Liberal Arts Math Exam 3 Study Guide: Probability, Counting, and Combinatorics

Q1. In how many different ways can a new car be ordered given 8 possible colors and 5 binary options (air conditioning, heated seats, anti-lock brakes, power windows, CD player)?

Background

Topic: Counting Principles (Multiplication Principle)

This question tests your understanding of how to count the total number of possible outcomes when making independent choices for each feature of a product.

Key Terms and Formulas

• Multiplication Principle: If there are $m$ ways to do one thing and $n$ ways to do another, there are $m \times n$ ways to do both.

• Binary Option: A choice with two possibilities (e.g., with or without a feature).

Step-by-Step Guidance

1. Identify the number of choices for each feature: 8 colors, and each of the 5 features has 2 options (with or without).

2. Use the multiplication principle to combine the choices: Multiply the number of options for each feature together.

3. Set up the expression: $8 \times 2 \times 2 \times 2 \times 2 \times 2$

4. Simplify the powers of 2: $2^5$

Try solving on your own before revealing the answer!

Final Answer: 256 different ways

There are $8 \times 2^5 = 8 \times 32 = 256$ possible ways to order the car with these options.

Q2a. Evaluate the factorial expression: $\frac{322!}{320!}$

Background

Topic: Factorials and Simplification

This question tests your ability to simplify expressions involving factorials.

Key Terms and Formulas

• Factorial ($n!$): The product of all positive integers up to $n$.

• Factorial Division: $\frac{n!}{(n-k)!} = n \times (n-1) \times \ldots \times (n-k+1)$

Step-by-Step Guidance

1. Write out the definition of $322!$ and $320!$.

2. Notice that $322! = 322 \times 321 \times 320!$.

3. Substitute this into the expression and cancel $320!$ from numerator and denominator.

Try solving on your own before revealing the answer!

Final Answer: 103,362

$\frac{322!}{320!} = 322 \times 321 = 103,362$

Q2b. Evaluate the expression: $\binom{21}{7}$

Background

Topic: Combinations

This question tests your ability to compute the number of ways to choose $k$ items from $n$ without regard to order.

Key Terms and Formulas

• Combination Formula: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Step-by-Step Guidance

1. Identify $n = 21$ and $k = 7$.

2. Plug these values into the combination formula.

3. Set up the factorials: $\frac{21!}{7! \times 14!}$

Try solving on your own before revealing the answer!

Final Answer: 116,280

$\binom{21}{7} = 116,280$

Q3. A bell tower has 12 bells, each with a different note. 6 bells are rung in sequence, no bell more than once. How many unique sequences are there?

Background

Topic: Permutations

This question tests your understanding of how to count the number of ways to arrange a subset of items in order.

Key Terms and Formulas

• Permutation Formula: $P(n, k) = \frac{n!}{(n-k)!}$

Step-by-Step Guidance

1. Identify $n = 12$ (bells) and $k = 6$ (bells to be rung).

2. Plug these values into the permutation formula.

3. Set up the expression: $\frac{12!}{(12-6)!} = \frac{12!}{6!}$

Try solving on your own before revealing the answer!

Final Answer: 665,280

There are $\frac{12!}{6!} = 665,280$ unique sequences.

Q4. To win at LOTTO, select 5 numbers from 50 (order does not matter). How many different selections are possible?

Background

Topic: Combinations

This question tests your ability to count the number of ways to choose a subset of items when order does not matter.

Key Terms and Formulas

• Combination Formula: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Step-by-Step Guidance

1. Identify $n = 50$ and $k = 5$.

2. Plug these values into the combination formula.

3. Set up the expression: $\frac{50!}{5! \times 45!}$

Try solving on your own before revealing the answer!

Final Answer: 2,118,760

There are $\binom{50}{5} = 2,118,760$ possible selections.

Q5. Suppose a fair, six-sided die is rolled. Find the probability of:

a) Rolling a prime number.

Background

Topic: Basic Probability

This question tests your ability to count favorable outcomes and calculate probability.

Key Terms and Formulas

• Probability: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Step-by-Step Guidance

1. List the numbers on a die: 1, 2, 3, 4, 5, 6.

2. Identify which are prime: 2, 3, 5.

3. Count the number of favorable outcomes (prime numbers).

4. Set up the probability formula with the number of favorable and total outcomes.

Try solving on your own before revealing the answer!

Final Answer: $\frac{1}{2}$

There are 3 prime numbers out of 6, so $P = \frac{3}{6} = \frac{1}{2}$.

b) Rolling a number more than 4.

Background

Topic: Basic Probability

This question tests your ability to count outcomes greater than a given value.

Key Terms and Formulas

• Probability formula as above.

Step-by-Step Guidance

1. List numbers greater than 4: 5, 6.

2. Count the number of favorable outcomes.

3. Set up the probability formula.

Try solving on your own before revealing the answer!

Final Answer: $\frac{1}{3}$

There are 2 numbers greater than 4 out of 6, so $P = \frac{2}{6} = \frac{1}{3}$.

Q6. In roulette, betting $10 on red (18 red, 18 black, 2 neither). If you win, you keep your $10 and win $10. If not, you lose your $10. What is the expected value for playing?

Background

Topic: Expected Value in Probability

This question tests your ability to calculate expected value based on possible outcomes and their probabilities.

Key Terms and Formulas

• Expected Value: $E = \sum (\text{Value of outcome}) \times (\text{Probability of outcome})$

Step-by-Step Guidance

1. Calculate the probability of winning (landing on red): $\frac{18}{38}$.

2. Calculate the probability of losing (landing on black or neither): $\frac{20}{38}$.

3. Determine the net gain/loss for each outcome: win = +$10, lose = -$10.

4. Set up the expected value formula using these probabilities and outcomes.

Try solving on your own before revealing the answer!

Final Answer: -$0.53

The expected value is negative, meaning you lose about 53 cents per $10 bet on average.

Q7. A box contains 40 widgets, 5 defective. If 5 are sold at random, find the probability that:

a) All are defective.

Background

Topic: Probability with Combinations (Hypergeometric Probability)

This question tests your ability to calculate probabilities when sampling without replacement.

Key Terms and Formulas

• Hypergeometric Probability: $P = \frac{\binom{D}{k} \binom{N-D}{n-k}}{\binom{N}{n}}$

Step-by-Step Guidance

1. Identify $N = 40$ (total), $D = 5$ (defective), $n = 5$ (chosen), $k = 5$ (all defective).

2. Plug values into the formula: $\frac{\binom{5}{5} \binom{35}{0}}{\binom{40}{5}}$

Try solving on your own before revealing the answer!

Final Answer: 1 in 658,008

The probability is $\frac{1}{658,008}$, a very rare event.

b) None are defective.

Background

Same as above, but $k = 0$.

Step-by-Step Guidance

1. Plug values into the formula: $\frac{\binom{5}{0} \binom{35}{5}}{\binom{40}{5}}$

Try solving on your own before revealing the answer!

Final Answer: 0.237

The probability is about 23.7% that none are defective.

Q8. Use the table of movers in Country X to find probabilities:

a) Probability a randomly selected mover was an owner.

Background

Topic: Probability from Contingency Tables

This question tests your ability to use data tables to find probabilities.

Key Terms and Formulas

• Probability: $P = \frac{\text{Number in category}}{\text{Total}}$

Step-by-Step Guidance

1. Find the total number of owners who moved (sum across all columns for owners).

2. Find the total number of people who moved (sum all entries).

3. Set up the probability formula.

Try solving on your own before revealing the answer!

Final Answer: 0.386

Probability is $\frac{16.2}{42.0} \approx 0.386$.

b) Probability a mover was a renter or moved to a different continent.

Background

Topic: Probability with "or" (Union of Events)

Key Terms and Formulas

• Union Formula: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$

Step-by-Step Guidance

1. Find the total number of renters and the total who moved to a different continent.

2. Find the overlap (renters who moved to a different continent).

3. Apply the union formula and divide by the total.

Try solving on your own before revealing the answer!

Final Answer: 0.645

Probability is $\frac{25.8 + 1.1 - 0.9}{42.0} = 0.645$.

c) Probability a mover was an owner given they moved to a different country on the same continent.

Background

Topic: Conditional Probability

Key Terms and Formulas

• Conditional Probability: $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$

Step-by-Step Guidance

1. Find the number of owners who moved to a different country on the same continent.

2. Find the total number who moved to a different country on the same continent.

3. Set up the conditional probability formula.

Try solving on your own before revealing the answer!

Final Answer: 0.383

Probability is $\frac{3.6}{9.4} \approx 0.383$.

Q9. Suppose you draw one card from a 52-card deck. Find the probability that:

a) It is not a face card.

Background

Topic: Probability with Playing Cards

This question tests your knowledge of the composition of a standard deck and basic probability.

Key Terms and Formulas

• Face cards: Jack, Queen, King (3 per suit, 12 total).

• Probability formula as above.

Step-by-Step Guidance

1. Calculate the number of non-face cards: 52 total - 12 face cards.

2. Set up the probability formula.

Try solving on your own before revealing the answer!

Final Answer: $\frac{40}{52} = \frac{10}{13}$

There are 40 non-face cards in a deck of 52.

b) It is either a club or a black card.

Background

Topic: Probability with Overlapping Events

Key Terms and Formulas

• Clubs: 13 cards, all black.

• Black cards: 26 cards (clubs + spades).

• Union formula as above.

Step-by-Step Guidance

1. Count the number of clubs (13) and black cards (26).

2. Notice that all clubs are black, so the overlap is 13.

3. Apply the union formula: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$.

Try solving on your own before revealing the answer!

Final Answer: $\frac{26}{52} = \frac{1}{2}$

All clubs are black, so the probability is $\frac{26}{52}$.

Q10. The winner of a raffle will receive a new car. If 25,000 tickets were sold and you purchased 20, what are the odds against your winning?

Background

Topic: Odds in Probability

This question tests your understanding of how to calculate odds against an event.

Key Terms and Formulas

• Odds against: $\frac{\text{Number of ways to lose}}{\text{Number of ways to win}}$

Step-by-Step Guidance

1. Calculate the number of ways to win (tickets you have).

2. Calculate the number of ways to lose (total tickets - your tickets).

3. Set up the odds against formula.

Try solving on your own before revealing the answer!

Final Answer: 1,249 to 1

Odds against winning are $\frac{24,980}{20} = 1,249$ to 1.

Q11. There are 38 chocolates: 12 with nuts, 12 with caramel, 14 solid. You select one, eat it, then select a second. Find the probability both are solid chocolate.

Background

Topic: Probability without Replacement

This question tests your ability to calculate probabilities when the sample size decreases after each selection.

Key Terms and Formulas

• Probability without replacement: $P = \frac{\text{favorable outcomes for first}}{\text{total}} \times \frac{\text{favorable for second}}{\text{remaining total}}$

Step-by-Step Guidance

1. Find the probability the first chocolate is solid: $\frac{14}{38}$.

2. After eating one, there are 13 solid left and 37 total.

3. Find the probability the second is solid: $\frac{13}{37}$.

4. Multiply the two probabilities together.

Try solving on your own before revealing the answer!

Final Answer: 0.132

The probability is $\frac{14}{38} \times \frac{13}{37} \approx 0.132$.

Q12. An unfair coin is flipped repeatedly. The probability of heads is 0.48.

a) Probability of heads two flips in a row.

Background

Topic: Probability of Independent Events

This question tests your ability to multiply probabilities for independent events.

Key Terms and Formulas

• Independent Events: $P(A \text{ and } B) = P(A) \times P(B)$

Step-by-Step Guidance

1. Probability of heads on one flip is 0.48.

2. Multiply the probability for two flips: $0.48 \times 0.48$.

Try solving on your own before revealing the answer!

Final Answer: 0.2304

Probability is $0.48 \times 0.48 = 0.2304$.

b) Probability of tails four flips in a row.

Background

Probability of tails is $1 - 0.48 = 0.52$.

Step-by-Step Guidance

1. Probability of tails on one flip is 0.52.

2. Multiply for four flips: $0.52^4$.

Try solving on your own before revealing the answer!

Final Answer: 0.073

Probability is $0.52^4 = 0.073$.

c) Probability of a tail at least once in ten flips.

Background

Topic: Complement Rule

Key Terms and Formulas

• Complement Rule: $P(\text{at least one tail}) = 1 - P(\text{no tails})$

Step-by-Step Guidance

1. Probability of heads on one flip is 0.48.

2. Probability of all heads in 10 flips: $0.48^{10}$.

3. Subtract this from 1 to get the probability of at least one tail.

Try solving on your own before revealing the answer!

Final Answer: 0.999

Probability is $1 - 0.48^{10} \approx 0.999$.

Q13. The data in the table are based on 145 Americans tested for tuberculosis (TB). Use the data to find the following probabilities:

a) Probability a randomly selected person has TB.

Background

Topic: Probability from Contingency Tables

Key Terms and Formulas

• Probability: $P = \frac{\text{Number with TB}}{\text{Total tested}}$

Step-by-Step Guidance

1. Find the total number of people with TB (sum across the TB row).

2. Find the total number of people tested.

3. Set up the probability formula.

Try solving on your own before revealing the answer!

Final Answer: 0.069

Probability is $\frac{10}{145} \approx 0.069$.

b) Suppose a person has a positive test screening. Find the probability they have TB.

Background

Topic: Conditional Probability

Key Terms and Formulas

• Conditional Probability: $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$

Step-by-Step Guidance

1. Find the number of people with TB and a positive test.

2. Find the total number of people with a positive test.

3. Set up the conditional probability formula.

Try solving on your own before revealing the answer!

Final Answer: 0.5

Probability is $\frac{5}{10} = 0.5$.