Question 1

List the outcome(s)of the stated event.
-The odds against winning in a horse race are shown in the following table. $$\begin{array} { l | r r r r r r r } 
\text { Horse } & \# 1 & \# 2 & \# 3 & \# 4 & \# 5 & \# 6 & \# 7 \
\hline \text { Odds } & 8 & 15 & 2 & 22 & 12 & 15 & 22
\end{array}$$ Based on these odds, which horses comprise: A = event one of the top two favorites wins the race?

Accepted Answer

The answer of List the outcome(s)of the stated event.
-The odds...

Question 2

Use Bayes's rule to find the indicated probability.
-34% of the workers at Motor Works are female, while 61% of the workers at City Bank are female. If one of these companies is selected at random (assume a 50-50 chance for each), and then a worker Is selected at random, what is the probability that the worker is female, given that the worker Comes from City Bank?

Accepted Answer

The probability that a worker is female given that they come from City Bank is simply the probability of selecting a female worker from City Bank, which is 61% or 0.61. However, the choices provided do not include 0.61, so the correct choice is the one that matches the calculation of the probability of selecting a female worker from City Bank, which is 0.305.

Question 3

Use the basic counting rule to solve the problem.
-A shirt company has 3 designs each of which can be made with short or long sleeves. There are 5 color patterns available. How many different types of shirts are available from this company?

Accepted Answer

The basic counting rule states that if there are multiple independent choices, the total number of combinations is the product of the number of choices for each option. Here, there are 3 designs, 2 sleeve options (short or long), and 5 color patterns. Therefore, the total number of different types of shirts is 3 * 2 * 5 = 30.

Question 4

Describe the specified event in words.
-When a quarter is tossed four times, 16 outcomes are possible. $\begin{array} { l l l l } \text { HHHH } & \text { HHHT } & \text { HHTH } & \text { HHTT } \ \text { HTHH } & \text { HTHT } & \text { HTTH } & \text { HTTT } \ \text { THHH } & \text { THHT } & \text { THTH } & \text { THTT } \ \text { TTHH } & \text { TTHT } & \text { TTTH } & \text { TTTT } \end{array}$
Here, for example, HTTH represents the outcome that the first toss is heads, the next two tosses are tails, and the fourth toss is heads. The events $A$ and $B$ are defined as follows.
$A =$ event exactly two tails are tossed
$B =$ event the last two tosses are heads
Describe the event (A & B) in words.

Accepted Answer

The answer of Describe the specified event in words.
-When a...

Question 5

Find the indicated probability.
-A class consists of 54 women and 77 men. If a student is randomly selected, what is the probability that the student is a woman?

Accepted Answer

The probability that a student is a woman is calculated by dividing the number of women by the total number of students, which is $54 / (54 + 77) = 54 / 131$.

Question 6

List the outcomes comprising the specified event.
-In a competition, two people will be selected from four finalists to receive the first and second prizes. The prize winners will be selected by drawing names from a hat. The names of the four Finalists are Jim, George, Helen, and Maggie. The possible outcomes can be represented as follows. $$\begin{array} { c c c c c c } 
J G & J H & J M & \text { GJ } & \text { GH } & \text { GM } \
\text { HJ } & \text { HG } & \text { HM } & \text { MJ } & \text { MG } & \text { MH }
\end{array}$$ Here, for example, JG represents the outcome that Jim receives the first prize and George receives The second prize. The events A and B are defined as follows. 
$A =$ event that Helen gets first prize
$B =$ event that George gets a prize
List the outcomes that comprise the event (A or B).

Accepted Answer

Event A includes outcomes where Helen gets the first prize (HJ, HG, HM), and event B includes outcomes where George gets any prize (JG, GJ, GH, GM, HG, MG). Combining these without repetition gives us JG, GJ, GH, GM, HJ, HG, HM, MG.

Question 7

List the outcomes comprising the specified event.
-Three board members for a nonprofit organization will be selected from a group of five people. The board members will be selected by drawing names from a hat. The names of the five possible board Members are Allison, Betty, Charlie, Dave, and Emily. The possible outcomes can be represented as Follows. 
$\begin{array} { l l l l l } \mathrm { ABC } & \mathrm { ABD } & \mathrm { ABE } & \mathrm { ACD } & \mathrm { ACE } \ \mathrm { ADE } & \mathrm { BCD } & \mathrm { BCE } & \mathrm { BDE } & \mathrm { CDE } \end{array}$
Here, for example, ABC represents the outcome that Allison, Betty, and Charlie are selected to be on the board. List the outcomes that comprise the following event.
$A =$ event that fewer than two men are selected

Accepted Answer

Assuming Allison and Betty are women, and Charlie, Dave, and Emily are men (based on common name associations and the context of the question focusing on gender balance), the correct outcomes where fewer than two men are selected include combinations with at least two women (Allison, Betty, Emily). Thus, all combinations excluding those with two or more men (Charlie, Dave) are listed in option D.

Question 8

Determine the number of outcomes that comprise the specified event.
-The number of hours needed by sixth grade students to complete a research project was recorded with the following results. $\begin{array}{cc}
\text { Hours } & \text { Number of students (f) } \
\hline 4 & 24 \
5 & 16 \
6 & 27 \
7 & 13 \
8 & 11 \
9 & 4 \
10+ & 8
\end{array}$

A student is selected at random. The events A and B are defined as follows.
$\mathrm { A } =$ the event the student took at most 8 hours
$B =$ the event the student took at least 8 hours
Determine the number of outcomes that comprise the event (A or B).

Accepted Answer

The answer of Determine the number of outcomes that comprise...

Question 9

Use Bayes's rule to find the indicated probability.
-The first two columns of the table below give a percentage distribution for adults in one city by income group. The third column gives the percentage of people in each income group who plan to Buy a new car next year. $$\begin{array} { c c c } 
\begin{array} { l } 
\text { Income } \
\text { (dollars) }
\end{array} & \begin{array} { c } 
\text { Percentage } \
\text { of population }
\end{array} & \begin{array} { c } 
\text { Percentage that will } \
\text { buy new car next year }
\end{array} \
\hline 0 - 4999 & 5.2 & 2 \
5000 - 9999 & 6.4 & 3 \
10,000 - 14,999 & 5.4 & 6 \
15,000 - 19,999 & 8.7 & 7 \
20,000 - 24,999 & 9.4 & 9 \
25,000 - 29,999 & 10.2 & 10 \
30,000 - 34,999 & 13.8 & 11 \
35,000 - 39,999 & 10.7 & 13 \
40,000 - 49,999 & 15.5 & 15 \
50,000 \text { and over } & 14.7 & 19
\end{array}$$ An adult is picked at random from the city. Given that the person selected plans to buy a new car Next year, what is the probability that the person's income is between $5,000 and $9,999?

Accepted Answer

P(5000<X<9999|Buy new car next year) = P(Buy new car next year|5000<X<9999) * P(5000<X<9999) / P(Buy new car next year)= 0.03 * 0.064 / 0.15= 0.02

Question 10

List the outcomes comprising the specified event.
-When a quarter is tossed four times, 16 outcomes are possible. $\begin{array} { l l l l } \text { HHHH } & \text { HHHT } & \text { HHTH } & \text { HHTT } \ \text { HTHH } & \text { HTHT } & \text { HTTH } & \text { HTTT } \ \text { THHH } & \text { THHT } & \text { THTH } & \text { THTT } \ \text { TTHH } & \text { TTHT } & \text { TTTH } & \text { TTTT } \end{array}$
Here, for example, HTTH represents the outcome that the first toss is heads, the next two tosses are tails, and the fourth toss is heads. List the outcomes that comprise the following event.
$\mathrm { A } =$ event the last toss is heads

Accepted Answer

Each outcome listed in option C ends with an H, which corresponds to the event where the last toss results in heads.

Question 11

Describe the specified event in words.
-When a quarter is tossed four times, 16 outcomes are possible. $$\begin{array} { l l l l } 
\text { НHHH } & \text { HНHT } & \text { HHTH } & \text { HHTT } \
\text { HTHH } & \text { HTHT } & \text { HTTH } & \text { HTTT } \
\text { THHH } & \text { THHT } & \text { THTH } & \text { THTT } \
\text { TTHH } & \text { TTHT } & \text { TTTH } & \text { TTTT }
\end{array}$$ Here, for example, HTTH represents the outcome that the first toss is heads, the next two tosses are tails, and the fourth toss is heads. The event A is defined as follows.
A = event the first two tosses are heads
Describe the event (not A) in words.

Accepted Answer

The event not A refers to all outcomes where the first two tosses are not both heads, which includes any combination where at least one of the first two tosses is a tail.

Question 12

Find the indicated probability.
-On a multiple choice test, each question has 5 possible answers. If you make a random guess on the first question, what is the probability that you are correct?

Accepted Answer

Since there are 5 possible answers and only one of them is correct, the probability of guessing the correct answer randomly is 1 out of 5, or $\frac{1}{5}$.

Question 13

Use the basic counting rule to solve the problem.
-A singer-songwriter wishes to compose a melody. Each note in the melody must be one of the 12 notes in her vocal range. How many different sequences of 4 notes are possible?

Accepted Answer

The basic counting rule states that if there are n ways to do something and m ways to do another thing, there are n * m ways to do both. Here, there are 12 choices for each of the 4 notes, so the total number of sequences is 12^4 = 20,736.

Question 14

List the outcomes comprising the specified event.
-Three board members for a nonprofit organization will be selected from a group of five people. The board members will be selected by drawing names from a hat. The names of the five possible board Members are Allison, Bob, Charlie, Dave, and Emily. The possible outcomes can be represented as Follows.
 $\begin{array}{lllll}
\mathrm{ABC} & \mathrm{ABD} & \mathrm{ABE} & \mathrm{ACD} & \mathrm{ACE} \
\mathrm{ADE} & \mathrm{BCD} & \mathrm{BCE} & \mathrm{BDE} & \mathrm{CDE}
\end{array}$

Here, for example, ABC represents the outcome that Allison, Bob, and Charlie are selected to be on the board. The events $A$ and $B$ are defined as follows.
$A =$ event that Dave is selected
$B =$ event that fewer than two men are selected
List the outcomes that comprise the event (A & B).

Accepted Answer

Event A requires Dave to be selected, and event B requires fewer than two men to be selected. The outcomes that satisfy both conditions (Dave being selected and fewer than two men being selected) are ABD, ADE, and BDE, as these combinations include Dave and at most one other man, meeting the criteria for both events A and B.

Question 15

Find the conditional probability.
-If two cards are drawn without replacement from a deck, find the probability that the second card is a spade, given that the first card was a spade.

Accepted Answer

The answer of Find the conditional probability.
-If two cards are...

Question 16

Find the indicated probability by using the general addition rule.
-For a person selected randomly from a certain population, events A and B are defined as follows. A = event the person is male
B = event the person is a smoker For this particular population, it is found that $\mathrm { P } ( \mathrm { A } ) = 0.47 , \mathrm { P } ( \mathrm { B } ) = 0.31$, and $\mathrm { P } ( \mathrm { A } \& \mathrm {~B} ) = 0.12$. Find $\mathrm { P } ( \mathrm { A }$ or $\mathrm { B } )$. Round approximations to two decimal places.

Accepted Answer

The general addition rule is $ P(A \cup B) = P(A) + P(B) - P(A \cap B) $. Substituting the given values: $ 0.47 + 0.31 - 0.12 = 0.66 $.

Question 17

Use counting rules to determine the probability.
-8 basketball players are to be selected to play in a special game. The players will be selected from a list of 27 players. If the players are selected randomly, what is the probability that the 8 tallest Players will be selected?

Accepted Answer

The probability is calculated by dividing the number of ways to choose the 8 tallest players (which is 1 way, since there's only one group of the 8 tallest) by the total number of ways to choose any 8 players from 27, which is calculated using the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$. Therefore, the probability is $\frac{1}{C(27, 8)} = \frac{1}{2,220,075}$.

Question 18

Determine the number of outcomes that comprise the specified event.
-Suppose you are playing a game of chance. If you bet $5 on a certain event, you will collect $150 (including your $5 bet)if you win. Find the odds used for determining the payoff.

Accepted Answer

The payoff is $150, including the original $5 bet, meaning the profit is $145. The odds are calculated based on the profit versus the original bet. Since you bet $5 to win $145, the casino is essentially saying the odds of the event happening are 29 to 1 against you. This is because for every $5 bet, the payout is as if you had bet on something 29 times and won once, hence 29 to 1.

Question 19

Find the indicated probability by using the special addition rule.
-A percentage distribution is given below for the size of families in one U.S. city. 
$\begin{array} { l c } 
\text { Size } & \text { Percentage } \
\hline 2 & 49.6 \
3 & 24.6 \
4 & 13.4 \
5 & 7.3 \
6 & 3.1 \
7 + & 2.0
\end{array}$ 
A family is selected at random. Find the probability that the size of the family is at most 3. Round approximations to three decimal places.

Accepted Answer

The probability that the size of the family is at most 3 is the sum of the probabilities of having a family size of 2 or 3, which is 49.6% + 24.6% = 74.2% or 0.742 when expressed as a decimal.

Question 20

Use the special multiplication rule to find the indicated probability.
-When a pair of dice is rolled there are 36 different possible outcomes: 1-1, 1-2, ... 6-6. If a pair of dice is rolled 3 times, what is the probability of getting a sum of 5 every time?

Accepted Answer

The probability of getting a sum of 5 on one roll is 4/36 (or 1/9), since there are 4 outcomes that result in a sum of 5: (1,4), (2,3), (3,2), and (4,1). Using the special multiplication rule for independent events, the probability of this happening 3 times in a row is (1/9) * (1/9) * (1/9) = 1/729, which equals approximately 0.00137174.

Quiz 4: Probability Concepts

Use the basic counting rule to solve the problem. -A shirt company has 3 designs each of which can be made with short or long sleeves. There are 5 color patterns available. How many different types of shirts are available from this company?

Find the indicated probability. -A class consists of 54 women and 77 men. If a student is randomly selected, what is the probability that the student is a woman?

Find the indicated probability. -On a multiple choice test, each question has 5 possible answers. If you make a random guess on the first question, what is the probability that you are correct?

Use the basic counting rule to solve the problem. -A singer-songwriter wishes to compose a melody. Each note in the melody must be one of the 12 notes in her vocal range. How many different sequences of 4 notes are possible?

Find the conditional probability. -If two cards are drawn without replacement from a deck, find the probability that the second card is a spade, given that the first card was a spade.

Use counting rules to determine the probability. -8 basketball players are to be selected to play in a special game. The players will be selected from a list of 27 players. If the players are selected randomly, what is the probability that the 8 tallest Players will be selected?

Determine the number of outcomes that comprise the specified event. -Suppose you are playing a game of chance. If you bet $5 on a certain event, you will collect $150 (including your $5 bet) if you win. Find the odds used for determining the payoff.

Use the special multiplication rule to find the indicated probability. -When a pair of dice is rolled there are 36 different possible outcomes: 1-1, 1-2, ... 6-6. If a pair of dice is rolled 3 times, what is the probability of getting a sum of 5 every time?