A letter is chosen at random from the word 'ASSASSINATION'. Find the probability that letter is (i) a vowel (i

Exercise 14.2 · Q10

Q. A letter is chosen at random from the word 'ASSASSINATION'. Find the probability that letter is

(i) a vowel

(ii) a consonant.

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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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The problem is a direct application of classical probability: count the total letters, count the favourable outcomes (vowels or consonants), then divide. For the word 'ASSASSINATION', the probability of picking a vowel is $\frac{6}{13}$ and the probability of picking a consonant is $\frac{7}{13}$ .

Concept and Intuition

Classical probability works when each outcome is equally likely. Here, "a letter is chosen at random" means every letter in the word has the same chance of being selected. So the probability of an event is simply:

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

The "total number of possible outcomes" is just the total number of letters in the word. The "favourable outcomes" are the number of letters that satisfy the condition (vowel or consonant). The trick is to count carefully — the word has repeated letters, so you must count each occurrence, not just distinct letters.

Watch out

A common mistake is to count only distinct letters (e.g., saying there are 4 distinct vowels: A, I, O). But the problem asks for the probability when a letter is chosen at random from the word, so each occurrence matters. The word 'ASSASSINATION' has 13 letters in total, and you must count every A, every I, etc.

Step-by-Step Solution

Count the total number of letters in the word.

Write out the word: A S S A S S I N A T I O N.

Count them: A (1), S (2), S (3), A (4), S (5), S (6), I (7), N (8), A (9), T (10), I (11), O (12), N (13).

So total letters = $13$ .
Identify the vowels in the word.

Vowels in English are A, E, I, O, U. In 'ASSASSINATION', the vowels present are A, I, O.

Now count how many times each appears:
- A appears at positions 1, 4, 9 → 3 times.
- I appears at positions 7, 11 → 2 times.
- O appears at position 12 → 1 time. Total vowels = $3 + 2 + 1 = 6$ .
Probability of picking a vowel.

P(\text{vowel}) = \frac{\text{Number of vowels}}{\text{Total letters}} = \frac{6}{13}

Identify the consonants in the word. …

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Q.A letter is chosen at random from the word 'ASSASSINATION'. Find the probability that letter is

Concept and Intuition

Step-by-Step Solution

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