Compound Events Mastery Worksheet
Lesson 8.5 • SP8 Probability & Statistics
Compound Events Mastery
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Core Concept: Independent vs. Dependent Compound Events
With Replacement (Independent): The item is returned. The total outcomes and favorable counts stay the same for the second pick.
Without Replacement (Dependent): The item is NOT returned. Total outcomes decrease by 1, and the count for that item's color also drops if picked.
FOCUS LESSON - PART 1 Without Replacement
Bag A: 5 Purple, 1 Orange, 1 Brown (7 Total)
Two marbles in a row are picked at random. Without putting the first marble back, a second marble is picked at random. Find the probability that the first marble picked is orange and the second marble picked is purple.
Generate a multiplication expression to determine the probability. Simplify to a percentage.
BAG A CONTENTS
P
P
P
P
P
O
B
Total = 7 Marbles
Show your work below (Multiplication Expression → Simplified Fraction → Percentage): P(Orange, then Purple)
FOCUS LESSON - PART 2 With Replacement
Bag A: 5 Purple, 1 Orange, 1 Brown (7 Total)
Two marbles in a row are picked at random from a bag. The first marble is replaced back into the bag of marbles, then a second marble is picked at random. Find the probability that the first marble picked is brown and the second marble picked is purple.
Generate a multiplication expression to determine the probability. Simplify to a percentage.
BAG A CONTENTS
P
P
P
P
P
O
B
Total = 7 Marbles
Show your work below (Multiplication Expression → Simplified Fraction → Percentage): P(Brown, then Purple)
Lesson 8.5 • Compound Probability Page 1 of 5
Lesson 8.5 • SP8
Sample Spaces & Tree Diagrams
Total Outcomes: \(6 \times 6 = 36\) Possible Pairs
TREE DIAGRAM: TWO ROLLS OF A 6-SIDED NUMBER CUBE Roll 1 (Top Nodes 1–6) → Roll 2 (Leaves 1–6)
FIRST ROLL (OUTCOMES 1 TO 6) 1 1 2 3 4 5 6 2 1 2 3 4 5 6 3 1 2 3 4 5 6 4 1 2 3 4 5 6 5 1 2 3 4 5 6 6 1 2 3 4 5 6 Second Roll Outcomes (1 through 6 for each First Roll)
FOCUS LESSON - PART 3a Event: Sum of Two Rolls
Scenario: A teacher has a number cube with faces numbered 1 through 6. He will roll the number cube two times. Each time the teacher rolls the number cube, he will record the number shown on the top face. Then he will find the sum of the two numbers he records.
What is the probability that the sum of the numbers he records will be 8? Write your answer as a fraction in simplest form.
List all favorable pairs \((r_1, r_2)\) that add to 8:
Probability Calculation:
Favorable: _____
Total: 36
P(Sum = 8) = ______
FOCUS LESSON - PART 3b Event: Product of Two Rolls
Scenario: Using the same tree diagram of 36 outcomes, the teacher rolls the number cube two times and records the two numbers. This time, he calculates the product (multiplication) of the two numbers.
What is the probability that the product of the numbers he records will be greater than 10? Write your answer as a simplified fraction and percentage.
Show organized work (list or count favorable products > 10):
Final Probability:
Favorable: _____ / 36
Fraction: _______
Percent: ≈ ______%
Lesson 8.5 • Tree Diagrams & Sample Spaces Page 2 of 5
Guided Practice • SP8
Tree Diagram Mastery & Analysis
Total Outcomes: \(6 \times 6 = 36\) Possible Pairs
SAMPLE SPACE TREE DIAGRAM: TWO ROLLS OF A NUMBER CUBE Use the branches below to count outcomes
FIRST ROLL (OUTCOMES 1 TO 6) 1 1 2 3 4 5 6 2 1 2 3 4 5 6 3 1 2 3 4 5 6 4 1 2 3 4 5 6 5 1 2 3 4 5 6 6 1 2 3 4 5 6 Second Roll Outcomes (1 through 6 for each First Roll)
PRACTICE - PART 4a Event: Sum is a Multiple of 4
Scenario: A teacher rolls a 6-sided number cube two times and records the top face for each roll. He calculates the sum of the two numbers.
What is the probability that the sum of the numbers he records will be a multiple of 4 ? Write your answer as a simplified fraction and percentage.
List all favorable pairs \((r_1, r_2)\) for Sum = 4, 8, or 12:
Probability Calculation:
Favorable: _____ / 36
Fraction: _______
Percent: ______%
PRACTICE - PART 4b Event: Product is an Odd Number
Scenario: Using the tree diagram, the teacher rolls the number cube two times. This time, he calculates the product (multiplication) of the two numbers.
What is the probability that the product of the numbers will be an odd number? Write your answer as a simplified fraction and percentage.
Show organized work (list odd pairs from {1, 3, 5}):
Final Probability:
Favorable: _____ / 36
Fraction: _______
Percent: ______%
Lesson 8.5 • Tree Diagrams & Sample Spaces Page 3 of 5
Guided Practice
Marble Bag Compound Events
Carefully note: replacement vs. non-replacement
Bag A Reference 7 Total
5 Purple, 1 Orange/Yellow, 1 Brown
Bag A Marbles
P
P
P
P
P
O
B
Bag B Reference 6 Total
3 Green, 2 Red, 1 Brown
Bag B Marbles
G
G
G
R
R
B
GP1 WITHOUT REPLACEMENT
From Bag A, two marbles in a row are picked at random. Without putting the first marble back, a second marble is picked.
Find \(P(\text{Purple, then Yellow/Orange})\).
Write multiplication expression & percentage:
Expr: ___________ %: _______%
GP2 WITH REPLACEMENT
From Bag B, two marbles in a row are picked at random. The first marble is replaced back, then a second marble is picked.
Find \(P(\text{Red, then Green})\).
Write multiplication expression & percentage:
Expr: ___________ %: _______%
GP3 WITH REPLACEMENT
From Bag A, two marbles in a row are picked. The first marble is replaced back into the bag, then a second marble is picked at random.
Find \(P(\text{Purple, then Purple})\).
Write multiplication expression & percentage:
Expr: ___________ %: _______%
GP4 WITHOUT REPLACEMENT
From Bag B, two marbles in a row are picked. Without replacement, a second marble is picked at random.
Find \(P(\text{Brown, then Green})\).
Write multiplication expression & percentage:
Expr: ___________ %: _______%
Key Checkpoint: When picking without replacement, remember that the second fraction's denominator must be reduced by 1 because total marbles went from \(n\) to \(n-1\)!
Lesson 8.5 • Marble Compound Probability Page 4 of 5
Guided Practice
Multi-Stage Independent Events
Coins, Dice & Number Cards
Multiplication Rule for Independent Events: \(P(A \text{ and } B) = P(A) \times P(B)\) | Events do NOT affect one another
GP6 Coin & Standard 6-Sided Die
A coin is flipped and then a standard 6-sided die is rolled. What is the probability of landing on tails and then rolling an even number?
T
⚁
Show work: \(P(\text{Tails}) \times P(\text{Even})\)
Answer P = ______
GP7 Complementary Event (Not a 3)
A coin is flipped and then a die is rolled. What is the probability of landing on tails and then rolling a number that is not a 3?
T
≠3
Show work: \(P(\text{Tails}) \times P(\text{Not 3})\)
Answer P = ______
GP8 Number Cards & Coin Flip
A card is picked from the set shown below and then a coin is flipped. What is the probability of picking a card that is not even (odd) and then landing on heads?
3 4 5 6 7 8
H
Show work: Identify odd cards in {3,4,5,6,7,8} → Multiply by \(P(\text{Heads})\)
Answer P = ______
Self-Check & Reflection:
Explain in one sentence why flipping a coin and rolling a die is considered an independent compound event:
Lesson 8.5 • Independent Events Practice Page 5 of 5
Compound Events Answer Key
Teacher Solutions Lesson 8.5 • SP8
Compound Events Answer Key
Standard: CCSS.MATH.CONTENT.7.SP.C.8
Full Solutions & Teacher Notes
Key Conceptual Teaching Points: With vs. Without Replacement Bag A: 5 Purple, 1 Orange, 1 Brown (7 Total)
Without Replacement (Dependent): Denominator shrinks from 7 to 6. The second event depends on the outcome of the first draw.
With Replacement (Independent): Denominator remains 7. The marble is put back, preserving initial probabilities for both draws.
FOCUS LESSON - PART 1 Without Replacement
Target: \(P(\text{Orange, then Purple})\)
Step-by-Step Solution: Answer: \(\frac{5}{42} \approx 11.90\%\)
Step 1: First draw: There is 1 orange marble out of 7 total marbles → \(P(\text{Orange}) = \mathbf{\frac{1}{7}}\)
Step 2: Second draw (without replacement): Total marbles drop to \(7 - 1 = 6\). All 5 purple marbles remain → \(P(\text{Purple}) = \mathbf{\frac{5}{6}}\)
Step 3: Multiplication rule for dependent events:
\(P(\text{Orange, then Purple}) = \frac{1}{7} \times \frac{5}{6} = \mathbf{\frac{5}{42}} \approx \mathbf{0.119047...} \approx \mathbf{11.90\%}\)
Common Error: Students often keep the denominator as 7 for both draws (\(\frac{1}{7} \times \frac{5}{7} = \frac{5}{49}\)). Emphasize checking whether the marble was replaced.
FOCUS LESSON - PART 2 With Replacement
Target: \(P(\text{Brown, then Purple})\)
Step-by-Step Solution: Answer: \(\frac{5}{49} \approx 10.20\%\)
Step 1: First draw: There is 1 brown marble out of 7 total marbles → \(P(\text{Brown}) = \mathbf{\frac{1}{7}}\)
Step 2: Second draw (with replacement): The brown marble is replaced, so total remains 7 → \(P(\text{Purple}) = \mathbf{\frac{5}{7}}\)
Step 3: Multiplication rule for independent events:
\(P(\text{Brown, then Purple}) = \frac{1}{7} \times \frac{5}{7} = \mathbf{\frac{5}{49}} \approx \mathbf{0.10204...} \approx \mathbf{10.20\%}\)
Comparison Note: Notice how with replacement has a slightly smaller probability (\(10.20\%\)) than without replacement (\(11.90\%\)) because the denominator is larger (49 vs 42).
Lesson 8.5 • Compound Probability Solutions Page 1 of 5 (Answer Key)
Teacher Solutions Lesson 8.5 • SP8
Sample Spaces & Tree Diagrams (Solutions)
Total Sample Space: \(6 \times 6 = 36\) Equally Likely Outcomes
SAMPLE SPACE GRID SUMMARY \((r_1, r_2)\) Each cell represents 1 of 36 equally likely outcomes
(1,1)..(1,6)
(2,1)..(2,6)
(3,1)..(3,6)
(4,1)..(4,6)
(5,1)..(5,6)
(6,1)..(6,6)
FOCUS LESSON - PART 3a SOLUTION Event: Sum of Two Rolls = 8
Favorable Pairs & Calculation: P(Sum = 8) = \(\frac{5}{36}\)
Systematically list every pair \((r_1, r_2)\) where \(r_1 + r_2 = 8\) from faces \(\{1, 2, 3, 4, 5, 6\}\):
(2, 6)
(3, 5)
(4, 4)
(5, 3)
(6, 2)
Number of Favorable Outcomes: 5 pairs Total Possible Outcomes: 36 pairs
\(P(\text{Sum} = 8) = \mathbf{\frac{5}{36}}\) \(\approx 0.1389 \approx \mathbf{13.89\%}\)
Key Misconception: Students often count \((2, 6)\) and \((6, 2)\) as the same single outcome. Emphasize that rolling a 2 first and 6 second is a distinct branch on the tree diagram from rolling a 6 first and 2 second.
FOCUS LESSON - PART 3b SOLUTION Event: Product > 10
Systematic Count by First Roll: P(Product > 10) = \(\frac{17}{36} \approx 47.22\%\)
First Roll = 1: None (\(\max = 1 \times 6 = 6\)) → 0
First Roll = 2: (2, 6) = 12 → 1 pair
First Roll = 3: (3,4)=12, (3,5)=15, (3,6)=18 → 3 pairs
First Roll = 4: (4,3)=12, (4,4)=16, (4,5)=20, (4,6)=24 → 4 pairs
First Roll = 5: (5,3)=15, (5,4)=20, (5,5)=25, (5,6)=30 → 4 pairs
First Roll = 6: (6,2)=12, (6,3)=18, (6,4)=24, (6,5)=30, (6,6)=36 → 5 pairs
Total Favorable: \(0 + 1 + 3 + 4 + 4 + 5 = \mathbf{17}\) outcomes
Fraction: \(\mathbf{\frac{17}{36}}\) Percent: \(\approx \mathbf{47.22\%}\)
Watch Out: Notice that \((2, 5)\) and \((5, 2)\) equal exactly 10, but the condition states strictly greater than 10, so they are excluded!
Lesson 8.5 • Tree Diagrams & Sample Spaces Page 2 of 5 (Answer Key)
Teacher Solutions Guided Practice • SP8
Tree Diagram Mastery & Analysis (Solutions)
Sample Space: 36 Outcomes
Probability Formula: \(P(\text{Event}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Outcomes in Sample Space (36)}}\)
Simplify to lowest terms
PRACTICE - PART 4a SOLUTION Event: Sum is a Multiple of 4
Favorable Pairs Breakdown: Answer: \(\frac{1}{4} = 25\%\)
Possible sums from two dice range from 2 to 12. The multiples of 4 in this range are 4, 8, and 12:
Sum = 4 (3 pairs): (1,3), (2,2), (3,1)
Sum = 8 (5 pairs): (2,6), (3,5), (4,4), (5,3), (6,2)
Sum = 12 (1 pair): (6,6)
Total Favorable: \(3 + 5 + 1 = \mathbf{9}\) outcomes Calculation: \(\frac{9}{36} = \frac{1}{4}\)
\(\mathbf{\frac{1}{4}}\) \(\mathbf{25.0\%}\)
Check: 9 out of 36 outcomes exactly equals one-fourth of the sample space.
PRACTICE - PART 4b SOLUTION Event: Product is an Odd Number
Mathematical Logic & Outcomes: Answer: \(\frac{1}{4} = 25\%\)
Key Number Property: For a product to be odd, both numbers must be odd (\(\text{Odd} \times \text{Odd} = \text{Odd}\)). If either number is even, the product is even!
Odd faces available: \(\{1, 3, 5\}\) → \(3 \times 3 = 9\) pairs:
(1,1), (1,3), (1,5) (3,1), (3,3), (3,5) (5,1), (5,3), (5,5)
Method 1 (Counting): \(\frac{9 \text{ odd pairs}}{36 \text{ total}} = \mathbf{\frac{1}{4}}\) Method 2 (Multiplication Rule): \(P(\text{Odd}) \times P(\text{Odd}) = \frac{3}{6} \times \frac{3}{6} = \frac{1}{2} \times \frac{1}{2} = \mathbf{\frac{1}{4}}\)
\(\mathbf{\frac{1}{4}}\) \(\mathbf{25.0\%}\)
Teacher Note: Having students compare both methods provides a wonderful connection between sample space counting and the multiplication rule for independent events!
Lesson 8.5 • Tree Diagrams & Sample Spaces Page 3 of 5 (Answer Key)
Teacher Solutions Guided Practice
Marble Bag Compound Events (Solutions)
Bag A: 7 Marbles | Bag B: 6 Marbles
Bag A Summary (7 Total) 5 Purple \((5/7)\), 1 Yellow \((1/7)\), 1 Brown \((1/7)\)
Bag B Summary (6 Total) 3 Green \((3/6)\), 2 Red \((2/6)\), 1 Brown \((1/6)\)
GP1 SOLUTION WITHOUT REPLACEMENT
Bag A: \(P(\text{Purple, then Yellow/Orange})\)
Draw 1: \(P(\text{Purple})\) \(\frac{5}{7}\)
Draw 2: \(P(\text{Yellow} \mid \text{Purple})\) \(\frac{1}{6}\) (6 remaining)
Expression: \(\frac{5}{7} \times \frac{1}{6} = \mathbf{\frac{5}{42}}\)
Final Result:
\(\frac{5}{42}\) \(\approx 11.90\%\)
GP2 SOLUTION WITH REPLACEMENT
Bag B: \(P(\text{Red, then Green})\)
Draw 1: \(P(\text{Red})\) \(\frac{2}{6} = \frac{1}{3}\)
Draw 2: \(P(\text{Green})\) \(\frac{3}{6} = \frac{1}{2}\)
Expression: \(\frac{2}{6} \times \frac{3}{6} = \frac{6}{36} = \mathbf{\frac{1}{6}}\)
Final Result:
\(\frac{1}{6}\) \(\approx 16.67\%\)
GP3 SOLUTION WITH REPLACEMENT
Bag A: \(P(\text{Purple, then Purple})\)
Draw 1: \(P(\text{Purple})\) \(\frac{5}{7}\)
Draw 2: \(P(\text{Purple})\) \(\frac{5}{7}\) (Replaced)
Expression: \(\frac{5}{7} \times \frac{5}{7} = \mathbf{\frac{25}{49}}\)
Final Result:
\(\frac{25}{49}\) \(\approx 51.02\%\)
GP4 SOLUTION WITHOUT REPLACEMENT
Bag B: \(P(\text{Brown, then Green})\)
Draw 1: \(P(\text{Brown})\) \(\frac{1}{6}\)
Draw 2: \(P(\text{Green} \mid \text{Brown})\) \(\frac{3}{5}\) (5 remaining)
Expression: \(\frac{1}{6} \times \frac{3}{5} = \frac{3}{30} = \mathbf{\frac{1}{10}}\)
Final Result:
\(\frac{1}{10}\) \(10.0\%\)
Grading Quick Check: In GP4, confirm students simplified \(\frac{3}{30}\) to \(\frac{1}{10}\). In GP2, students may multiply unreduced fractions (\(\frac{2}{6} \times \frac{3}{6} = \frac{6}{36}\)) before simplifying to \(\frac{1}{6}\).
Lesson 8.5 • Marble Compound Probability Page 4 of 5 (Answer Key)
Teacher Solutions Guided Practice
Multi-Stage Independent Events (Solutions)
Coins, Dice & Number Cards
Independent Events Multiplication Law: \(P(A \text{ and } B) = P(A) \times P(B)\) | Prior outcome never alters next outcome
GP6 SOLUTION Coin & Standard 6-Sided Die: Tails & Even
\(P(\text{Tails}) = \frac{1}{2}\) | Even numbers \(\{2, 4, 6\} \rightarrow P(\text{Even}) = \frac{3}{6} = \frac{1}{2}\)
Answer \(\mathbf{\frac{1}{4}} = \mathbf{25\%}\)
Calculation: \(P(\text{Tails and Even}) = \frac{1}{2} \times \frac{1}{2} = \mathbf{\frac{1}{4}}\)
Equally likely pairs: 3 out of 12 = 1/4
GP7 SOLUTION Complementary Event: Tails & Not a 3
\(P(\text{Tails}) = \frac{1}{2}\) | Not a 3 \(\{1, 2, 4, 5, 6\} \rightarrow P(\text{Not 3}) = 1 - \frac{1}{6} = \frac{5}{6}\)
Answer \(\mathbf{\frac{5}{12}} \approx \mathbf{41.67\%}\)
Calculation: \(P(\text{Tails and Not 3}) = \frac{1}{2} \times \frac{5}{6} = \mathbf{\frac{5}{12}} \approx \mathbf{41.67\%}\)
Equally likely pairs: 5 out of 12 = 5/12
GP8 SOLUTION Number Cards & Coin Flip: Odd Card & Heads
Cards \(\{3, 4, 5, 6, 7, 8\}\) (6 cards) → Odd cards \(\{3, 5, 7\}\) (3 cards) → \(P(\text{Odd}) = \frac{3}{6} = \frac{1}{2}\)
Answer \(\mathbf{\frac{1}{4}} = \mathbf{25\%}\)
Calculation: \(P(\text{Odd and Heads}) = \frac{1}{2} \times \frac{1}{2} = \mathbf{\frac{1}{4}}\)
3 odd cards \(\times\) 1 head = 3 out of 12 outcomes
Self-Check & Reflection — Teacher Grading Rubric & Exemplar:
Exemplar Student Response: "Flipping a coin and rolling a die are independent events because the physical outcome of the coin toss (heads or tails) has no impact or influence on what number lands face up on the die; the probabilities for each trial remain completely unchanged."
Full Credit Criteria: States that the first event does not change/affect the second event's outcomes or probability. 2 / 2 pts
Lesson 8.5 • Independent Events Practice Page 5 of 5 (Answer Key)
