A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and siz

Worked Examples · Example 6

Q. A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be

(i) red,

(ii) yellow,

(iii) blue,

(iv) not blue,

(v) either red or blue.

Sikkim CbseNCERTSubjective· 2mImportance★★★★★est

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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.

Classical probability: count favourable outcomes and divide by total outcomes. With 9 discs (4 red, 3 blue, 2 yellow), we compute the probability of each event by forming the ratio of discs matching the condition to the total number of discs.

Understanding Classical Probability

When every outcome in a sample space is equally likely, the probability of an event is simply the ratio of favourable outcomes to total outcomes. Here we have 9 discs, each with an equal chance of being drawn. The key is to identify how many discs satisfy each condition, then divide by 9.

The sample space has $n(S) = 9$ equally likely outcomes (one for each disc).

Solution

1. Probability that the disc is red

We have 4 red discs out of 9 total discs. The number of favourable outcomes is 4.

P(\text{red}) = \frac{\text{Number of red discs}}{\text{Total number of discs}} = \frac{4}{9}

2. Probability that the disc is yellow

There are 2 yellow discs in the bag.

P(\text{yellow}) = \frac{2}{9}

3. Probability that the disc is blue

The bag contains 3 blue discs.

P(\text{blue}) = \frac{3}{9} = \frac{1}{3}

4. Probability that the disc is not blue

"Not blue" means the disc is either red or yellow. We can approach this in two ways:

Method 1 (Direct counting): Red discs + Yellow discs = $4 + 2 = 6$ discs.

P(\text{not blue}) = \frac{6}{9} = \frac{2}{3}

Method 2 (Complement rule): Since $P(\text{blue}) = \frac{1}{3}$ , we have

P(\text{not blue}) = 1 - P(\text{blue}) = 1 - \frac{1}{3} = \frac{2}{3}

Tip

The complement rule $P(A^c) = 1 - P(A)$ is often faster when computing "not" probabilities, especially when the original event is simpler to count.

5. Probability that the disc is either red or blue

"Either red or blue" means we count all red discs and all blue discs (these events are mutually exclusive—a disc cannot be both red and blue).

Number of red or blue discs = $4 + 3 = 7$ .

P(\text{red or blue}) = \frac{7}{9}

Alternatively, using the addition rule for mutually exclusive events:

P(\text{red or blue}) = P(\text{red}) + P(\text{blue}) = \frac{4}{9} + \frac{3}{9} = \frac{7}{9}

✓Final answer

\boxed{\frac{4}{9}}

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Q.A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be

Understanding Classical Probability

Solution

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