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Probability: theoretical probability — KCSE Mathematics

KCSE Mathematics · 99 practice questions · 3 syllabus objectives · 3 revision lessons

35 easy34 medium30 hard

Last updated · Aligned to the KNEC KCSE syllabus

What You'll Learn

Key learning outcomes for this topic, aligned to the KNEC KCSE syllabus.

Define the sample space, event and probability; calculate P(E) = n(favourable outcomes)/n(sample space) for equally likely outcomes

Apply the complementary rule P(E') = 1 – P(E) and identify mutually exclusive and independent events

Probability: theoretical probability

Revision Notes

Concise lesson notes for Probability: theoretical probability, written to the KCSE Mathematics marking standard. Read the first lesson free below.

Understanding Probability Basics

In probability, we define sample space, event, and probability.

  1. Sample Space (S): This is the set of all possible outcomes of a random experiment. For example, when tossing a coin, the sample space is {Heads, Tails}.

  2. Event (E): An event is a subset of the sample space. For instance, if we define the event of getting a Head when tossing a coin, E = {Heads}.

  3. Probability (P): The probability of an event is calculated using the formula:

    P(E) = n(favourable outcomes) / n(sample space)

    where n(favourable outcomes) is the number of outcomes that make the event true, and n(sample space) is the total number of outcomes in the sample space.

Example: Consider rolling a die. The sample space is S = {1, 2, 3, 4, 5, 6}. If we want to find the probability of rolling a 4:

  • n(favourable outcomes) = 1 (only the outcome 4)
  • n(sample space) = 6

Thus, P(rolling a 4) = 1/6.

Key points to remember

  • Sample space includes all possible outcomes of an experiment.
  • An event is a specific outcome or set of outcomes.
  • Probability is calculated as favorable outcomes over total outcomes.
  • Use P(E) = n(favourable outcomes) / n(sample space) for calculations.
  • Equally likely outcomes simplify probability calculations.

Worked example

What is the probability of drawing a red card from a standard deck of cards?

  • Sample space, S = 52 cards.
  • Favorable outcomes, E = 26 red cards.
  • Therefore, P(red card) = 26/52 = 1/2.

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Lesson 2: Understanding Probability and Complementary Events

Objective: Apply the complementary rule P(E') = 1 – P(E) and identify mutually exclusive and independent events

In probability, the complementary rule states that the probability of an event not occurring (E') is equal to 1 minus the probability of the event occurring (P(E)). This can be expressed as:

P(E') = 1 - P(E)

Mutually Exclusive Events are events that cannot occur at the same time. For example, when flipping a coin, the outcomes 'heads' and 'tails' are mutually exclusive. If one occurs, the other cannot.

Independent Events are events where the occurrence of one does not affect the occurrence of the other. For instance, rolling a die and flipping a coin are independent events. The outcome of one does not influence the other.

To apply the complementary rule, first calculate P(E), then subtract from 1 to find P(E').

Example: If the probability of raining tomorrow (P(E)) is 0.3, then:

  • P(E') = 1 - P(E) = 1 - 0.3 = 0.7 This means there is a 70% chance it will not rain tomorrow.
  • Complementary rule: P(E') = 1 - P(E)
  • Mutually exclusive events cannot occur simultaneously.
  • Independent events do not affect each other's occurrence.
  • Calculate P(E') after finding P(E).
  • Use examples to illustrate concepts clearly.

If the probability of drawing a red card from a deck (P(E)) is 0.26, find P(E').

  • P(E') = 1 - P(E) = 1 - 0.26 = 0.74
Lesson 3: Understanding Theoretical Probability

Objective: Probability: theoretical probability

Theoretical probability is calculated based on the possible outcomes of an event. It is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes. To express this mathematically, the formula is:

Probability (P) = Number of favorable outcomes / Total number of possible outcomes

For example, when rolling a fair six-sided die, the probability of rolling a 3 is:

  • Favorable outcomes: 1 (only the outcome of 3)
  • Total outcomes: 6 (1, 2, 3, 4, 5, 6)

Thus, the probability P of rolling a 3 is:

P(rolling a 3) = 1/6.

Another example is drawing a red card from a standard deck of 52 playing cards. The number of favorable outcomes is 26 (since there are 26 red cards). Therefore, the probability P of drawing a red card is:

P(drawing a red card) = 26/52 = 1/2.

Understanding theoretical probability helps in predicting outcomes in various scenarios, making it a crucial concept in mathematics.

  • Theoretical probability is based on possible outcomes.
  • Formula: P = Favorable outcomes / Total outcomes.
  • Example: P(rolling a 3) = 1/6.
  • Example: P(drawing a red card) = 1/2.
  • Helps in predicting outcomes in real-life situations.

Question: What is the theoretical probability of flipping a coin and getting heads? Answer: P(getting heads) = 1 favorable outcome / 2 total outcomes = 1/2.

Sample Questions

Read 3 questions and answers free. Sign up to access all 99 questions with full KNEC-style marking schemes and a personalised study plan.

1
easySHORT ANSWER3 marks

State whether the events of flipping a coin and rolling a die are independent. Justify your answer. (3 marks)

Answer & marking scheme

Part (a) — 1 mark
The events are independent. (1 mk)
Part (b) — 2 marks
The outcome of flipping a coin does not affect the outcome of rolling a die. (1 mk)
Mathematically, P(A and B) = P(A) * P(B) for independent events. (1 mk)
2
easySHORT ANSWER2 marks

State the probability of not rolling a six when a fair die is thrown once. (2 marks)

Answer & marking scheme

Part (a) — 2 marks
P(not rolling a six) = 1 - P(rolling a six) (1 mk)
P(rolling a six) = 1/6, thus P(not rolling a six) = 5/6 (1 mk)
3
easySHORT ANSWER4 marks

A spinner is divided into 8 equal sections numbered from 1 to 8. (a) Describe the sample space when the spinner is spun once. (2 marks) (b) Find the probability of landing on an even number. (2 marks)

Answer & marking scheme

Part (a) — 2 marks
Sample space: {1, 2, 3, 4, 5, 6, 7, 8} (1 mk)
Total outcomes = 8 (1 mk)
Part (b) — 2 marks
Favourable outcomes = 4 (Even numbers: 2, 4, 6, 8) (1 mk)
P(Even) = 4/8 = 1/2 (1 mk)
4

A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. (a) State the sample space when one marble is drawn. (2 marks) (b) Calculate the probability of drawing a blue marble. (2 marks)

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Frequently asked questions

What does the KCSE Mathematics topic "Probability: theoretical probability" cover?

Probability: theoretical probability covers Define the sample space, event and probability; calculate P(E) = n(favourable outcomes)/n(sample space) for equally likely outcomes; Apply the complementary rule P(E') = 1 – P(E) and identify mutually exclusive and independent events; Probability: theoretical probability, all aligned to the official KNEC KCSE Mathematics syllabus.

How many practice questions are available for Probability: theoretical probability?

HighMarks has 99 Probability: theoretical probability practice questions for KCSE Mathematics, each with a full marking scheme. The first 3 are free; sign up to access the rest, plus all KCSE mock exams and past papers.

Are these aligned with the KNEC KCSE syllabus?

Yes. Every objective on this page is taken directly from the official KNEC KCSE Mathematics syllabus. Practice questions match the KCSE exam format and are graded against the standard KNEC marking scheme.

How should I revise Probability: theoretical probability for the KCSE exam?

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