A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability

Exercise 14.2 · Q5

Q. A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is

(i) 3

(ii) 12.

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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 coin has faces 1 and 6; the die has faces 1–6. The total number of equally likely outcomes is $2 \times 6 = 12$ . The probability of sum 3 is $\frac{1}{12}$ ; the probability of sum 12 is also $\frac{1}{12}$ .

Why classical probability works here

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

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

This is the classical definition of probability. It applies because the coin is fair (both faces equally likely) and the die is fair (all six faces equally likely). The coin and die are tossed independently, so each combination of coin face and die face is equally likely.

The key is to first list all possible outcomes systematically, then count how many give the required sum.

Step-by-step solution

1. Determine the sample space

The coin shows either $1$ or $6$ . The die shows any of $1, 2, 3, 4, 5, 6$ . Each pair (coin, die) is an outcome.

Total outcomes = $2 \times 6 = 12$ .

We can list them:

Coin	Die	Sum
1	1	2
1	2	3
1	3	4
1	4	5
1	5	6
1	6	7
6	1	7
6	2	8
6	3	9
6	4	10
6	5	11
6	6	12

2. Find outcomes where sum = 3

From the table, only one outcome gives sum 3: coin = 1, die = 2.

Favourable outcomes = 1.

P(\text{sum}=3) = \frac{1}{12}

3. Find outcomes where sum = 12

Only one outcome gives sum 12: coin = 6, die = 6.

Favourable outcomes = 1.

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Q.A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is

Why classical probability works here

Step-by-step solution

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