Probability

Syllabus Identity

• Curriculum: Mathematics
• Topic ID: topic-csee-basic-mathematics-2005-[[probability|probability]]
• Form: Form IV
• Hub: Statistics and Probability
• Competence grouping: Statistics and probability

This is a current Mathematics syllabus topic. It preserves the 2005 Basic Mathematics identity and order for exam-facing mapping. Do not merge it into the 2023 Mathematics transition topic page even when the learning idea overlaps.

Official Scope

Current Mathematics syllabus topic covering probability of an event; combined events; simple combinations of probabilities.

Subtopics

Core Concepts

Probability is the branch of mathematics that quantifies the likelihood of an event occurring. It forms the foundation for statistics and is widely applicable in various real-world scenarios such as games of chance, weather forecasting, and risk assessment.

1. Probability of an Event

To understand the probability of a single event, we must define three fundamental terms:

• Experiment: Any procedure that can be infinitely repeated and has a well-defined set of possible outcomes (e.g., tossing a coin, rolling a die).
• Sample Space ($S$): The set of all possible outcomes of an experiment. For example, when rolling a standard six-sided die, the sample space is $S = \{1, 2, 3, 4, 5, 6\}$.
• Event ($E$): Any subset of the sample space. An event could be rolling an even number: $E = \{2, 4, 6\}$.

The theoretical probability of an event $E$ occurring is given by the ratio of the number of favorable outcomes to the total number of possible outcomes in the sample space, provided all outcomes are equally likely: $$ P(E) = \frac{n(E)}{n(S)} $$ where:

• $n(E)$ is the number of favorable outcomes.
• $n(S)$ is the total number of outcomes in the sample space.

Key Properties of Probability:

1. The probability of any event $E$ lies between $0$ and $1$ inclusive:

$$ 0 \le P(E) \le 1 $$

• If $P(E) = 0$, the event is impossible.
• If $P(E) = 1$, the event is certain.
2. The sum of the probabilities of all mutually exclusive and exhaustive events in a sample space is always equal to $1$.
3. Complementary Events: The complement of an event $E$, denoted as $E'$, is the event that $E$ does not occur. Since $E$ and $E'$ together make up the entire sample space:

$$ P(E) + P(E') = 1 \implies P(E') = 1 - P(E) $$

2. Combined Events

Combined events involve finding the probability of two or more events happening either simultaneously or in sequence.

Mutually Exclusive Events: Two events $A$ and $B$ are mutually exclusive if they cannot happen at the same time (i.e., their intersection is empty, $A \cap B = \emptyset$).

• Addition Rule: To find the probability of either event $A$ or event $B$ occurring:

$$ P(A \text{ or } B) = P(A \cup B) = P(A) + P(B) $$

Non-Mutually Exclusive Events: If events $A$ and $B$ can occur at the same time, we must account for the overlap to avoid double counting. This concept is closely related to set theory operations.

• General Addition Rule:

$$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $$

Independent Events: Two events are independent if the occurrence of one event does not affect the probability of the other event occurring. A classic example is tossing a coin and rolling a die, or drawing items from a bag with replacement.

• Multiplication Rule: To find the probability of both event $A$ and event $B$ occurring:

$$ P(A \text{ and } B) = P(A \cap B) = P(A) \times P(B) $$

Dependent Events: Two events are dependent if the occurrence of the first event alters the probability of the second event. A common example is drawing items from a bag without replacement.

• We use conditional probability, denoted as $P(B|A)$, which is the probability of $B$ occurring given that $A$ has already occurred:

$$ P(A \text{ and } B) = P(A) \times P(B|A) $$

Visual Tools for Combined Events:

1. Tree Diagrams: Excellent for sequential events (e.g., drawing three shirts in succession, or germinating three seeds). Branches represent different possible outcomes at each stage.
• Multiply probabilities along a continuous path (branches moving from left to right) to find the probability of that specific combination.
• Add the probabilities of all paths that satisfy the event condition.
2. Venn Diagrams: Highly effective for non-mutually exclusive events or when probability questions are intertwined with set theory. By finding the number of elements in specific regions (e.g., "only A", "both A and B", "neither"), you can easily calculate probabilities using $P(E) = \frac{n(E)}{n(U)}$, where $U$ is the universal set.

Worked Examples

Example 1: Basic Probability of an Event A class has $50$ students of which $35$ are boys and $15$ are girls. If a student is chosen at random, find the probability that the student is a boy.

Solution: Total number of students in the sample space, $n(S) = 50$. Number of boys (favorable outcomes), $n(B) = 35$. $$ P(\text{Boy}) = \frac{n(B)}{n(S)} = \frac{35}{50} $$ Simplify the fraction by dividing the numerator and denominator by $5$: $$ P(\text{Boy}) = \frac{7}{10} $$ The probability that the chosen student is a boy is $\frac{7}{10}$.

Example 2: Probability from a Frequency Table The following table shows the number of tables in twenty offices from a certain company:

| Number of Tables | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | Number of Offices | $4$ | $5$ | $6$ | $2$ | $1$ | $2$ |

If one office is picked randomly, find the probability that the office has: (a) Two tables. (b) At least five tables.

Solution: The total number of offices is $n(S) = 20$.

(a) Offices with exactly two tables: Number of offices with $2$ tables, $n(\text{Two}) = 5$. $$ P(\text{Two tables}) = \frac{5}{20} = \frac{1}{4} $$

(b) "At least five tables" means the office can have $5$ tables or $6$ tables. These are mutually exclusive events. Number of offices with $5$ tables $= 1$. Number of offices with $6$ tables $= 2$. Total favorable outcomes $n(\ge 5 \text{ tables}) = 1 + 2 = 3$. $$ P(\text{At least five tables}) = \frac{3}{20} $$

Example 3: Independent Events using a Tree Diagram Jonika has $2$ shirts (blue and red) and $3$ trousers (black, green and yellow). Draw a tree diagram and find the probability of choosing a blue shirt and a black trouser.

Solution: Let Shirts be: Blue ($B_s$), Red ($R_s$). Let Trousers be: Black ($B_t$), Green ($G_t$), Yellow ($Y_t$).

Since choosing a shirt does not affect choosing a trouser, these are independent events. $P(B_s) = \frac{1}{2}$, $P(R_s) = \frac{1}{2}$. $P(B_t) = \frac{1}{3}$, $P(G_t) = \frac{1}{3}$, $P(Y_t) = \frac{1}{3}$.

Tree Diagram Construction:

• First set of branches: $B_s$ (probability $\frac{1}{2}$) and $R_s$ (probability $\frac{1}{2}$).
• Second set of branches (from both $B_s$ and $R_s$): $B_t$, $G_t$, and $Y_t$, each with probability $\frac{1}{3}$.

To find the probability of choosing a blue shirt and a black trouser, we follow the path $B_s \to B_t$. Multiply the probabilities along the branches: $$ P(B_s \text{ and } B_t) = P(B_s) \times P(B_t) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} $$ The probability is $\frac{1}{6}$.

Example 4: Probability using Set Theory and Venn Diagrams In a class of $45$ students, $30$ study Chemistry, $20$ study Physics and $5$ study neither of the two subjects. Find the probability that a student selected at random from the class will be studying Chemistry only.

Solution: Let $U$ be the universal set of all students, $n(U) = 45$. Let $C$ be the set of students studying Chemistry, $n(C) = 30$. Let $P$ be the set of students studying Physics, $n(P) = 20$. Students studying neither, $n(C \cup P)' = 5$.

First, find the total number of students studying at least one subject: $$ n(C \cup P) = n(U) - n(C \cup P)' = 45 - 5 = 40 $$

Use the general addition rule for sets to find the intersection (students studying both): $$ n(C \cup P) = n(C) + n(P) - n(C \cap P) $$ $$ 40 = 30 + 20 - n(C \cap P) $$ $$ 40 = 50 - n(C \cap P) \implies n(C \cap P) = 10 $$

Now, find the number of students studying Chemistry only: $$ n(\text{Chemistry only}) = n(C) - n(C \cap P) = 30 - 10 = 20 $$

Finally, calculate the probability: $$ P(\text{Chemistry only}) = \frac{n(\text{Chemistry only})}{n(U)} = \frac{20}{45} = \frac{4}{9} $$

Example 5: Combined Events with Replacement A bag contains $6$ white shirts and $3$ blue shirts. Three shirts are picked at random one after another with replacement. Determine the probability that two shirts are white and one shirt is blue.

Solution: Total number of shirts $= 6 \text{ (White)} + 3 \text{ (Blue)} = 9 \text{ shirts}$. Probability of drawing White on any draw, $P(W) = \frac{6}{9} = \frac{2}{3}$. Probability of drawing Blue on any draw, $P(B) = \frac{3}{9} = \frac{1}{3}$. Because the drawing is with replacement, the events are independent, and the probabilities remain constant for every draw.

We want exactly two white shirts and one blue shirt in three draws. This can happen in different sequences:

1. White, White, Blue ($WWB$)
2. White, Blue, White ($WBW$)
3. Blue, White, White ($BWW$)

Calculate the probability for each path: $$ P(WWB) = P(W) \times P(W) \times P(B) = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27} $$ $$ P(WBW) = P(W) \times P(B) \times P(W) = \frac{2}{3} \times \frac{1}{3} \times \frac{2}{3} = \frac{4}{27} $$ $$ P(BWW) = P(B) \times P(W) \times P(W) = \frac{1}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{4}{27} $$

Since these sequences are mutually exclusive, we add their probabilities: $$ P(2W \text{ and } 1B) = P(WWB) + P(WBW) + P(BWW) = \frac{4}{27} + \frac{4}{27} + \frac{4}{27} = \frac{12}{27} $$ Simplify the fraction: $$ P(2W \text{ and } 1B) = \frac{4}{9} $$

Example 6: Tree Diagrams and "At Least" Conditions A farmer was given three seeds to germinate in a nursery. The probability that a seed will germinate is $\frac{1}{3}$. Using a tree diagram, find the probability that at least two seeds will germinate.

Solution: Let $G$ be the event that a seed germinates. $P(G) = \frac{1}{3}$. Let $G'$ be the event that a seed does not germinate. $P(G') = 1 - \frac{1}{3} = \frac{2}{3}$. There are three independent trials (the three seeds).

Paths for 3 trials:

• $G \to G \to G$ (3 germinated)
• $G \to G \to G'$ (2 germinated)
• $G \to G' \to G$ (2 germinated)
• $G \to G' \to G'$ (1 germinated)
• $G' \to G \to G$ (2 germinated)
• $G' \to G \to G'$ (1 germinated)
• $G' \to G' \to G$ (1 germinated)
• $G' \to G' \to G'$ (0 germinated)

"At least two seeds will germinate" means either $2$ seeds germinate or all $3$ seeds germinate. We identify the relevant paths from the tree diagram:

1. $GGG$: $P(GGG) = \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$
2. $GGG'$: $P(GGG') = \frac{1}{3} \times \frac{1}{3} \times \frac{2}{3} = \frac{2}{27}$
3. $GG'G$: $P(GG'G) = \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{2}{27}$
4. $G'GG$: $P(G'GG) = \frac{2}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{2}{27}$

Summing the probabilities of these mutually exclusive paths: $$ P(\text{At least } 2) = \frac{1}{27} + \frac{2}{27} + \frac{2}{27} + \frac{2}{27} = \frac{7}{27} $$

Common Pitfalls & Misconceptions

1. Confusing "With Replacement" and "Without Replacement": This is the most frequent source of error in combined events.
• With replacement: The events are independent. The denominator (and numerator if the same item is drawn again) remains the same for every draw.
• Without replacement: The events are dependent. The total sample size (denominator) decreases by $1$ after each draw, and the number of favorable items (numerator) decreases if an item of that type was just drawn.
2. Misinterpreting "At least" and "At most": Students often calculate the exact value rather than the range.
• "At least two" means $2, 3, 4, \dots$ up to the maximum possible.
• "At most two" means $0, 1, \text{ or } 2$.

Tip: Using the complement rule is sometimes faster. For example, $P(\text{at least } 1) = 1 - P(\text{none})$.

3. Adding Instead of Multiplying in Tree Diagrams: A common mistake is adding the probabilities along a branch path. Remember: Multiply horizontally along the branches to find the outcome of a specific path. Add vertically when combining the probabilities of different successful paths.
4. Venn Diagram Universal Set Oversights: When translating a word problem into a Venn diagram to find probabilities, students sometimes forget to account for the elements that lie outside the intersecting circles (the "neither" category) when writing the equation for the universal set. Always ensure that the sum of all distinct regions equals $n(U)$.

NECTA Exam Focus

An analysis of recent NECTA CSEE past papers reveals several recurring themes in how Probability is tested:

• Set Theory Integration: A significant portion of Section B probability questions require students to first construct a Venn diagram from a word problem. You must correctly deduce the number of elements in specific regions (such as intersections or "only A") before calculating the final probability.
• Tree Diagrams for Multi-Stage Events: Drawing tree diagrams is explicitly requested in many exams. NECTA frequently tests $2$-stage or $3$-stage events, such as picking items consecutively or independent repeated trials (e.g., germinating seeds).
• Frequency Tables: Data is occasionally presented in a simple frequency distribution table, where students must aggregate frequencies to find probabilities for specific conditions (e.g., "exactly 2" or "at least 5").
• Language of Probability: Mastery of mathematical language is essential. NECTA rigorously uses phrases like "with replacement", "at least", "neither", and "only". Misreading these keywords almost always leads to incorrect models.

Practice Problems

Category: Basic

1. [2023 Paper 1] A class has $50$ students of which $35$ are boys and $15$ are girls. If a student is chosen at random, find the probability that the student is a girl.
2. [2019 Paper 1] Use the following information about a Venn diagram: The universal set $U = \{a, b, c, d, e, f, g, h\}$. Set $A = \{c, f, g\}$ and Set $B = \{d, b, h\}$. The elements $\{a, e\}$ are outside both sets. If an element is picked at random from the universal set, find the probability that it is not an element of set $B$.

Category: Intermediate

1. [2024 Paper 1] The following table shows the number of tables in twenty offices from a certain company:

| Number of Tables | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | Number of Offices | $4$ | $5$ | $6$ | $2$ | $1$ | $2$ |

If one office is picked randomly, find the probability that the office has exactly three tables.

1. [2025 Paper 1] In a sample of $35$ animal keepers, $18$ keep goats, $20$ keep cows and $3$ keep both goats and cows. By using a Venn diagram, find the probability of getting a person who keeps goats only.
2. [2018 Paper 1] A bag contains $6$ white shirts and $3$ blue shirts. Three shirts are picked at random one after another with replacement. Determine the probability that all three shirts are blue in colour.
3. [2018 Paper 1] From the same bag of $6$ white and $3$ blue shirts, if three shirts are picked at random one after another with replacement, determine the probability that one shirt is white and two shirts are blue.

Category: Advanced (NECTA Section B Style)

1. [2020 Paper 1] In a certain school, $40$ students were asked about whether they like tennis or football or both. It was found that the number of students who like both tennis and football was three times the number of students who like tennis only. Furthermore, the number of students who like football only was $6$ more than twice the number of students who like tennis only. However, $4$ students like neither tennis nor football. Determine the probability that a student selected at random likes football only.
2. [2020 Paper 1] Using the results from the school scenario above, determine the probability that a student selected at random likes both football and tennis.
3. [2022 Paper 1] A farmer was given three seeds to germinate in a nursery. The probability that a seed will germinate is $\frac{1}{3}$. Using a tree diagram, find the probability that exactly one seed will germinate.

Crosswalk Notes

Cross-version relationships are drafted in data/curricula/crosswalks/csee-basic-mathematics-2005-to-mathematics-2023.json. Partial and 2005-only mappings remain reviewable.