Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multipl

NCERT Exemplar · Q3

Q. Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.

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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.

By inclusion–exclusion the count is $500+111-55=556$ , so $P=\dfrac{556}{1000}=\dfrac{139}{250}$ .

Solution

Let $A$ be the set of multiples of $2$ and $B$ the set of multiples of $9$ among $\{1,2,\dots,1000\}$ . We want $P(A\cup B)$ .

Count each set:

|A|=\left\lfloor\frac{1000}{2}\right\rfloor=500,\qquad |B|=\left\lfloor\frac{1000}{9}\right\rfloor=111.

A number divisible by both $2$ and $9$ is divisible by $\mathrm{lcm}(2,9)=18$ :

|A\cap B|=\left\lfloor\frac{1000}{18}\right\rfloor=55.

By the inclusion–exclusion principle,

|A\cup B|=|A|+|B|-|A\cap B|=500+111-55=556.

With $1000$ equally likely integers,

P(A\cup B)=\frac{556}{1000}=\frac{139}{250}.

✓Final answer

2

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 mark
$Q. One card is drawn from a well shuffled deck of 52 cards. If each outcomes is equally likely, then what is the probability that the card will not be a diamond.$
›Reveal solutionSolution

With 13 diamonds out of 52 cards, the probability of not drawing a diamond is 3/4.

$P(\text{diamond}) = \dfrac{13}{52} = \dfrac14$

$P(\text{not diamond}) = 1 - P(\text{diamond}) = 1-\dfrac14 = \dfrac34$

✓Final answer
$P(not a diamond) = 3/4.$
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2025Set ANNUAL1 mark
$Q. Two dice are thrown. What is the probability of getting a doublet?$
›Reveal solutionSolution

Out of 36 equally likely outcomes when two dice are thrown, 6 are doublets, giving probability $1/6$ .

When two dice are thrown, the total number of equally likely outcomes is $6 \times 6 = 36$ .

A "doublet" means both dice show the same face: $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$ — that's $6$ favourable outcomes.

Probability $= \dfrac{\text{favourable outcomes}}{\text{total outcomes}} = \dfrac{6}{36} = \dfrac{1}{6}$ .

✓Final answer
$\dfrac{1}{6}$
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2024Set ANNUAL1 mark
$Q. A bag contains 9 red and 12 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.$
›Reveal solutionSolution

$P(\text{white ball}) = \dfrac{4}{7}$ .

Total number of balls $= 9+12 = 21$ . Number of white balls $=12$ . Since one ball is drawn at random, each ball is equally likely, so $P(\text{white}) = \dfrac{\text{number of white balls}}{\text{total balls}} = \dfrac{12}{21} = \dfrac{4}{7}.$

✓Final answer
$P(\text{white}) = \dfrac{4}{7}$

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Q.Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.

Solution

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