While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards to be of different colours (A) $\frac{29}{52}$ (B) $\frac{1}{2}$ (C) $\frac{26}{51}$ (D) $\frac{27}{51}$

While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing

NCERT Exemplar · Q20

\frac{29}{52}

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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: …

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We use classical probability and combinations to find the probability of drawing two cards of different colours from a 52-card deck. The total number of ways to draw two cards is $C(52, 2)$ , and the number of ways to draw one red and one black card is $C(26, 1) \times C(26, 1)$ . The probability is $\boxed{\frac{26}{51}}$ .

When dealing with probability problems like this, where we're selecting items from a group and the order of selection doesn't matter, the concept of classical probability combined with combinations is key.

Classical probability states that if all outcomes of an experiment are equally likely, the probability of an event $E$ occurring is given by the ratio:

P(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes}}

In this problem, our "experiment" is accidentally dropping two cards from a shuffled deck. We want to find the probability that these two missing cards are of different colours.

Let's break this down step-by-step.

Understand the Deck Composition:

A standard deck of 52 playing cards consists of:
- 26 red cards (13 Hearts and 13 Diamonds)
- 26 black cards (13 Clubs and 13 Spades) The problem asks for cards of "different colours", meaning one red card and one black card.
Calculate the Total Number of Possible Outcomes:

This is the total number of ways to choose any 2 cards from the 52 cards in the deck. Since the order in which the cards are dropped doesn't matter, we use combinations.

The number of ways to choose $r$ items from a set of $n$ items is given by the combination formula: $C(n, r) = \frac{n!}{r!(n-r)!}$ .

Here, $n=52$ (total cards) and $r=2$ (cards dropped).

C(52, 2) = \frac{52!}{2!(52-2)!} = \frac{52!}{2!50!} = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326

So, there are 1326 different pairs of cards that could be dropped.

3. Calculate the Number of Favorable Outcomes:

We want the two missing cards to be of different colours. This means we need to choose one red card AND one black card.

* Number of ways to choose 1 red card from the 26 red cards:

C(26, 1) = \frac{26!}{1!(26-1)!} = \frac{26!}{1!25!} = 26

*   Number of ways to choose 1 black card from the 26 black cards:
    $$C(26, 1) = \frac{26!}{1!(26-1)!} = \frac{26!}{1!25!} = 26$$ …

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Q.While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards to be of different colours (A) 2952\frac{29}{52}5229​ (B) 12\frac{1}{2}21​ (C) 2651\frac{26}{51}5126​ (D) 2751\frac{27}{51}5127​

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Q.While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards to be of different colours
(A) $\frac{29}{52}$
(B) $\frac{1}{2}$
(C) $\frac{26}{51}$
(D) $\frac{27}{51}$

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