Determine the probability $p$, for each of the following events. (a) An odd number appears in a single toss of a fair die. (b) At least one head appears in two tosses of a fair coin. (c) A king, 9 of hearts, or 3 of spades appears in drawing a single card from a well shuffled ordinary deck of 52 cards. (d) The sum of 6 appears in a single toss of a pair of fair dice.

> [!TLDR] > This problem applies the fundamental principle of classical probability, $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$, to four distinct scenarios, yielding probabilities of $\frac{1}{2}$, $\frac{3}{4}$, $\frac{3}{26}$, and $\frac{5}{36}$ respectively. Probability is a measure of the likelihood of an event occurring. When all possible outcomes of an experiment are equally likely, we use the concept of **classical probability**. This is the most intuitive way to understand probability and forms the basis for many problems in competitive exams. The core idea is simple: To find the probability of an event $E$, we count how many ways that event can happen (favorable outcomes) and divide it by the total number of distinct ways the experiment can turn out (total possible outcomes). > [!FORMULA] > The probability of an event $E$, denoted $P(E)$, is given by: > $$P(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes}}$$ Let's apply this principle to each part of the problem. ### (a) An odd number appears in a single toss of a fair die. 1. **Identify the experiment and its total possible outcomes.** When a fair die is tossed, the possible outcomes are the numbers on its faces: $1, 2, 3, 4, 5, 6$. The total number of possible outcomes is $N = 6$. 2. **Identify the event and its favorable outcomes.** The event $E$ is "an odd number appears". The odd numbers among the possible outcomes are $1, 3, 5$. The number of favorable outcomes is $n(E) = 3$. 3. **Calculate the probability.** Using the classical probability formula: $P(E) = \frac{n(E)}{N} = \frac{3}{6} = \frac{1}{2}$. ### (b) At least one head appears in two tosses of a fair coin. 1. **Identify the experiment and its total possible outcomes.** When a fair coin is tossed twice, we can list all possible sequences of outcomes. Let H denote a Head and T denote a Tail. The possible

Determine the probability p, for each of the following events. (a) An odd number appears in a single toss of a

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(a) An odd number appears in a single toss of a fair die.

(b) At least one head appears in two tosses of a fair coin.

(d) The sum of 6 appears in a single toss of a pair of fair dice.

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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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This problem applies the fundamental principle of classical probability, $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$ , to four distinct scenarios, yielding probabilities of $\frac{1}{2}$ , $\frac{3}{4}$ , $\frac{3}{26}$ , and $\frac{5}{36}$ respectively.

Probability is a measure of the likelihood of an event occurring. When all possible outcomes of an experiment are equally likely, we use the concept of classical probability. This is the most intuitive way to understand probability and forms the basis for many problems in competitive exams.

The core idea is simple:

To find the probability of an event $E$ , we count how many ways that event can happen (favorable outcomes) and divide it by the total number of distinct ways the experiment can turn out (total possible outcomes).

E

Let's apply this principle to each part of the problem.

(a) An odd number appears in a single toss of a fair die.

Identify the experiment and its total possible outcomes.

When a fair die is tossed, the possible outcomes are the numbers on its faces: $1, 2, 3, 4, 5, 6$ .

The total number of possible outcomes is $N = 6$ .
Identify the event and its favorable outcomes.

The event $E$ is "an odd number appears". The odd numbers among the possible outcomes are $1, 3, 5$ .

The number of favorable outcomes is $n(E) = 3$ .
Calculate the probability.

Using the classical probability formula:

$P(E) = \frac{n(E)}{N} = \frac{3}{6} = \frac{1}{2}$ .

(b) At least one head appears in two tosses of a fair coin.

Identify the experiment and its total possible outcomes.

When a fair coin is tossed twice, we can list all possible sequences of outcomes. Let H denote a Head and T denote a Tail.

The possible outcomes are:
- HH (Head on first toss, Head on second toss)
- HT (Head on first toss, Tail on second toss)
- TH (Tail on first toss, Head on second toss)
- TT (Tail on first toss, Tail on second toss) The total number of possible outcomes is $N = 4$ .
Identify the event and its favorable outcomes.

The event $E$ is "at least one head appears". This means we are looking for outcomes where there is one head or two heads.

The favorable outcomes are: HH, HT, TH.

The number of favorable outcomes is $n(E) = 3$ .
Calculate the probability.

Using the classical probability formula:

$P(E) = \frac{n(E)}{N} = \frac{3}{4}$ .

Tip

For "at least one" problems, it's often easier to calculate the probability of the complementary event ("none") and subtract it from 1.

The complement of "at least one head" is "no heads" (i.e., both tails, TT).

$P(\text{no heads}) = P(\text{TT}) = \frac{1}{4}$ .

So, $P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - \frac{1}{4} = \frac{3}{4}$ . This confirms our result.

(c) A king, 9 of hearts, or 3 of spades appears in drawing a single card from a well shuffled ordinary deck of 52 cards.

Identify the experiment and its total possible outcomes.

A single card is drawn from a standard deck of 52 cards.

The total number of possible outcomes is $N = 52$ .
Identify the event and its favorable outcomes.

The event $E$ is "drawing a king, a 9 of hearts, or a 3 of spades".

Let's count the favorable outcomes:
- There are 4 kings in a deck (King of Spades, King of Clubs, King of Hearts, King of Diamonds).
- There is only 1 "9 of hearts".
- There is only 1 "3 of spades". These are all distinct cards, meaning they are mutually exclusive events (drawing a King does not prevent drawing a 9 of hearts, and a 9 of hearts is not a King). So, the number of favorable outcomes is $n(E) = 4 + 1 + 1 = 6$ .
Calculate the probability.

Using the classical probability formula:

$P(E) = \frac{n(E)}{N} = \frac{6}{52}$ .

This fraction can be simplified by dividing both the numerator and denominator by 2. …

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Q.Determine the probability ppp, for each of the following events.

(a) An odd number appears in a single toss of a fair die.

(b) At least one head appears in two tosses of a fair coin.

(c) A king, 9 of hearts, or 3 of spades appears in drawing a single card from a well shuffled ordinary deck of 52 cards.

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Q.Determine the probability $p$ , for each of the following events.

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