One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the prob

Worked Examples · Example 5

Q. One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be

(i) a diamond

(ii) not an ace

(iii) a black card (i.e., a club or a spade)

(iv) not a diamond

(v) not a black card.

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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 is the ratio of favourable outcomes to total outcomes. For a standard 52-card deck, the probabilities are: (i) diamond = $\frac{1}{4}$ ,

(ii) not an ace = $\frac{12}{13}$ ,

(iii) black card = $\frac{1}{2}$ ,

(iv) not a diamond = $\frac{3}{4}$ ,

(v) not a black card = $\frac{1}{2}$ .

The Core Idea: Classical Probability

When every outcome in a random experiment is equally likely, the probability of an event is simply:

P(\text{event}) = \frac{\text{Number of outcomes favourable to the event}}{\text{Total number of possible outcomes}}

This is the classical definition of probability. It works perfectly here because the deck is well-shuffled — each of the 52 cards has the same chance of being drawn. So the denominator for every part is 52. The numerator changes depending on what we're counting.

Let’s recall the composition of a standard deck:

4 suits: spades (♠), hearts (♥), diamonds (♦), clubs (♣)
Each suit has 13 cards: Ace, 2–10, Jack, Queen, King
Spades and clubs are black (26 cards total); hearts and diamonds are red (26 cards total)
There are 4 aces (one per suit)

Now we work through each part.

(i) Probability of drawing a diamond

There are 13 diamonds in the deck. Favourable outcomes = 13. Total outcomes = 52.

P(\text{diamond}) = \frac{13}{52} = \frac{1}{4}

Tip

Since each suit has the same number of cards, the probability of any specific suit is always $\frac{1}{4}$ .

(ii) Probability of not drawing an ace

There are 4 aces. So the number of cards that are not aces = $52 - 4 = 48$ .

P(\text{not an ace}) = \frac{48}{52} = \frac{12}{13}

Watch out

A common mistake is to think "not an ace" means 51 cards (removing just one ace). But there are 4 aces, so you must subtract all 4.

(iii) Probability of a black card

Black cards are clubs and spades — 13 each, total 26.

P(\text{black card}) = \frac{26}{52} = \frac{1}{2}

(iv) Probability of not drawing a diamond

If diamonds are 13 cards, then non-diamonds are $52 - 13 = 39$ .

P(\text{not a diamond}) = \frac{39}{52} = \frac{3}{4}

Note

Notice that "not a diamond" is the complement of "diamond". So $P(\text{not diamond}) = 1 - P(\text{diamond}) = 1 - \frac{1}{4} = \frac{3}{4}$ . This is a faster way.

(v) Probability of not drawing a black card

Black cards are 26. So non-black (i.e., red cards) = $52 - 26 = 26$ .

P(\text{not a black card}) = \frac{26}{52} = \frac{1}{2}

Again, this is the complement of part (iii): $1 - \frac{1}{2} = \frac{1}{2}$ .

✓Final answer

\frac{1}{4}

←Previous section · 14.2.4Probability of Event ‘not A’→Next question · 14.2.4Example 6 — A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow…

Q.One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be