Without repetition of the numbers, four digit numbers are formed with the numbers $0, 2, 3, 5$. The probability of such a number divisible by 5 is (A) $\frac{1}{5}$ (B) $\frac{4}{5}$ (C) $\frac{1}{30}$ (D) $\frac{5}{9}$

Without repetition of the numbers, four digit numbers are formed with the numbers 0, 2, 3, 5. The probability

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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 count all four-digit numbers formed without repetition from {0,2,3,5}, then count those divisible by 5 (last digit 0 or 5). The probability is $\frac{5}{9}$ , which corresponds to option (D).

Concept and Intuition

Classical probability is simply:

P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}

Here, the "outcomes" are all distinct four-digit numbers we can form using the digits 0, 2, 3, 5 exactly once each. The key constraint: a four-digit number cannot start with 0 — that would make it a three-digit number. So the total count isn't just $4!$ ; we must exclude numbers beginning with 0.

For divisibility by 5, a number must end in 0 or 5. That's the core condition. We'll count favourable cases carefully, watching out for the "first digit can't be 0" rule in each scenario.

Step-by-step solution

1. Count total four-digit numbers (without repetition)

We have four distinct digits: 0, 2, 3, 5.

Total permutations of all four digits = $4! = 24$ .

But numbers starting with 0 are invalid (they'd be three-digit numbers).

How many start with 0? Fix 0 in the first place; the remaining three digits (2,3,5) can be arranged in $3! = 6$ ways.

So total valid four-digit numbers:

24 - 6 = 18

Tip

A faster way: choose the first digit from {2,3,5} (3 choices), then arrange the remaining 3 digits in any order in the last three places ( $3! = 6$ ). So $3 \times 6 = 18$ . Same result.

2. Count favourable numbers (divisible by 5)

A number is divisible by 5 iff its last digit is 0 or 5. We handle these two cases separately.

Case 1: Last digit = 0

If the last digit is fixed as 0, the first three digits must be a permutation of {2,3,5}.

No restriction on the first digit here — 2, 3, and 5 are all non-zero, so every arrangement is valid.

Number of ways: $3! = 6$ .

Case 2: Last digit = 5 …

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

GUJCET 2024Set 131 markMCQ
$\frac{1}{36}$
›Reveal solutionSolution

The only even prime number is $2$ , so both dice must show $2$ — probability $\tfrac16\times\tfrac16=\tfrac1{36}$ .

Concept. A prime that is even can only be $2$ (every other even number is divisible by $2$ ). So on a single die, the favourable outcome is exactly $\{2\}$ , giving $P = \tfrac16$ . …
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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. The probability of obtaining an even number on each die, when a pair of dice is rolled is ______.$
(a) $\frac{1}{9}$
(b) $\frac{1}{2}$
(c) $\frac{1}{4}$
(d) $\frac{1}{36}$
›Reveal solutionSolution

For independent events, multiply the individual probabilities.

On one die, $P(\text{even})=\frac{3}{6}=\frac{1}{2}$ (outcomes $2,4,6$ ).

…

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GUJCET 2023Set 091 markMCQ
$\frac{5}{16}$
›Reveal solutionSolution

Binomial: $P(X=k)=\binom{n}{k}p^k q^{n-k}$ with $p=\tfrac12$ .

Concept: For 5 tosses of a fair coin, …
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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. If three balanced six-faced dice are thrown together, what is the total number of sample points in the resulting sample space?$
(a) $6^2$
(b) $3^6$
(c) $6 \times 3$
(d) $6^3$
›Reveal solutionSolution

For independent trials, the sample-space size multiplies: 3 dice, each with 6 faces,

give $6^3$ sample points.

When a single fair six-faced die is thrown, its sample space has $6$ equally likely

outcomes $\{1,2,3,4,5,6\}$ . When three such dice are thrown together, each die's outcome

is independent of the others, so every outcome of the combined experiment is an ordered …

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GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$Q. The probability of obtaining an even prime number on each dice when a pair of dice is rolled is ___.$
(a) 1
(b) 0
(c) $1/36$
(d) $35/36$
›Reveal solutionSolution

The only even prime number is $2$ ; find the probability that both dice independently show $2$ .

The only prime that is also even is $2$ (every other prime is odd). So we need $P(\text{both dice show }2)$ .

…

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Q.Without repetition of the numbers, four digit numbers are formed with the numbers 0,2,3,50, 2, 3, 50,2,3,5. The probability of such a number divisible by 5 is (A) 15\frac{1}{5}51​ (B) 45\frac{4}{5}54​ (C) 130\frac{1}{30}301​ (D) 59\frac{5}{9}95​

Concept and Intuition

Step-by-step solution

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Q.Without repetition of the numbers, four digit numbers are formed with the numbers $0, 2, 3, 5$ . The probability of such a number divisible by 5 is
(A) $\frac{1}{5}$
(B) $\frac{4}{5}$
(C) $\frac{1}{30}$
(D) $\frac{5}{9}$

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