Mathematics • Year 8 • Unit 4 • Lesson 18
Tree Diagrams — Mixed Challenge
Pull every idea from Lesson 18 together: 2-stage and 3-stage trees, with and without replacement, biased coins, the complement shortcut. Six mixed problems, one "find the mistake", one open-ended challenge.
1. Mixed problems — choose the right move
Each question uses a different idea from Lesson 18. Show working. 3 marks each
1.1 A biased coin has P(H) = 0.7 and P(T) = 0.3. The coin is flipped twice. Find P(HH), P(at least one H), and verify that all four endpoint probabilities sum to 1.
1.2 A bag has 5 Red and 3 Blue marbles (8 total). Two are drawn with replacement. Find P(both same colour).
1.3 Same bag (5 R, 3 B, 8 total). Two are drawn without replacement. Find P(both same colour) and compare your answer to 1.2 — is it larger or smaller?
1.4 Three coins are flipped. Use a tree (or the 8-outcome sample space) to find P(at least one tail). Use the complement.
1.5 A test has two parts. P(student passes part 1) = 0.9. P(student passes part 2) = 0.8. Pass events are independent. Find P(passes both), P(passes only one), and P(fails both).
1.6 A jar has 4 toffees and 6 mints. Two lollies are drawn without replacement. Find P(one of each kind). Show both paths.
2. Find the mistake
A student attempted this without-replacement problem. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Problem: A bag has 4 Red and 2 Blue marbles (6 total). Two marbles are drawn without replacement. Find P(both Red).
Line 1: P(R on first draw) = 4/6 = 2/3.
Line 2: P(R on second draw) = 4/6 = 2/3.
Line 3: P(both R) = 2/3 × 2/3 = 4/9.
Line 4: So the probability is 4/9.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full and state the correct probability.
Stuck? Without replacement, after a R is drawn there are only 3 R and 2 B left (5 marbles total). So the second-stage P(R | R) ≠ 4/6.3. Open-ended challenge — design a two-stage experiment
This question has many valid answers. 4 marks
3.1 Your job: design a two-stage random experiment with a clear real-world context (e.g., two penalty kicks, two days of weather, two free throws). The two stages must be independent and use unequal branch probabilities (not 1/2 and 1/2).
Your answer must include:
(i) A clear context and the two stages.
(ii) Branch probabilities (must add to 1 at each stage and not be 1/2-1/2).
(iii) A drawn (or listed) tree diagram with all four endpoint probabilities calculated.
(iv) Use your tree to find P(both occur), P(only one occurs), and P(neither occurs). Verify they sum to 1.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Biased coin, two flips
P(HH) = 0.7 × 0.7 = 0.49. P(HT) = 0.7 × 0.3 = 0.21. P(TH) = 0.3 × 0.7 = 0.21. P(TT) = 0.3 × 0.3 = 0.09. P(at least one H) = 1 − P(TT) = 1 − 0.09 = 0.91. Check: 0.49 + 0.21 + 0.21 + 0.09 = 1.00 ✓.
1.2 — 5 R, 3 B with replacement, same colour
P(R) = 5/8, P(B) = 3/8 at every draw.
P(both R) = 25/64. P(both B) = 9/64.
P(same) = 25/64 + 9/64 = 34/64 = 17/32.
1.3 — 5 R, 3 B without replacement, same colour
P(both R) = 5/8 × 4/7 = 20/56 = 5/14.
P(both B) = 3/8 × 2/7 = 6/56 = 3/28.
P(same) = 5/14 + 3/28 = 10/28 + 3/28 = 13/28. Converting 1.2 to a 56-denominator for comparison: 17/32 = 29.75/56 ≈ 0.531, and 13/28 = 26/56 ≈ 0.464. Without-replacement value (≈ 0.464) is smaller.
1.4 — Three coins, at least one T
P(no tails) = P(HHH) = 1/2 × 1/2 × 1/2 = 1/8. P(at least one T) = 1 − 1/8 = 7/8.
1.5 — Two-part test
P(passes both) = 0.9 × 0.8 = 0.72.
P(only one passes) = (0.9 × 0.2) + (0.1 × 0.8) = 0.18 + 0.08 = 0.26.
P(fails both) = 0.1 × 0.2 = 0.02. Check: 0.72 + 0.26 + 0.02 = 1.00 ✓.
1.6 — Jar of toffees and mints (without replacement)
P(T then M) = 4/10 × 6/9 = 24/90.
P(M then T) = 6/10 × 4/9 = 24/90.
P(one of each) = 48/90 = 8/15.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) Without replacement, after one R is drawn there are only 3 R and 2 B left (5 marbles total), so P(R on the second draw, given R first) = 3/5, NOT 4/6. The denominator must decrease by 1, and the numerator decreases too because a red was removed.
(c) Corrected working:
P(R first) = 4/6 = 2/3. P(R | R) = 3/5.
P(both R) = 2/3 × 3/5 = 6/15 = 2/5.
3 — Open-ended design (sample solution)
(i) Context: A striker takes two penalty kicks; the events are independent.
(ii) Branch probabilities: at each stage, P(Goal) = 0.8 and P(Miss) = 0.2 (sums to 1 at each stage, not 1/2-1/2). ✓
(iii) Tree endpoints:
P(GG) = 0.8 × 0.8 = 0.64.
P(GM) = 0.8 × 0.2 = 0.16.
P(MG) = 0.2 × 0.8 = 0.16.
P(MM) = 0.2 × 0.2 = 0.04.
(iv) Combined events:
P(both occur, ie both goals) = 0.64.
P(only one occurs) = 0.16 + 0.16 = 0.32.
P(neither, ie both misses) = 0.04.
Check sum: 0.64 + 0.32 + 0.04 = 1.00 ✓.
Marking: 1 mark for clear real-world context with two independent stages; 1 mark for non-trivial branch probabilities that sum to 1; 1 mark for correct endpoint probabilities; 1 mark for the three combined events with the sum-to-1 check.