Mathematics • Year 10 • Unit 4 • Lesson 16
Compound Events and Tree Diagrams, Skill Drill
Build fluency with the two compound-event rules from Lesson 16: for independent events, P(A and B) = P(A) × P(B); for mutually exclusive events, P(A or B) = P(A) + P(B). Practise reading tree diagrams (multiply along a branch) and counting sample-space outcomes.
1. I do, fully worked example
Read every step. Notice which rule is being used and why.
Problem. A fair coin is flipped and a fair six-sided die is rolled. Find P(head AND a number greater than 4).
Step 1, Identify the events.
A = "coin shows head". B = "die shows a number > 4" (i.e. 5 or 6).
Reason: write the events in symbols so the rule is easy to apply.
Step 2, Are A and B independent?
The coin result cannot change the die result. → independent.
Reason: Lesson 16 Think First, "If you roll a die and flip a coin, does the die result affect the coin result?"
Step 3, Apply the multiplication rule.
P(A) = 1/2. P(B) = 2/6 = 1/3.
P(A and B) = P(A) × P(B) = 1/2 × 1/3 = 1/6.
Reason: Lesson 16 Key Term, "For independent events: P(A and B) = P(A) × P(B)."
Step 4, Check by counting the sample space.
Sample space = 2 × 6 = 12 equally likely outcomes.
Favourable: (H,5), (H,6) → 2 outcomes. So P = 2/12 = 1/6. ✓
Answer: P(head and number > 4) = 1/6.
2. We do, fill in the missing steps
Same structure as Section 1, but the working is faded. Fill the blanks. 4 marks
Problem. A bag has 3 red and 2 blue marbles. A marble is drawn, the colour noted, then the marble is replaced. A second marble is drawn. Find P(both red) and draw the first two levels of a tree diagram.
Step 1, With replacement → independent. The marble is returned, so the bag is unchanged. P(red on draw 2) does not depend on draw 1. → events are __________________.
Step 2, Probabilities on each branch.
P(red) = 3/5. P(blue) = ______.
Step 3, Use the tree-diagram rule. "On a tree diagram, probabilities along a branch are ____________________" (Lesson 16 MCQ Q5).
P(red and red) = 3/5 × ______ = ______.
Step 4, Final answer.
P(both red) = ______ (as a fraction in simplest form).
3. You do, independent practice
Show working for every question. The first four are foundation (one rule). The middle two are standard (two-step compound). The last two are extension (tree diagrams + mutually exclusive trap).
Foundation, pick the right rule
3.1 P(A) = 0.4 and P(B) = 0.5 with A and B independent. Find P(A and B). 1 mark
3.2 A and B are mutually exclusive with P(A) = 0.2 and P(B) = 0.35. Find P(A or B). 1 mark
3.3 A spinner has equal sectors numbered 1-8. A coin is also flipped. How many outcomes are in the sample space? 1 mark
3.4 Drawing two cards from a standard deck without replacement makes the two draws ______________ (independent / dependent). One word answer. 1 mark
Standard, combine the rules
3.5 A fair coin is flipped twice. Draw a tree diagram, then find P(two heads), P(exactly one head) and P(at least one head). 3 marks
3.6 A bag has 4 red and 6 blue marbles. Two marbles are drawn with replacement. Find P(red then blue) and P(both the same colour). 2 marks
Extension, tree diagrams + the rules together
3.7 A coin is flipped three times. Draw the full tree diagram. Find P(exactly two heads). 3 marks
3.8 P(A) = 0.3, P(B) = 0.5, and P(A and B) = 0.15. A student writes "they're mutually exclusive, so P(A or B) = 0.3 + 0.5 = 0.8". Is the student correct? Justify against Lesson 16's "Wrong: n(A or B) = n(A) + n(B)" misconception, and find the correct P(A or B). 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (with-replacement marbles)
Step 1: events are independent (with replacement).
Step 2: P(blue) = 2/5.
Step 3: probabilities along a branch are multiplied. P(red and red) = 3/5 × 3/5 = 9/25.
Step 4: P(both red) = 9/25.
3.1, P(A and B), independent
P(A and B) = P(A) × P(B) = 0.4 × 0.5 = 0.2.
3.2, P(A or B), mutually exclusive
P(A or B) = P(A) + P(B) = 0.2 + 0.35 = 0.55.
3.3, Spinner and coin sample space
2 × 8 = 16 outcomes.
3.4, Without replacement
Dependent (Lesson 16 MCQ Q4). The first card changes what is left for the second draw.
3.5, Two coin flips
Tree branches: H/T then H/T → outcomes HH, HT, TH, TT each with probability 1/4.
P(two heads) = 1/4.
P(exactly one head) = P(HT) + P(TH) = 1/4 + 1/4 = 1/2.
P(at least one head) = 1 − P(no heads) = 1 − 1/4 = 3/4.
3.6, Two marbles with replacement
P(red) = 4/10 = 2/5. P(blue) = 6/10 = 3/5. With replacement → independent.
P(red then blue) = 2/5 × 3/5 = 6/25.
P(both same colour) = P(RR) + P(BB) = (2/5)² + (3/5)² = 4/25 + 9/25 = 13/25.
3.7, Three coin flips
Eight outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT (each probability 1/8).
Exactly two heads: HHT, HTH, THH → 3 outcomes.
P(exactly two heads) = 3/8.
3.8, The mutually-exclusive trap
The student is wrong. P(A and B) = 0.15 ≠ 0, so A and B are not mutually exclusive. The simple addition P(A) + P(B) double-counts the overlap. Use the corrected rule from Lesson 16 (n(A or B) = n(A) + n(B) − n(A and B)):
P(A or B) = 0.3 + 0.5 − 0.15 = 0.65.