Question 1

Find the indicated probability. Round your answer to 6 decimal places when necessary.
-Find the probability of correctly answering the first 3 questions on a multiple choice test if random guesses are made and each question has 4 possible answers.

Accepted Answer

The probability of guessing one question correctly is $\frac{1}{4}$ since there are 4 options. For 3 questions, it's $\left(\frac{1}{4}\right)^3 = \frac{1}{64}$.

Question 2

Solve the problem. Round your answer to 2 decimal places when necessary.
-In a given year, the population of a certain country is about 155 million. The overall birth rate is 15.7 births per 1000. The overall death rate is 11.1 deaths per 1000. If the population of the country
Increased by 1.06 million, what is the approximate net immigration to the country during this
Year?

Accepted Answer

The net population increase due to birth and death rates is calculated as follows: Births = 155 million * (15.7/1000) = 2.4335 million. Deaths = 155 million * (11.1/1000) = 1.7205 million. Net increase due to births and deaths = 2.4335 million - 1.7205 million = 0.713 million. The actual population increase is 1.06 million, so the net immigration is 1.06 million - 0.713 million = 0.347 million or 347,000.

Question 3

Evaluate the factorial expression.
-18 !

Accepted Answer

The answer of Evaluate the factorial expression.
-18 !
...

Question 4

Solve the problem.
-How many different sequences of 4 digits are possible if the first digit must be 3, 4, or 5 and if the sequence may not end in 000? Repetition of digits is allowed.

Accepted Answer

First, calculate the total number of sequences without restrictions. With 3 choices for the first digit and 10 choices for each of the other three digits, there are 3 * 10 * 10 * 10 = 3,000 possible sequences. Subtract the sequences that end in 000, which are 3 (3000, 4000, 5000), resulting in 3,000 - 3 = 2,997 sequences.

Question 5

Solve the problem.
-In a certain lottery, 6 different numbers between 1 and 8 inclusive are drawn. These are the winning numbers. How many different selections are possible? Assume that the order in which the
Numbers are drawn is unimportant.

Accepted Answer

The number of different selections is given by the combination formula C(n, k) = n! / (k!(n-k)!), where n is the total number of options (8) and k is the number of selections (6). So, C(8, 6) = 8! / (6!2!) = 28.

Question 6

Solve the problem.
-A committee of 17 members is voting on a proposal. Each member casts a yea or nay vote. On a random voting basis, what is the probability that the final vote count is unanimous?

Accepted Answer

The probability of a unanimous vote (all yea or all nay) is $ \frac{2}{2^{17}} = \frac{1}{2^{16}} = \frac{1}{65,536} $.

Question 7

Find the expected value.
-You are given 12 to 1 odds against drawing two hearts when two cards are selected at random from a standard deck of 52 cards (with replacement of the first card before the second card is drawn).
This means that you win $12 if you succeed and you lose $1 if you fail. Find the expected value (to
You)of the game. Round to the nearest cent.

Accepted Answer

The probability of drawing a heart from a standard deck of 52 cards is 1/4 (since there are 13 hearts out of 52 cards). Since the drawing is with replacement, the probability of drawing two hearts in a row is (1/4) * (1/4) = 1/16. The probability of not drawing two hearts is therefore 1 - 1/16 = 15/16. The expected value (EV) is calculated as: EV = (Probability of winning * Amount won) + (Probability of losing * Amount lost) = (1/16 * $12) + (15/16 * -$1) = $0.75 - $0.9375 = -$0.1875, which rounds to -$0.19.

Question 8

Solve the problem.
-How many different 4-topping pizzas can be made if there are 13 individual toppings to choose from? Assume that no topping is used more than once and that the order of the toppings on the
Pizza is unimportant.

Accepted Answer

The number of different 4-topping pizzas that can be made from 13 individual toppings, assuming no topping is used more than once and the order of toppings is unimportant, is calculated using the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$, where $n$ is the total number of items to choose from (13 toppings), $k$ is the number of items to choose (4 toppings), and $n!$ denotes the factorial of $n$. Thus, $C(13, 4) = \frac{13!}{4!(13-4)!} = \frac{13!}{4!9!} = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 715$.

Question 9

Find the indicated probability.
-The following table displays a distribution for a group of retired people by career and age at retirement.  Suppose one of these people is selected at random. Compute the probability that the person
Selected was an attorney and that they retired between 61 and 65. Round your answer to the nearest
Thousandth.

Accepted Answer

The answer of Find the indicated probability.
-The following table displays...

Question 10

Find the indicated probability. Round your answer to 6 decimal places when necessary.
-A family has five children. The probability of having a girl is 0.5. What is the probability of having 3 girls followed by 2 boys?

Accepted Answer

The probability of having 3 girls followed by 2 boys, with each child's gender being independent and having a probability of 0.5, is calculated as $0.5^3 \times 0.5^2 = 0.5^5 = \frac{1}{32}$.

Question 11

Use the relative frequency method to estimate the probability. Round your answer to 2 decimal places when necessary.
-Of 1428 people who came into a blood bank to give blood, 303 people had high blood pressure. Estimate the probability that the next person who comes in to give blood will have high blood
Pressure.

Accepted Answer

The probability is estimated by dividing the number of people with high blood pressure by the total number of people: 303/1428 ≈ 0.21.

Question 12

Find the expected value.
-The following table gives the distribution of heights for female students at one U.S. college. $$\begin{array}{l}
\text { Height category }\
\begin{array} { l | l } 
\text { (inches) } & \text { Percentage } \
\hline 57 - 59 & 7.2 \
60 - 62 & 20.5 \
63 - 65 & 36.9 \
66 - 68 & 19.7 \
69 - 71 & 10.5 \
72 \text { and over } & 5.2
\end{array}
\end{array}$$ Find the expected value of height for women at this college. Use the midpoint of each category, (for example use 58 for the 57-59 category). Use 74 for the 72 and over category.

Accepted Answer

To find the expected value of height, we calculate the weighted average using the midpoint of each category and the corresponding percentage. The calculation is as follows:- For 57-59 inches, the midpoint is 58 inches. The expected value contribution is $58 \times 0.072 = 4.176$.- For 60-62 inches, the midpoint is 61 inches. The expected value contribution is $61 \times 0.205 = 12.505$.- For 63-65 inches, the midpoint is 64 inches. The expected value contribution is $64 \times 0.369 = 23.616$.- For 66-68 inches, the midpoint is 67 inches. The expected value contribution is $67 \times 0.197 = 13.199$.- For 69-71 inches, the midpoint is 70 inches. The expected value contribution is $70 \times 0.105 = 7.35$.- For 72 and over, we use 74 inches as the midpoint. The expected value contribution is $74 \times 0.052 = 3.848$.Adding these contributions together gives the expected value of height:$$4.176 + 12.505 + 23.616 + 13.199 + 7.35 + 3.848 = 64.694$$Rounded to one decimal place, the expected value is 64.7 inches.

Question 13

Find the indicated probability. Round your answer to 6 decimal places when necessary.
-A bag contains 19 balls numbered 1 through 19. What is the probability that a randomly selected ball has an even number?

Accepted Answer

In a set of numbers from 1 through 19, there are 9 even numbers (2, 4, 6, 8, 10, 12, 14, 16, 18). Therefore, the probability of selecting an even number is 9 out of 19, or $\frac{9}{19}$.

Question 14

Solve the problem.
-Mr. Larsen's third grade class has 22 students, 12 girls and 10 boys. Two students must be selected at random to be in the fall play. What is the probability that no boys will be chosen? Order is not
Important.

Accepted Answer

The probability that no boys will be chosen is calculated by finding the probability of selecting 2 girls out of 12, divided by the total ways of selecting 2 students out of 22. This is $\frac{{12C2}}{{22C2}} = \frac{{12! / (10! \cdot 2!)}}{{22! / (20! \cdot 2!)}} = \frac{{66}}{{231}} = \frac{2}{7}$.

Question 15

Solve the problem.
-How many ways can an IRS auditor select 5 of 10 tax returns for an audit?

Accepted Answer

The number of ways to select 5 of 10 tax returns for an audit is calculated using the combination formula, which is $C(n, k) = \frac{n!}{k!(n-k)!}$. Here, $n=10$ and $k=5$, so $C(10, 5) = \frac{10!}{5!5!} = 252$.

Question 16

Solve the problem.
-Ten thousand raffle tickets are sold. One first prize of $1600, 5 second prizes of $1000 each, and 10 third prizes of $200 each are to be awarded, with all winners selected randomly. If you are given
One ticket, what are your expected winnings?

Accepted Answer

The expected winnings can be calculated by multiplying the value of each prize by the probability of winning that prize and then summing these values. The probability of winning the first prize is 1/10,000, for the second prize (since there are 5 of them) it's 5/10,000, and for the third prize (with 10 of them) it's 10/10,000. Thus, the expected winnings are: (1/10,000 * $1600) + (5/10,000 * $1000) + (10/10,000 * $200) = $0.16 + $0.50 + $0.20 = $0.86 or 86 cents.

Question 17

Solve the problem.
-A pizza restaurant advertises 286 different 3-topping pizzas. How many individual toppings does the restaurant use? Assume that no topping is used more than once on a pizza and that the order of
The toppings on the pizza is unimportant.

Accepted Answer

The number of different 3-topping pizzas that can be made from n toppings is given by the combination formula $C(n, 3) = \frac{n!}{3!(n-3)!}$. We need to find the value of n that satisfies $C(n, 3) = 286$. Solving the equation $\frac{n!}{3!(n-3)!} = 286$ leads to $n = 13$, as 13 is the number of toppings that allows for 286 unique combinations of 3 toppings each.

Question 18

Find the expected value.
-Find the expected number of girls in a family of 7.

Accepted Answer

The expected number of girls in a family of 7, assuming the probability of having a girl is 0.5 (equal probability for boy or girl), is calculated by multiplying the number of trials (children) by the probability of the desired outcome (having a girl). Therefore, the expected number of girls = 7 * 0.5 = 3.5.

Question 19

Decide whether events A and B are overlapping or non-overlapping.
-You roll a red die and a blue die. Event A is that you get a sum of 11.
Event B is that you get a one on the blue die.

Accepted Answer

The answer of Decide whether events A and B are...

Question 20

Find the expected value.
-You are given 9 to 1 odds against rolling a sum of 8 with the roll of two fair dice, meaning you win $9 if you succeed and you lose $1 if you fail. Find the expected value (to you)of the game.

Accepted Answer

The probability of rolling a sum of 8 with two dice is 5/36, as there are 5 favorable outcomes (2-6, 3-5, 4-4, 5-3, 6-2) out of 36 possible outcomes. The probability of not rolling a sum of 8 is 31/36. The expected value is calculated as (Probability of winning * Amount won) + (Probability of losing * Amount lost) = (5/36 * $9) + (31/36 * -$1) = $0.39.

Quiz 7: Probability: Living With the Odds

Find the indicated probability. Round your answer to 6 decimal places when necessary. -Find the probability of correctly answering the first 3 questions on a multiple choice test if random guesses are made and each question has 4 possible answers.

Evaluate the factorial expression. -18 !

Solve the problem. -How many different sequences of 4 digits are possible if the first digit must be 3, 4, or 5 and if the sequence may not end in 000? Repetition of digits is allowed.

Solve the problem. -In a certain lottery, 6 different numbers between 1 and 8 inclusive are drawn. These are the winning numbers. How many different selections are possible? Assume that the order in which the Numbers are drawn is unimportant.

Solve the problem. -A committee of 17 members is voting on a proposal. Each member casts a yea or nay vote. On a random voting basis, what is the probability that the final vote count is unanimous?

Solve the problem. -How many different 4-topping pizzas can be made if there are 13 individual toppings to choose from? Assume that no topping is used more than once and that the order of the toppings on the Pizza is unimportant.

Find the indicated probability. Round your answer to 6 decimal places when necessary. -A family has five children. The probability of having a girl is 0.5. What is the probability of having 3 girls followed by 2 boys?

Find the indicated probability. Round your answer to 6 decimal places when necessary. -A bag contains 19 balls numbered 1 through 19. What is the probability that a randomly selected ball has an even number?

Solve the problem. -Mr. Larsen's third grade class has 22 students, 12 girls and 10 boys. Two students must be selected at random to be in the fall play. What is the probability that no boys will be chosen? Order is not Important.

Solve the problem. -How many ways can an IRS auditor select 5 of 10 tax returns for an audit?

Solve the problem. -Ten thousand raffle tickets are sold. One first prize of $1600, 5 second prizes of $1000 each, and 10 third prizes of $200 each are to be awarded, with all winners selected randomly. If you are given One ticket, what are your expected winnings?

Solve the problem. -A pizza restaurant advertises 286 different 3-topping pizzas. How many individual toppings does the restaurant use? Assume that no topping is used more than once on a pizza and that the order of The toppings on the pizza is unimportant.

Find the expected value. -Find the expected number of girls in a family of 7.

Decide whether events A and B are overlapping or non-overlapping. -You roll a red die and a blue die. Event A is that you get a sum of 11. Event B is that you get a one on the blue die.

Find the expected value. -You are given 9 to 1 odds against rolling a sum of 8 with the roll of two fair dice, meaning you win $9 if you succeed and you lose $1 if you fail. Find the expected value (to you) of the game.