Exam 3 Review: Probability, Statistics, and Data Analysis Study Guide

Q1. What is the probability of spinning an even number on the spinner?

Background

Topic: Basic Probability

This question tests your understanding of probability using a spinner with numbered sections. You need to find the probability of landing on an even number.

Key Terms and Formulas:

• Probability ($P$): The likelihood of an event occurring, calculated as $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

• Even numbers: Numbers divisible by 2 (in this spinner, 2 and 4)

Step-by-Step Guidance

1. Count the total number of sections on the spinner.

2. Identify which sections are labeled with even numbers.

3. Count the number of sections with even numbers.

4. Set up the probability formula: $P(\text{even}) = \frac{\text{Number of even sections}}{\text{Total sections}}$

Try solving on your own before revealing the answer!

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

There are 4 sections labeled with even numbers (2 and 4), out of 8 total sections. So $P(\text{even}) = \frac{4}{8} = \frac{1}{2}$.

Q2. What is the probability of rolling a 5, an even number, or a number greater than 3 on a die?

Background

Topic: Probability with Dice

This question tests your ability to calculate probabilities for different outcomes when rolling a standard six-sided die.

Key Terms and Formulas:

• Standard die: Has numbers 1 through 6.

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

Step-by-Step Guidance

1. For each part, identify the favorable outcomes:

2. a) Rolling a 5: Only one outcome (5).

3. b) Rolling an even number: Outcomes are 2, 4, 6.

4. c) Rolling a number greater than 3: Outcomes are 4, 5, 6.

5. Set up the probability formula for each part.

Try solving on your own before revealing the answer!

Final Answers:

• a) $P(5) = \frac{1}{6}$

• b) $P(\text{even}) = \frac{3}{6} = \frac{1}{2}$

• c) $P(\text{greater than 3}) = \frac{3}{6} = \frac{1}{2}$

Each probability is calculated by counting the favorable outcomes and dividing by the total number of possible outcomes (6).

Q3. What is the probability of drawing a gold marble or not a black marble from a bag?

Background

Topic: Probability with Marbles

This question tests your ability to calculate probabilities for drawing specific colored marbles from a bag.

Key Terms and Formulas:

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

Step-by-Step Guidance

1. Count the total number of marbles in the bag.

2. For part a, count the number of gold marbles.

3. For part b, count the number of marbles that are not black (gold + silver).

4. Set up the probability formula for each part.

Try solving on your own before revealing the answer!

Final Answers:

• a) $P(\text{gold}) = \frac{3}{36}$

• b) $P(\text{not black}) = \frac{13}{36}$

There are 3 gold marbles and 13 marbles that are not black out of 36 total marbles.

Q4. What is the probability of spinning green and then yellow on the spinner?

Background

Topic: Compound Probability (Independent Events)

This question tests your ability to calculate the probability of two independent events occurring in sequence using a spinner.

Key Terms and Formulas:

• Compound probability: $P(A \text{ and } B) = P(A) \times P(B)$ for independent events.

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

Step-by-Step Guidance

1. Count the total number of sections on the spinner.

2. Count the number of green sections and yellow sections.

3. Calculate the probability of landing on green in the first spin.

4. Calculate the probability of landing on yellow in the second spin.

5. Set up the compound probability formula: $P(\text{green then yellow}) = P(\text{green}) \times P(\text{yellow})$

Try solving on your own before revealing the answer!

Final Answer: $\frac{1}{9}$

There are 2 green sections and 1 yellow section out of 6 total sections. $P(\text{green}) = \frac{2}{6}$, $P(\text{yellow}) = \frac{1}{6}$, so $P(\text{green then yellow}) = \frac{2}{6} \times \frac{1}{6} = \frac{1}{9}$.