This IB math probability practice question focuses on conditional probability with dependent events — a topic that looks straightforward until the question asks you to work backwards from a result, and suddenly the tree diagram becomes your best friend. Conditional probability appears on both Paper 1 and Paper 2, and the students who score full marks are the ones who draw a careful diagram before writing a single calculation. Before you open any hints, read the question slowly, draw your tree, and label every branch with a fraction. That one habit is worth more than any shortcut. 🟡 Medium ⏱ 9–12 minutes 📄 Paper 1 (no calculator) 📋 Jump To This Week’s IB Math Probability Practice Question What You Need to Know Hints Full Worked Solution Examiner Notes and Common Mistakes This Week’s IB Math Probability Practice Question 📝 Question of the Week A bag contains \( 5 \) red marbles and \( 3 \) blue marbles. Two marbles are selected at random from the bag, one after the other, without replacement. (a) Complete the tree diagram below to show all possible outcomes and their probabilities for the two selections. [3 marks] (b) Find the probability that both marbles selected are the same colour. [3 marks] (c) Given that the second marble selected is red, find the probability that the first marble selected was also red. [3 marks] Total: [9 marks] What You Need to Know 📖 Key Information Topic: Probability — Conditional probability and dependent events (AA SL 4.5–4.6, AA HL 4.5–4.6) Paper style: Paper 1 (no calculator) — all probabilities are exact fractions Estimated time: 9–12 minutes Key formulas from the data booklet: Multiplication rule for dependent events: $$P(A \cap B) = P(A) \times P(B \mid A)$$ Conditional probability formula: $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$ Key concepts: Without replacement: After the first marble is drawn, the total changes from 8 to 7, and the count of the relevant colour also changes. This is what makes the events dependent. Tree diagram branches multiply: The probability at each terminal node is the product of all branch probabilities along that path. Total probability: To find \( P(\text{second marble is red}) \), add the probabilities of all paths ending in red on the second draw. If you want to strengthen your understanding of how probability connects to statistics topics like distributions, visit our post on Question of the Week 3: Statistics Challenge for more exam-style practice. You can also review the official syllabus outcomes on the IB Mathematics curriculum page. Hints 💡 Hint 1 — Getting Started Begin by drawing the tree diagram carefully. For the first draw, there are 8 marbles total: 5 red and 3 blue. So the first-level branch probabilities are \( \frac{5}{8} \) for red and \( \frac{3}{8} \) for blue. Now think about what happens to the totals for the second draw — you have one fewer marble, and one fewer of whichever colour was drawn first. There are four possible paths: RR, RB, BR, BB. ⏸ Try working with this hint before opening Hint 2. 💡 Hint 2 — Second-Level Branches and Part (b) For the second draw, the denominators all become 7 (one marble has been removed). If the first marble was red, there are now 4 red and 3 blue left. If the first was blue, there are 5 red and 2 blue left. For part (b), “both the same colour” means either RR or BB. Calculate the probability of each path by multiplying along the branches, then add: $$P(\text{same colour}) = P(RR) + P(BB)$$ ⏸ Try completing parts (a) and (b) before opening Hint 3. 💡 Hint 3 — Conditional Probability in Part (c) Part (c) asks: given that the second marble is red, what is the probability the first was red? This is Bayes-style conditional probability. Use the formula: $$P(\text{first R} \mid \text{second R}) = \frac{P(\text{first R} \cap \text{second R})}{P(\text{second R})}$$ You already know \( P(RR) \) from part (b). For the denominator, find \( P(\text{second R}) \) by adding all paths where the second marble is red: that is \( P(RR) + P(BR) \). ⏸ Now try finishing the solution on your own. Full Worked Solution ✍️ Step-by-Step Solution Part (a) Step 1: Set Up the Tree Diagram For this IB math probability practice question, we begin with 5 red (R) and 3 blue (B) marbles — 8 total. The first-level branches are: $$P(\text{1st R}) = \frac{5}{8}, \qquad P(\text{1st B}) = \frac{3}{8}$$ After the first draw (without replacement), 7 marbles remain. The second-level conditional probabilities are: $$\begin{aligned} P(\text{2nd R} \mid \text{1st R}) &= \frac{4}{7}, \quad P(\text{2nd B} \mid \text{1st R}) = \frac{3}{7} \\[6pt] P(\text{2nd R} \mid \text{1st B}) &= \frac{5}{7}, \quad P(\text{2nd B} \mid \text{1st B}) = \frac{2}{7} \end{aligned}$$ Part (a) Step 2: Calculate All Four Path Probabilities Multiply along each branch path to find the probability of each outcome: $$\begin{aligned} P(RR) &= \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} \\[6pt] P(RB) &= \frac{5}{8} \times \frac{3}{7} = \frac{15}{56} \\[6pt] P(BR) &= \frac{3}{8} \times \frac{5}{7} = \frac{15}{56} \\[6pt] P(BB) &= \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28} \end{aligned}$$ Verification: \( \frac{20}{56} + \frac{15}{56} + \frac{15}{56} + \frac{6}{56} = \frac{56}{56} = 1 \checkmark \) Part (b): Probability That Both Marbles Are the Same Colour “Same colour” means either both red (RR) or both blue (BB). These are mutually exclusive outcomes, so we add their probabilities: $$\begin{aligned} P(\text{same colour}) &= P(RR) + P(BB) \\ &= \frac{20}{56} + \frac{6}{56} \\ &= \frac{26}{56} \\ &= \frac{13}{28} \end{aligned}$$ Part (c) Method 1: Using the Conditional Probability Formula We need \( P(\text{1st R} \mid \text{2nd R}) \). First, find the total probability that the second marble is red by summing all paths ending in R on the second draw: $$\begin{aligned} P(\text{2nd R}) &= P(RR) + P(BR) \\ &= \frac{20}{56} + \frac{15}{56} \\ &= \frac{35}{56} = \frac{5}{8} \end{aligned}$$ Now apply the conditional probability formula: $$P(\text{1st R} \mid \text{2nd R}) = \frac{P(RR)}{P(\text{2nd R})} = \frac{\dfrac{20}{56}}{\dfrac{35}{56}} = \frac{20}{35} = \frac{4}{7}$$ Part (c) Method 2: Intuitive Check Using Symmetry