**PROBABILITY: THEORETICAL APPROACH**
**Definition & Core Concept**
• Theoretical Probability P(E) = (Number of outcomes favourable to E) / (Total number of possible outcomes)
• Assumes all outcomes are equally likely
• Classical probability defined by Pierre Simon Laplace (1795)
• Different from experimental probability which uses: P(E) = (Trials where event happened) / (Total trials)
**When to Use Theoretical vs Experimental Probability**
**Essential Conditions for Using Theoretical Probability**
• Equally likely outcomes must be assumed or proven
• Outcomes must be mutually exclusive (cannot occur together)
• All possible outcomes must be identifiable in advance
• Fair/unbiased equipment (fair coin, fair die)
**Important Definitions**
• Elementary Event: An event with only one outcome of the experiment
• Compound Event: An event with more than one outcome
• Complement of Event E (denoted Ē or 'not E'): All outcomes not in E
**Fundamental Theorem on Complementary Events**
P(E) + P(Ē) = 1
Therefore: P(Ē) = 1 – P(E)
**Key Property**
• Sum of probabilities of all elementary events = 1
• Proof insight: All elementary events cover the entire sample space with no overlap, so their combined probability equals 1
**Range of Probability**
• For any event E: 0 ≤ P(E) ≤ 1
• P(E) = 0 → Impossible event (never occurs)
• P(E) = 1 → Certain/sure event (always occurs)
• 0 < P(E) < 1 → Possible event (may or may not occur)
**Step-by-Step Solution Method**
1. Identify the experiment clearly
2. List all possible outcomes (sample space)
3. Count total number of possible outcomes
4. Identify outcomes favourable to the given event
5. Count number of favourable outcomes
6. Apply formula: P(E) = Favourable outcomes / Total outcomes
7. Simplify the fraction if needed
**Common Problem Types & Solutions**
**Type 1: Simple Events (Coin/Single Die)**
• Coin toss: P(Head) = 1/2, P(Tail) = 1/2
• Die roll: P(any number 1-6) = 1/6
• Multiple outcomes: P(odd numbers) = 3/6 = 1/2
**Type 2: Drawing Balls from Bag**
• Total outcomes = Total number of balls
• Favorable outcomes = Number of balls matching condition
• Example: 4 red + 1 blue ball → P(red) = 4/5, P(blue) = 1/5
**Type 3: Conditional Probability (Implicit)**
• "Greater than" or "less than" conditions
• Example: Die showing >4 means outcomes {5,6} → P(E) = 2/6 = 1/3
**Type 4: Complement Events**
• If P(E) is given, find P(not E)
• Example: P(Head) = 1/2, then P(not Head) = P(Tail) = 1 – 1/2 = 1/2
**Common Mistakes to Avoid**
❌ Not assuming equally likely outcomes without checking
❌ Forgetting that sample space must include ALL possible outcomes
❌ Miscounting favorable outcomes (missing or duplicating)
❌ Writing probability > 1 or < 0 (immediately indicates error)
❌ Confusing experimental with theoretical probability
❌ Not simplifying fractions in final answer
❌ Assuming outcomes are equally likely when they are not (4 red + 1 blue ball → not equally likely outcomes)
❌ Forgetting P(E) + P(Ē) = 1 relationship
**Important Remarks**
• Probability is always expressed as a fraction, decimal, or percentage between 0 and 1
• In equally likely outcomes, each elementary event has equal probability
• The complement relationship is powerful for complex events (easier to calculate 1 – P(E) sometimes)
• Historical note: Probability theory originated in 16th century with Cardan's work on games of chance; Laplace's 1812 work is foundational
**Quick Reference Formulas**
• P(E) = Favorable outcomes / Total outcomes
• P(Ē) = 1 – P(E)
• Total probability in sample space = 1
• If events are elementary: Sum of all P(E₁) + P(E₂) + ... + P(Eₙ) = 1
Q1. A bag contains 4 red balls and 1 blue ball. Kavya draws a ball without looking. She claims that the outcomes 'red ball' and 'blue ball' are equally likely. Using the chapter's definition of probability, her claim is —
Answer: B — The chapter explicitly states that outcomes are equally likely only when each has the same chance of occurring; here 4 red vs 1 blue means drawing red is more likely, so the outcomes (red or blue) are NOT equally likely, making option B the correct conceptual understanding.
Q2. Assertion (A): The sum of probabilities of all elementary events in any experiment equals 1. Reason (R): Each elementary event represents exactly one outcome, and the total probability must account for all possible outcomes. Choose the correct option:
Answer: A — Assertion A is the direct definition from the chapter (Example 1 and 2 demonstrate this); Reason R correctly explains why—because elementary events are mutually exclusive and exhaustive, their probabilities must sum to 1.
Q3. When a die is thrown, a student writes: 'The probability of getting a number greater than 4 is 1/3, and the probability of getting a number less than or equal to 4 is 2/3.' The student's reasoning is —
Answer: B — Numbers > 4 are {5, 6}, giving P = 2/6 = 1/3; numbers ≤ 4 are {1, 2, 3, 4}, giving P = 4/6 = 2/3; these sum to 1, so the student's statement is actually correct—the distractor B tests whether students recognize that complementary events must sum to 1.
Q4. The chapter defines theoretical probability assuming equally likely outcomes. A teacher asks: 'When is this assumption valid?' The most accurate answer is —
Answer: B — The chapter explicitly states this assumption is valid 'in many experiments' but not all—it must hold when outcomes are truly equally likely (fair coin, fair die), not for asymmetric situations like the 4 red + 1 blue ball example.
Q5. Assertion (A): If an event E is an elementary event, then P(E) must be less than 1. Reason (R): Theoretical probability is defined as the ratio of favourable outcomes to total possible outcomes. Choose the correct option:
Answer: B — A is true (an elementary event has 1 favourable outcome out of n total, so P < 1); R is also true (it restates the definition); but R does NOT explain why P(E) < 1 for an elementary event—the explanation is that one outcome cannot be all possible outcomes.
Q6. A bag contains 3 identical balls. Kritika draws one ball without looking. Which statement about this experiment is INCORRECT?
Answer: D — Options A, B, C are all correct by the chapter's definitions; option D is incorrect because empirical probability approaches theoretical probability only as the number of trials becomes very large, not exactly in finite repetitions.
Q7. Assertion (A): The probability of getting a head when a fair coin is tossed is 1/2. Reason (R): A fair coin is symmetrical, so there is no reason for it to come down more often on one side than the other. Choose the correct option:
Answer: A — Both statements are true and directly linked: R describes the assumption (fairness and symmetry) that justifies the equally likely outcomes assumption, which leads to A (P = 1/2).
Q8. A student claims: 'Since I got 3 heads in 5 coin tosses, the experimental probability of heads is 3/5, so the theoretical probability should also be 3/5.' This claim is —
Answer: B — The chapter distinguishes between empirical (experimental) probability from repeated trials and theoretical probability calculated from assumptions about equally likely outcomes; the student confuses the two—empirical approaches theoretical only as trials increase, not in small samples.
Q9. When drawing a ball from a bag with 1 red, 1 blue, and 1 yellow ball, a student correctly identifies P(red) = 1/3. Her reasoning is: 'There are 3 possible outcomes, and 1 is favourable to red.' Is this reasoning —
Answer: A — The student directly applies the theoretical probability definition: P(E) = number of favourable outcomes / total outcomes = 1/3, correctly assuming equally likely outcomes (all balls same size).
Q10. Assertion (A): The empirical probability of an event requires repeating the experiment many times to be accurate. Reason (R): Theoretical probability can be calculated directly from equally likely outcomes without repetition. Choose the correct option:
Answer: B — Both A and R are true per the chapter; however, R does not explain A—R explains the advantage of theoretical probability, not why empirical probability requires repetition (which is because it's computed from trial outcomes, not from prior assumptions).
Define theoretical probability P(E).
P(E) = Number of favourable outcomes ÷ Number of all possible outcomes, assuming equally likely outcomes.
What is an elementary event?
An elementary event has exactly one outcome of the experiment.
What does 'equally likely outcomes' mean?
Each outcome of an experiment has the same chance of occurring (e.g., heads and tails in a fair coin toss).
State the complement rule for probability.
P(not E) = 1 − P(E), where not E is the complement of event E.
What is the sum of probabilities of all elementary events?
The sum of probabilities of all elementary events in an experiment always equals 1.
Why can't we always use empirical probability?
Empirical probability requires repeating an experiment many times, which is impractical or expensive for events like satellite launches or earthquakes.
When throwing a fair die once, what is P(number ≤ 4)?
P(number ≤ 4) = 4/6 = 2/3, since favourable outcomes are 1, 2, 3, 4.
What assumption must hold for the Laplace definition of probability?
All outcomes of the experiment must be equally likely.
Are drawing outcomes 'red ball' and 'blue ball' equally likely from a bag with 4 red and 1 blue ball?
No; red ball is more likely because there are 4 red balls versus only 1 blue ball.
What does 'fair' or 'unbiased' mean for a coin or die?
Fair means symmetrical with no bias, so each outcome has an equal chance of occurring.
Define equally likely outcomes. Give one example where outcomes are equally likely and one where they are not. [2 marks]
Equally likely outcomes have the same chance of occurring. Example 1: coin toss (H, T both P = 1/2). Example 2: bag with 4 red and 1 blue ball (red more likely than blue).
A die is thrown once. Find the probability of getting (i) a number greater than 4, and (ii) the complement of this event. Verify that P(E) + P(not E) = 1. [3 marks]
For (i), favourable outcomes are 5, 6 (count = 2 out of 6), so P = 2/6 = 1/3. For (ii), not E is 'number ≤ 4' with outcomes 1, 2, 3, 4 (count = 4), so P(not E) = 4/6 = 2/3. Then verify 1/3 + 2/3 = 1.
A bag contains 2 red balls, 3 blue balls, and 1 yellow ball. A ball is drawn at random. Find the probability of each colour, verify that the sum of all elementary event probabilities equals 1, and explain why you can use theoretical probability here. [5 marks]
Total = 6 balls. P(red) = 2/6 = 1/3, P(blue) = 3/6 = 1/2, P(yellow) = 1/6. Sum = 1/3 + 1/2 + 1/6 = 1. Theoretical probability applies because outcomes (drawing each ball) are equally likely, and we don't need to repeat the experiment; Laplace's assumption of equally likely outcomes holds.
Practice with interactive flashcards, mind maps, upload your own chapters and get AI study kits instantly
Try StudyOS Free →