A die is thrown, find the probability of following events: (i) A prime number will appear, (ii) A number great

Exercise 14.2 · Q3

Q. A die is thrown, find the probability of following events:

(i) A prime number will appear,

(ii) A number greater than or equal to 3 will appear,

(iii) A number less than or equal to one will appear,

(iv) A number more than 6 will appear,

(v) A number less than 6 will appear.

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Concept understanding — Classical Probability

Classical Probability: The "Fair Game" Definition

Imagine you roll a fair six-sided die. Before it lands, you know there are exactly six possible outcomes — 1, 2, 3, 4, 5, or 6 — and you have no reason to believe any one face is more likely than another. That gut feeling of "all outcomes are equally likely" is the entire foundation of classical probability.

The Intuition

Classical probability was born from games of chance — dice, coins, cards. In these settings, the physical symmetry of the objects (a balanced die, a fair coin) guarantees that no outcome is favoured. So the probability of an event is simply:

Number of ways the event can happen, divided by the total number of possible outcomes.

If you want the chance of rolling an even number on a die, count the evens: 2, 4, 6 — that's 3 ways. Total outcomes: 6. So probability = $3/6 = 1/2$ .

This is the "counting" approach. It works beautifully when the underlying experiment is symmetric and finite.

The Precise Statement

P(E) = \frac{\text{Number of outcomes favourable to event } E}{\text{Total number of equally likely outcomes}}

This is called the classical definition (or a priori definition) of probability. It was formalised by Pierre-Simon Laplace in the 18th century.

Three conditions must hold for this definition to apply:

Finite sample space — there are only a fixed, countable number of possible outcomes.
Equally likely outcomes — each outcome has the same chance of occurring (the "fairness" condition).
Mutually exclusive outcomes — no two outcomes can happen at the same time.

Watch out

The biggest mistake students make is applying classical probability to situations where outcomes are not equally likely. For example: "I can either pass or fail the exam — two outcomes, so probability of passing is $1/2$ ." That's nonsense, because passing and failing are not equally likely. The die works only because the die is fair.

A Simple Example

Problem: A bag contains 3 red marbles and 2 blue marbles. You pick one marble at random. What is the probability it is red?

Step 1 — Identify the sample space: There are 5 marbles total. If the marbles are physically identical except for colour, and you pick without looking, each marble is equally likely to be chosen. So total outcomes = 5.

Step 2 — Count favourable outcomes: 3 marbles are red. So favourable outcomes = 3.

Step 3 — Apply the formula:

P(\text{red}) = \frac{3}{5}

That's it. No deeper theory needed for this case.

When Classical Probability Fails

Classical probability cannot handle:

Infinite outcomes (e.g., "pick any real number between 0 and 1")
Unequally likely outcomes (e.g., "will it rain tomorrow?")
Situations where "equally likely" is not physically justified

For those, we need other definitions — relative frequency (based on repeated experiments) or axiomatic probability (Kolmogorov's modern framework). But classical probability remains the cleanest starting point, and it's still the go-to method for most exam problems involving dice, coins, cards, and lotteries.

Tip

In exam problems, the phrase "at random" or "fair" is your signal that classical probability applies. If you see "randomly selected" without further qualification, assume equally likely outcomes unless told otherwise.

Classical Probability is the starting definition used throughout the NCERT Class 11 Mathematics chapter on Probability, matching searches like "classical probability: formula and examples" or "probability important questions class 11 maths". This equally-likely-outcomes approach to dice, coins, and cards is one of the most frequently tested question types in CBSE boards and competitive exams like JEE Main and state CETs.

A standard die has six equally likely outcomes: $\{1, 2, 3, 4, 5, 6\}$ . Count favorable outcomes for each event and divide by 6 to find the probability.

When we throw a fair die, each of the six faces has an equal chance of landing face-up. This is the essence of classical probability: when all outcomes are equally likely, the probability of an event is simply the ratio of favorable outcomes to total possible outcomes.

P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}

For a single die, the total number of outcomes is always 6. The sample space is $S = \{1, 2, 3, 4, 5, 6\}$ .

Let me work through each event systematically.

(i) A prime number will appear

Identify prime numbers on the die. A prime number has exactly two distinct divisors: 1 and itself. Among the numbers on a die:
- 1 is not prime (by definition, primes must be greater than 1)
- 2 is prime
- 3 is prime
- 4 = 2 × 2, not prime
- 5 is prime
- 6 = 2 × 3, not prime
Count favorable outcomes. The prime numbers are $\{2, 3, 5\}$ , giving us 3 favorable outcomes.
Calculate probability. $P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$

(ii) A number greater than or equal to 3 will appear

Identify qualifying numbers. We need numbers where $n \geq 3$ : these are $\{3, 4, 5, 6\}$ .
Count favorable outcomes. We have 4 favorable outcomes.
Calculate probability. $P(n \geq 3) = \frac{4}{6} = \frac{2}{3}$

(iii) A number less than or equal to one will appear

Identify qualifying numbers. We need $n \leq 1$ . On a standard die, only the number 1 satisfies this condition.
Count favorable outcomes. Just 1 favorable outcome: $\{1\}$ .
Calculate probability. $P(n \leq 1) = \frac{1}{6}$

(iv) A number more than 6 will appear

Check the sample space. A standard die shows only the numbers 1 through 6. There is no face showing 7 or any number greater than 6.
Count favorable outcomes. Zero favorable outcomes.
Calculate probability. $P(n > 6) = \frac{0}{6} = 0$

Note

This is an impossible event. Its probability is 0, meaning it can never occur when throwing a standard die.

(v) A number less than 6 will appear

Identify qualifying numbers. We need $n < 6$ : these are $\{1, 2, 3, 4, 5\}$ .
Count favorable outcomes. We have 5 favorable outcomes.
Calculate probability. $P(n < 6) = \frac{5}{6}$

Tip

Notice that events (iv) and (v) are nearly complementary. If we included "equal to 6" in event (v), we'd have $P(n \leq 6) = 1$ (a certain event), and together with $P(n > 6) = 0$ , they would sum to 1.

Event	Favorable Outcomes	Probability
(i) Prime number	$\{2, 3, 5\}$	$\frac{1}{2}$
(ii) $n \geq 3$	$\{3, 4, 5, 6\}$	$\frac{2}{3}$
(iii) $n \leq 1$	$\{1\}$	$\frac{1}{6}$
(iv) $n > 6$	$\{\}$	$0$
(v) $n < 6$	$\{1, 2, 3, 4, 5\}$	$\frac{5}{6}$

✓Final answer

\boxed{\frac{1}{2}}

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Q.A die is thrown, find the probability of following events:

(i) A prime number will appear

(ii) A number greater than or equal to 3 will appear

(iii) A number less than or equal to one will appear

(iv) A number more than 6 will appear

(v) A number less than 6 will appear

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