Statistics Chapter 4 Activity: Step-by-Step Study Guidance

Q2. You roll two 6-sided dice. Make a table/visual to write the sample space for the sum of the two dice. You should have 36 outcomes. Find the probability of getting the following:

• (a) A sum of 5 or 6

• (b) A sum more than 6

• (c) A sum that is divisible by 4

Background

Topic: Probability with Discrete Sample Spaces

This question tests your understanding of sample spaces, counting outcomes, and calculating probabilities for events involving rolling two dice.

Key Terms and Formulas

• Sample Space: The set of all possible outcomes. For two dice, there are $6 \times 6 = 36$ possible outcomes.

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

Step-by-Step Guidance

1. List all possible sums for two dice (from 2 to 12) and count how many ways each sum can occur. You can do this by making a table or listing pairs.

2. For part (a), identify all pairs that sum to 5 or 6. Count the number of outcomes for each sum and add them together.

3. For part (b), identify all sums greater than 6 (i.e., 7, 8, 9, 10, 11, 12). Count the number of outcomes for each of these sums and add them up.

4. For part (c), determine which sums are divisible by 4 (i.e., 4, 8, 12). Count the number of outcomes for each of these sums.

5. For each part, set up the probability as $P = \frac{\text{number of favorable outcomes}}{36}$.

Try solving on your own before revealing the answer!

Q3. In a survey, 18 people preferred milk, 29 people preferred coffee, and 13 people preferred juice as their primary drink for breakfast. If a person is selected at random, find the probability that the person preferred juice or milk as their primary drink.

Background

Topic: Basic Probability (Addition Rule for Mutually Exclusive Events)

This question tests your ability to calculate the probability of the union of two mutually exclusive events.

Key Terms and Formulas

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

• For mutually exclusive events A and B: $P(A \text{ or } B) = P(A) + P(B)$

Step-by-Step Guidance

1. Find the total number of people surveyed by adding the number of people who preferred each drink.

2. Identify the number of people who preferred juice and the number who preferred milk.

3. Since a person cannot prefer both as their primary drink, these events are mutually exclusive.

4. Add the number of people who preferred juice and milk to get the total number of favorable outcomes.

5. Set up the probability as $P = \frac{\text{number who preferred juice or milk}}{\text{total surveyed}}$.

Try solving on your own before revealing the answer!

Q4. A coin is tossed and a card is drawn from a standard 52-card deck. Find the probability of:

• (a) landing on heads then getting a 6

• (b) landing on tails then getting a red card

• (c) landing on heads then getting a club

Background

Topic: Probability of Compound (Independent) Events

This question tests your understanding of calculating probabilities for independent events (coin toss and card draw).

Key Terms and Formulas

• Probability of independent events: $P(A \text{ and } B) = P(A) \times P(B)$

• Probability of heads: $P(\text{heads}) = \frac{1}{2}$

• Probability of drawing a 6: $P(6) = \frac{4}{52}$ (since there are 4 sixes in a deck)

• Probability of drawing a red card: $P(\text{red}) = \frac{26}{52}$

• Probability of drawing a club: $P(\text{club}) = \frac{13}{52}$

Step-by-Step Guidance

1. For each part, identify the probability of the coin outcome and the probability of the card outcome.

2. Multiply the probabilities for the coin and the card, since the events are independent.

3. Set up the probability for each scenario as $P = P(\text{coin outcome}) \times P(\text{card outcome})$.

Try solving on your own before revealing the answer!

Q5. If 89% of households have at least one television set and 4 households are selected, find the probability that at least one household has a television set.

Background

Topic: Probability (Complement Rule, Binomial Probability)

This question tests your understanding of the complement rule and independent events.

Key Terms and Formulas

• Probability that a household has a TV: $P(\text{TV}) = 0.89$

• Probability that a household does not have a TV: $P(\text{no TV}) = 1 - 0.89 = 0.11$

• Probability that none of the 4 households have a TV: $P(\text{no TV in all 4}) = (0.11)^4$

• Probability that at least one has a TV: $P(\text{at least one}) = 1 - P(\text{none})$

Step-by-Step Guidance

1. Calculate the probability that a single household does not have a TV.

2. Since the households are selected independently, raise this probability to the 4th power to find the probability that none have a TV.

3. Use the complement rule: $P(\text{at least one}) = 1 - P(\text{none})$.

Try solving on your own before revealing the answer!

Q6. A candy store allows customers to select 3 different candies to be packaged and mailed. If there are 13 varieties available, how many possible selections can be made?

Background

Topic: Combinations (Counting Principle)

This question tests your understanding of combinations, where order does not matter and no repeats are allowed.

Key Terms and Formulas

• Combination formula: $C(n, r) = \frac{n!}{r!(n - r)!}$

• Here, $n = 13$ (varieties), $r = 3$ (candies selected)

Step-by-Step Guidance

1. Identify the total number of varieties ($n = 13$) and the number to be selected ($r = 3$).

2. Plug these values into the combination formula: $C(13, 3)$.

3. Set up the calculation: $C(13, 3) = \frac{13!}{3! \times 10!}$.

Try solving on your own before revealing the answer!

Q7. Daichi is making a quilt for their grandma. First, they need to select a color, then decide on a size, followed by the type of fabric, and lastly, which flower design they’ll include. How many quilt designs are possible? Answer as a whole number.

Background

Topic: Fundamental Counting Principle (Multiplication Rule)

This question tests your ability to use the multiplication rule to count the number of possible outcomes when making several independent choices.

Key Terms and Formulas

• Multiplication Rule: If there are $n_1$ ways to make the first choice, $n_2$ ways to make the second, etc., then the total number of outcomes is $n_1 \times n_2 \times \ldots$

• Number of options: Color (4), Size (2), Fabric (2), Flower (5)

Step-by-Step Guidance

1. List the number of options for each trait: Color (4), Size (2), Fabric (2), Flower (5).

2. Multiply the number of options for each trait together: $4 \times 2 \times 2 \times 5$.

Try solving on your own before revealing the answer!

Q8. There are ten people working in a particular department within a company. That department needs to create a three-person task force for a certain project. Answer as a whole number.

• (a) How many ways is there to select the task force if order does NOT matter?

• (b) How many ways is there to select the task force if each person will be given a different task and thus order DOES matter?

Background

Topic: Combinations and Permutations

This question tests your understanding of the difference between combinations (order doesn't matter) and permutations (order matters).

Key Terms and Formulas

• Combinations: $C(n, r) = \frac{n!}{r!(n - r)!}$

• Permutations: $P(n, r) = \frac{n!}{(n - r)!}$

• Here, $n = 10$, $r = 3$

Step-by-Step Guidance

1. For (a), use the combination formula: $C(10, 3)$.

2. Set up the calculation: $C(10, 3) = \frac{10!}{3! \times 7!}$.

3. For (b), use the permutation formula: $P(10, 3)$.

4. Set up the calculation: $P(10, 3) = \frac{10!}{7!}$.

Try solving on your own before revealing the answer!

Q9. How many ways is there for 9 runners to finish in 1st, 2nd, 3rd, and 4th place?

Background

Topic: Permutations (Order Matters)

This question tests your understanding of permutations, where order is important.

Key Terms and Formulas

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

• Here, $n = 9$, $r = 4$

Step-by-Step Guidance

1. Identify the total number of runners ($n = 9$) and the number of places ($r = 4$).

2. Plug these values into the permutation formula: $P(9, 4)$.

3. Set up the calculation: $P(9, 4) = \frac{9!}{5!}$.

Try solving on your own before revealing the answer!

Q10. At the beach there are 9 yellow towels, 5 red towels, 2 blue towels, 6 green towels, and 12 white towels. Find the following probabilities rounded to four decimal places or as a percent with 2-digits past the decimal. Make sure to show the setup.

• (a) You randomly select a green towel?

• (b) You randomly select a towel that is not red?

• (c) You randomly select a red towel, replace it, then select a green towel.

• (d) You randomly select 3 green towels without replacement.

• (e) You randomly select a blue towel, then a blue towel, and then a white towel without replacement.

Background

Topic: Probability (Simple, Compound, With and Without Replacement)

This question tests your ability to calculate probabilities for single and multiple events, with and without replacement.

Key Terms and Formulas

• Total towels: $9 + 5 + 2 + 6 + 12 = 34$

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

• For compound events with replacement: Multiply probabilities for each draw (probabilities stay the same).

• For compound events without replacement: Adjust the numerator and denominator after each draw.

Step-by-Step Guidance

1. For each part, identify the number of favorable towels and the total number of towels.

2. For (a), set up $P = \frac{6}{34}$ for green towels.

3. For (b), set up $P = \frac{\text{number of towels not red}}{34}$.

4. For (c), multiply the probability of selecting a red towel (first draw) by the probability of selecting a green towel (second draw, after replacement).

5. For (d), set up the probability for selecting 3 green towels in a row without replacement, adjusting the numerator and denominator each time.

6. For (e), set up the probability for selecting a blue, then a blue, then a white towel without replacement, adjusting the counts each time.

Try solving on your own before revealing the answer!

Q11. Suppose one card is drawn at random from a standard 52-card deck. What is the probability that the card drawn is a black card or a face card?

Background

Topic: Probability (Addition Rule, Overlapping Events)

This question tests your understanding of the addition rule for probabilities, especially when events are not mutually exclusive.

Key Terms and Formulas

• Number of black cards: 26

• Number of face cards: 12 (Jack, Queen, King in each suit)

• Number of black face cards: 6 (Jack, Queen, King in spades and clubs)

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

Step-by-Step Guidance

1. Find the probability of drawing a black card: $P(\text{black}) = \frac{26}{52}$.

2. Find the probability of drawing a face card: $P(\text{face}) = \frac{12}{52}$.

3. Find the probability of drawing a black face card: $P(\text{black face}) = \frac{6}{52}$.

4. Apply the addition rule: $P(\text{black or face}) = P(\text{black}) + P(\text{face}) - P(\text{black face})$.

Try solving on your own before revealing the answer!

Q12. Dallace won 5 extra tickets to Disneyland. He’s debating who to take since many of us are Disney Adults. Suppose he has 25 people to choose from. How many ways can Dallace select the 5 people if order DOES NOT matter?

Background

Topic: Combinations (Order Does Not Matter)

This question tests your understanding of combinations, where order does not matter and no repeats are allowed.

Key Terms and Formulas

• Combination formula: $C(n, r) = \frac{n!}{r!(n - r)!}$

• Here, $n = 25$, $r = 5$

Step-by-Step Guidance

1. Identify the total number of people ($n = 25$) and the number to be selected ($r = 5$).

2. Plug these values into the combination formula: $C(25, 5)$.

3. Set up the calculation: $C(25, 5) = \frac{25!}{5! \times 20!}$.

Try solving on your own before revealing the answer!

Q13. The following table are the results of 75 transfer students. It shows which school they’re transferring to and how they’re paying for tuition. You can leave results as fractions, decimals with places, or percents to 2 decimal places.

	UCLA	USC	CSULB	Total
Grants	7	5	20	32
Student Loans	12	6	11	29
Mixed Sources	4	1	9	14
Total	23	12	40	75

• (a) Find the probability of a student paying their tuition with Mixed Sources.

• (b) Find the probability of a student transferring to CSULB or paying for tuition with grants.

• (c) Find the probability of a student transferring to UCLA or USC.

• (d) Find the probability of a student transferring to USC given that they’re paying for their tuition with student loans.

• (e) Find the probability of a student paying for tuition with Mixed Sources given that they’re transferring to UCLA.

Background

Topic: Probability with Contingency Tables (Joint, Marginal, Conditional Probability)

This question tests your ability to read and interpret a contingency table and calculate various types of probabilities.

Key Terms and Formulas

• Marginal probability: Probability of a single event (e.g., Mixed Sources)

• Joint probability: Probability of two events both occurring (e.g., CSULB and Grants)

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

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

Step-by-Step Guidance

1. For (a), find the total number of students paying with Mixed Sources and divide by the total number of students.

2. For (b), find the number transferring to CSULB, the number paying with grants, and the overlap (students who are both). Use the addition rule.

3. For (c), add the number transferring to UCLA and USC, then divide by the total.

4. For (d), use the conditional probability formula: $P(\text{USC} | \text{Student Loans}) = \frac{\text{Number with Student Loans at USC}}{\text{Total with Student Loans}}$.

5. For (e), use the conditional probability formula: $P(\text{Mixed Sources} | \text{UCLA}) = \frac{\text{Number at UCLA with Mixed Sources}}{\text{Total at UCLA}}$.

Try solving on your own before revealing the answer!

Q14. What are the formulas/concepts for “or” and “replacement” scenarios? When do we use each? What do we have to watch out for in regards to the types of events we have?

Background

Topic: Probability Rules (Addition Rule, Replacement vs. Non-Replacement)

This question tests your understanding of the addition rule for probabilities and the difference between sampling with and without replacement.

Key Terms and Formulas

• Addition Rule for mutually exclusive events: $P(A \text{ or } B) = P(A) + P(B)$

• Addition Rule for non-mutually exclusive events: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$

• With replacement: Probabilities remain the same for each draw.

• Without replacement: Probabilities change after each draw (denominator decreases).

Step-by-Step Guidance

1. Recall the addition rule and when to use the basic vs. the general form (mutually exclusive vs. not).

2. Understand the difference between sampling with replacement (independent events) and without replacement (dependent events).

3. Be careful to check if events overlap (not mutually exclusive) and adjust the formula accordingly.

4. For replacement scenarios, probabilities for each draw stay the same; for non-replacement, adjust the numerator and denominator after each draw.

Try summarizing these concepts in your own words before checking the answer!