multi-vehicle accidents involving alcohol by the total number of multi-vehicle accidents. Multi-vehicle accidents are those involving 2 or 3 or more vehicles. From the table, the number of accidents involving alcohol with 2 vehicles is 94, and with 3 or more vehicles is 21, making a total of 115 accidents involving alcohol and multiple vehicles. The total number of multi-vehicle accidents is 264 (2 vehicles) + 56 (3 or more vehicles) = 320. Therefore, the probability is $ \frac{115}{320} = \frac{23}{64} $.

Quiz 5: Probability

Numbered disks are placed in a box and one disk is selected at random. If there are 6 red disks numbered 1through 6, and 4 yellow disks numbered 7 through 10, find the probability of selecting a yellow disk, given thatthe number selected is less than or equal to 3 or greater than or equal to 8.

Find the probability that of 25 randomly selected students, no two share the same birthday.

You are playing roulette at a casino in the United States. The wheel has 18 red slots, 18 black slots, and twogreen slots. In 4 spins of the wheel what is the probability of at least one red? Round to the nearestten-thousandth.

You toss a fair coin 5 times. What is the probability of at least one head? Round to the nearest ten-thousandth.

The probability that a region prone to hurricanes will be hit by a hurricane in any single year is 101 . What is theprobability of a hurricane at least once in the next 5 years?

Find the probability that of 25 randomly selected students, at least two share the same birthday. Round to thenearest thousandth.