":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"ed20d383-53b1-42bc-90c4-9878fc5e814c","label":"a","content":"Draw a tree diagram to show this information.","markscheme":"- Three initial branches (A, B, C) with probabilities 0.35, 0.45, 0.2 B1\n- Two branches from each variety for Fails (F) and Germinates (F') B1\n- Correct conditional probabilities: A(0.025, 0.975), B(0.015, 0.985), C(0.05, 0.95) B1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Correct seed variety branches with probabilities.","code":"B"},{"id":"p2","marks":1,"description":"Two branches for outcome (Fails/Germinates) for each variety.","code":"B"},{"id":"p3","marks":1,"description":"Accurate placement of failure rates on branches.","code":"B"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:03:36.410Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"treediagram","label":"Standard Tree Diagram","steps":[{"type":"text","order":0,"title":"First Stage: Seed Varieties","content":"To construct our tree diagram, we start with the first event: choosing a variety of tomato seed.\n\nThe three varieties are **Variety A**, **Variety B**, and **Variety C**.\n\nWe convert their given percentages into decimal probabilities:\n- $\\text{P}(A) = 35\\% = 0.35$\n- $\\text{P}(B) = 45\\% = 0.45$\n- $\\text{P}(C) = 20\\% = 0.20$\n\nThese three branches stem from our starting point.","highlights":[{"text":"Variety A makes up 35% of the crop, Variety B makes up 45%, and Variety C makes up 20%","location":"specification"}]},{"type":"text","order":1,"title":"Second Stage: Germination Outcomes","content":"For each variety, there are two possible germination outcomes:\n- **Fails to germinate** ($F$)\n- **Germinates** ($F'$)\n\nWe convert the given percentage probabilities for failure into decimals:\n- For Variety A: $\\text{P}(F|A) = 2.5\\% = 0.025$\n- For Variety B: $\\text{P}(F|B) = 1.5\\% = 0.015$\n- For Variety C: $\\text{P}(F|C) = 5.0\\% = 0.05$\n\nSince the branches from any node must add up to $1$, we calculate the probability that a seed germinates ($F'$) for each variety using $P(F') = 1 - P(F)$:\n- For Variety A: $\\text{P}(F'|A) = 1 - 0.025 = 0.975$\n- For Variety B: $\\text{P}(F'|B) = 1 - 0.015 = 0.985$\n- For Variety C: $\\text{P}(F'|C) = 1 - 0.05 = 0.95$","highlights":[{"text":"2.5% for Variety A","location":"specification"},{"text":"1.5% for Variety B","location":"specification"},{"text":"5.0% for Variety C","location":"specification"}]},{"type":"text","order":2,"title":"Assemble the Tree Diagram","content":"Now we combine these branches into a single tree diagram:\n\n$$\\begin{array}{c} \n\\text{\\bf Stage 1 (Variety)} & \\text{\\bf Stage 2 (Germination)} & \\text{\\bf Outcomes} \\\\\n\\\\ \n& \\xrightarrow{0.025} F & (A \\cap F) \\\\\n\\xrightarrow{0.35} A & & \\\\\n& \\xrightarrow{0.975} F' & (A \\cap F') \\\\\n\\\\ \n& \\xrightarrow{0.015} F & (B \\cap F) \\\\\n\\xrightarrow{0.45} B & & \\\\\n& \\xrightarrow{0.985} F' & (B \\cap F') \\\\\n\\\\ \n& \\xrightarrow{0.05} F & (C \\cap F) \\\\\n\\xrightarrow{0.2} C & & \\\\\n& \\xrightarrow{0.95} F' & (C \\cap F') \\\\\n\\end{array}$$\n\n**Marks awarded:**\n- **1 mark (B1)** for drawing the three initial branches $A$, $B$, and $C$ with correct probabilities ($0.35$, $0.45$, $0.20$).\n- **1 mark (B1)** for drawing two sub-branches ($F$ and $F'$) from each variety branch.\n- **1 mark (B1)** for correctly writing the six conditional probabilities."}]}],"generatedAt":"2026-06-11T03:41:57.403Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"5ad0af37-b0f5-4ce6-b6c1-28d9de362aeb","label":"b","content":"A seed is selected at random. Find the probability it fails to germinate.","markscheme":"- $P(F) = (0.35 \\times 0.025) + (0.45 \\times 0.015) + (0.2 \\times 0.05)$ M1\n- $0.00875 + 0.00675 + 0.01 = 0.0255$ A1","marks":2,"order":1,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Method to calculat

Conditional Probability

Question 40

Conditional Probability Questions

Conditional Probability

Question 40

Conditional Probability Questions