Probability -- Diagnostic Tests

Probability is like a weather forecast for uncertainty — it quantifies how likely events are, from coin flips to complex systems: Probability theory provides a rigorous framework for reasoning about uncertainty, turning vague intuition into precise calculation

Why it matters: From gambling odds to medical diagnoses to climate models, probability governs how we make decisions under uncertainty

The key insight: Probability theory provides a rigorous framework for reasoning about uncertainty, turning vague intuition into precise calculation

Tests edge cases, boundary conditions, and misconceptions for probability.

Question:

A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement.

(a) Find the probability that both balls are red.

(b) Find the probability that the second ball is red.

(c) Let $A$ be “the first ball is red” and $B$ be “the second ball is red.” A student claims that since $P(A) = \frac{5}{8}$ and $P(B) = \frac{5}{8}$The events are independent. Explain the error.

[Difficulty: hard. Tests the distinction between independent and dependent events in without-replacement scenarios.]

Solution:

(a) $P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$.

(b) By the law of total probability:

$P(B) = P(B|A)P(A) + P(B|A")P(A') = \frac{4}{7} \cdot \frac{5}{8} + \frac{5}{7} \cdot \frac{3}{8}$

$= \frac{20}{56} + \frac{15}{56} = \frac{35}{56} = \frac{5}{8}$

(c) The student’s error is applying the independence test $P(A \cap B) = P(A)P(B)$ without verifying it. In fact:

$P(A \cap B) = \frac{5}{14} \neq \frac{5}{8} \times \frac{5}{8} = \frac{25}{64}$

So $A$ and $B$ are not independent. The fact that $P(B) = P(A)$ is coincidental — it does not imply independence. The correct test for independence is $P(A \cap B) = P(A)P(B)$Not $P(B|A) = P(B)$.

Question:

A factory produces items using Machine $X$ (60% of output) and Machine $Y$ (40% of output). Machine $X$ has a defect rate of 3% and Machine $Y$ has a defect rate of 5%.

(a) A randomly selected item is found to be defective. What is the probability it was produced by Machine $X$?

(b) A student reasons: “60% of items are from Machine $X$ and 3% are defective, so the probability is $0.60 \times 0.03 = 0.018$.” Identify the error.

[Difficulty: hard. Tests correct application of Bayes’ theorem versus the common prior error.]

Solution:

(a) Let $D$ be “the item is defective” and $X$ be “produced by Machine $X$.”

$P(X|D) = \frac{P(D|X)P(X)}{P(D)}$

$P(D) = P(D|X)P(X) + P(D|Y)P(Y) = 0.03 \times 0.60 + 0.05 \times 0.40 = 0.018 + 0.020 = 0.038$

$P(X|D) = \frac{0.03 \times 0.60}{0.038} = \frac{0.018}{0.038} = \frac{9}{19} \approx 0.474$

(b) The student computed $P(X \cap D)$ instead of $P(X|D)$. Bayes’ theorem asks for the posterior probability $P(X|D)$ (probability the item is from $X$ given that it is defective), but the student computed the joint probability $P(X \cap D)$ (probability the item is both from $X$ and defective, without the “given” condition). The correct computation divides by the total probability of the evidence, $P(D) = 0.038$.

Tests synthesis of probability with other topics.

Question:

A committee of 5 people is to be selected from 7 men and 5 women. The committee must contain at least 2 men and at least 2 women.

(a) In how many ways can the committee be formed?

(b) If the committee must also include exactly one person as chairperson, how many different committees are possible?

[Difficulty: hard. Combines combinatorial counting with selection under multiple constraints.]

Solution:

(a) Total ways to choose 5 from 12 people: $\binom{12}{5} = 792$.

Subtract committees with fewer than 2 men or fewer than 2 women:

• 0 men, 5 women: $\binom{7}{0}\binom{5}{5} = 1$
• 1 man, 4 women: $\binom{7}{1}\binom{5}{4} = 35$
• 2 men, 3 women: $\binom{7}{2}\binom{5}{3} = 210$
• 3 men, 2 women: $\binom{7}{3}\binom{5}{2} = 350$
• 4 men, 1 woman: $\binom{7}{4}\binom{5}{1} = 175$
• 5 men, 0 women: $\binom{7}{5}\binom{5}{0} = 21$

Total excluded: $1 + 35 + 210 + 350 + 175 + 21 = 792$.

The constraint “at least 2 men and at least 2 women” only excludes the cases with 0 or 1 of one gender:

Excluded: 0 men, 5 women (1 way) + 1 man, 4 women (35 ways) + 5 men, 0 women (21 ways) + 4 men, 1 woman (175 ways) = 232 ways.

$\text{Valid committees} = 792 - 232 = 560$

(b) For each valid committee of 5, there are 5 choices for chairperson. So: $560 \times 5 = 2800$ different chaired committees.

This content page provides comprehensive coverage of Maths content for the Ib qualification, with detailed explanations, worked examples, and practice questions aligned to the specification.

This page includes:

• Key Definitions: Precise explanations of essential concepts
• Core Concepts: Detailed treatment of fundamental principles
• Worked Examples: Step-by-step solutions demonstrating application
• Practice Questions: Examination-style questions with mark schemes
• Common Pitfalls: Frequent errors and how to avoid them
• Exam Tips: Strategies for maximising marks

1. Read through the introductory material to establish context
2. Study the definitions and core concepts carefully
3. Work through the worked examples, following each step
4. Attempt the practice questions independently
5. Review your answers against the provided solutions
6. Note any areas requiring further revision

flowchart TD
    A[Diag Probability] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

• Foundational definitions and terminology
• Application of principles to examination contexts
• Connections to related topics within the specification
• Assessment objective alignment

• Active Recall: Test yourself on the material rather than passively re-reading
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Practise applying these concepts under timed conditions. Focus on understanding what each question is asking and how marks are allocated. Review examiner reports to learn from common mistakes made by other students.

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