IB Math AI HL Topic 4.5: Trial and outcome | Notes & Flashcards | Aimnova

Key concepts in Trial and outcome

Key Idea: Probability quantifies how likely an event is, on a scale from 0 (impossible) to 1 (certain). Topic 4.5 builds the core rules: addition rule for combined events, the product rule for independent events, and conditional probability for events that depend on each other. These rules underpin everything in Topics 4.6, 4.8, and 4.9.

✅ Core probability rules

Example: P(A) = 0.4, P(B) = 0.3, P(A∩B) = 0.12 P(A∪B) = 0.4 + 0.3 − 0.12 = 0.58 Are A and B independent? Check: P(A)×P(B) = 0.4×0.3 = 0.12 = P(A∩B). Yes, independent. P(A|B) = P(A∩B)/P(B) = 0.12/0.3 = 0.4 (= P(A), confirming independence) Tree diagram: A bag has 3 red, 2 blue balls. Draw without replacement. P(both red) = (3/5) × (2/4) = 6/20 = 3/10

Use tree diagrams for sequential events (drawing one then another). Multiply along branches (AND) and add across outcomes (OR). When drawing without replacement: the denominator for the second draw decreases by 1, and the available items change depending on the first draw.

Paper 1: Show the formula before substituting. Writing 'P(A∪B) = P(A) + P(B) − P(A∩B)' then substituting earns a method mark separately from the answer. Paper 2: For multi-stage problems, draw a tree diagram even if it takes time — it organises the calculation and prevents combining the wrong branches.

IB-style question [7 marks]

In a group of 90 students, 40 like Mathematics, 30 like Physics and 12 like both subjects. One student is chosen at random. (a) Find the probability that the student likes Mathematics or Physics (or both). (b) Given that the student likes Mathematics, find the probability that the student also likes Physics. (c) Determine whether 'likes Mathematics' and 'likes Physics' are independent events.

Step by step:

Write each basic probability as the count over the total of 90.
$P(M) = \frac{40}{90},\quad P(Ph) = \frac{30}{90},\quad P(M\cap Ph) = \frac{12}{90}$
(a) Use the addition rule and subtract the overlap so it is not counted twice.
$P(M\cup Ph) = P(M) + P(Ph) - P(M\cap Ph)$
Substitute the values.
$= \frac{40}{90} + \frac{30}{90} - \frac{12}{90} = \frac{58}{90} = \frac{29}{45} \approx 0.644$
(b) Use conditional probability: divide the overlap by the probability of the condition.
$P(Ph\mid M) = \frac{P(M\cap Ph)}{P(M)} = \frac{12/90}{40/90}$
Simplify.
$= \frac{12}{40} = \frac{3}{10} = 0.3$
(c) Events are independent only if P(M)×P(Ph) equals P(M∩Ph). Compare the two.
$P(M)\times P(Ph) = \frac{40}{90}\times\frac{30}{90} = \frac{4}{27} \approx 0.148$
Since 0.148 ≠ 12/90 ≈ 0.133, the product does not match the overlap.
$\frac{4}{27} \ne \frac{12}{90} \;\Rightarrow\; \text{not independent}$

Final answer:

(a) 29/45 ≈ 0.644. (b) 0.3. (c) Not independent (0.148 ≠ 0.133).

Key concepts in Trial and outcome

Key Idea: Probability quantifies how likely an event is, on a scale from 0 (impossible) to 1 (certain). Topic 4.5 builds the core rules: addition rule for combined events, the product rule for independent events, and conditional probability for events that depend on each other. These rules underpin everything in Topics 4.6, 4.8, and 4.9.

✅ Core probability rules

Example: P(A) = 0.4, P(B) = 0.3, P(A∩B) = 0.12 P(A∪B) = 0.4 + 0.3 − 0.12 = 0.58 Are A and B independent? Check: P(A)×P(B) = 0.4×0.3 = 0.12 = P(A∩B). Yes, independent. P(A|B) = P(A∩B)/P(B) = 0.12/0.3 = 0.4 (= P(A), confirming independence) Tree diagram: A bag has 3 red, 2 blue balls. Draw without replacement. P(both red) = (3/5) × (2/4) = 6/20 = 3/10

Use tree diagrams for sequential events (drawing one then another). Multiply along branches (AND) and add across outcomes (OR). When drawing without replacement: the denominator for the second draw decreases by 1, and the available items change depending on the first draw.

Paper 1: Show the formula before substituting. Writing 'P(A∪B) = P(A) + P(B) − P(A∩B)' then substituting earns a method mark separately from the answer. Paper 2: For multi-stage problems, draw a tree diagram even if it takes time — it organises the calculation and prevents combining the wrong branches.

IB-style question [7 marks]

Step by step:

Write each basic probability as the count over the total of 90.
$P(M) = \frac{40}{90},\quad P(Ph) = \frac{30}{90},\quad P(M\cap Ph) = \frac{12}{90}$
(a) Use the addition rule and subtract the overlap so it is not counted twice.
$P(M\cup Ph) = P(M) + P(Ph) - P(M\cap Ph)$
Substitute the values.
$= \frac{40}{90} + \frac{30}{90} - \frac{12}{90} = \frac{58}{90} = \frac{29}{45} \approx 0.644$
(b) Use conditional probability: divide the overlap by the probability of the condition.
$P(Ph\mid M) = \frac{P(M\cap Ph)}{P(M)} = \frac{12/90}{40/90}$
Simplify.
$= \frac{12}{40} = \frac{3}{10} = 0.3$
(c) Events are independent only if P(M)×P(Ph) equals P(M∩Ph). Compare the two.
$P(M)\times P(Ph) = \frac{40}{90}\times\frac{30}{90} = \frac{4}{27} \approx 0.148$
Since 0.148 ≠ 12/90 ≈ 0.133, the product does not match the overlap.
$\frac{4}{27} \ne \frac{12}{90} \;\Rightarrow\; \text{not independent}$

Final answer:

(a) 29/45 ≈ 0.644. (b) 0.3. (c) Not independent (0.148 ≠ 0.133).

IB Math AI HL — Trial and outcome

Key concepts in Trial and outcome

✅ Core probability rules

IB-style question [7 marks]

What you'll learn in Topic 4.5

Study resources — 4.5 Trial and outcome

Probability Fundamentals

Combined Events

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IB Math AI HL — Trial and outcome

Key concepts in Trial and outcome

✅ Core probability rules

IB-style question [7 marks]

What you'll learn in Topic 4.5

Study resources — 4.5 Trial and outcome

Probability Fundamentals

Combined Events

Ready to study Trial and outcome?

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