Two dice of different colours are thrown at the same time. Write down all the possible outcomes. What is the probability that :

(i) same number appears on both the dice ?

(ii) different number appears on both the dice ?

Probability

Probability with Two Dice

Official Answer

Sample space has 36 equally likely ordered outcomes (a,b), $a,b \in \{1,...,6\}$ .

(i) P(same number) = $\frac{6}{36} = \frac{1}{6}$ , since only (1,1),(2,2),...,(6,6) qualify.
(ii) P(different numbers) = $1 - \frac{1}{6} = \frac{5}{6}$ (equivalently $\frac{30}{36}$ ), as 'different' is the complement of 'same'.

sample spacetwo diceequally likely outcomesprobabilitycomplementary events1/65/6

Marking Scheme

11 mark: correctly listing/describing the full sample space of 36 ordered outcomes ( $6\times 6$ grid) since the dice are distinguishable by colour.
21 mark: correctly identifying 6 favourable outcomes for 'same number' and computing $P = \frac{6}{36} = \frac{1}{6}$ .
31 mark: correctly computing $P(\text{different number}) = 1 - \frac{1}{6} = \frac{5}{6}$ (or directly counting $\frac{30}{36}$ ), with proper justification.

Hint

List all 36 ordered pairs (since dice are different colours, (2,3) and (3,2) are distinct outcomes); count the 6 'same number' pairs for part (i), then use the complement rule for part (ii).

Quick Oral Answer

There are 36 equally likely outcomes when two distinguishable dice are thrown; 6 of these have matching numbers giving $P(\text{same})=\frac{1}{6}$ , so by the complement rule $P(\text{different})=1-\frac{1}{6}=\frac{5}{6}$ .

Analysis & Explanation

The classic two-different-coloured-dice probability question anchoring the CBSE probability chapter.

Concept

Different colours mean outcomes are ordered pairs (die1, die2), giving $6 \times 6 = 36$ equally likely outcomes.
'Same number' and 'different number' are complementary events: $P(\text{same})+P(\text{different}) = 1$ .

Key points

Favourable outcomes for same number: the 6 diagonal pairs (1,1)...(6,6).
P(different) can be found directly (30/36) or via 1 − P(same) — the faster route in an exam.

Common mistakes

Treating (2,3) and (3,2) as the same outcome, wrongly shrinking the sample space below 36.
Recounting all 30 'different number' outcomes instead of using the complement rule.

Real-world

Two-dice probability models are the basis for games of chance and risk-assessment problems in statistics.

Common Mistakes

1Treating the two dice as identical and undistinguishable, incorrectly reducing the sample space (e.g. treating (2,3) and (3,2) as one outcome instead of two).
2Miscounting the 'same number' outcomes as fewer or more than 6 by not listing all pairs (1,1) through (6,6) systematically.
3Forgetting the complement rule and trying to recount all 30 'different number' outcomes manually, wasting exam time and risking counting errors.

Interesting Facts

Dice-based probability problems date back to Gerolamo Cardano's 16th-century manuscript 'Liber de Ludo Aleae' (Book on Games of Chance), one of the earliest systematic studies of probability, motivated by gamblers' questions about dice games.

With two distinguishable dice, there are exactly 11 possible sums (2 through 12), but they are NOT equally likely — a sum of 7 has 6 ways to occur while a sum of 2 or 12 has only 1 way, a subtlety often tested in follow-up CBSE questions.

The specific phrase 'two dice of different colours' in CBSE papers is a deliberate hint telling students to treat the dice as distinguishable, ensuring the sample space has 36 (not fewer) equally likely outcomes.

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Frequently Asked Questions

Why are there 36 outcomes and not 21?

Because the two dice are of different colours, they are distinguishable, so (2,5) and (5,2) count as two different outcomes. This gives $6\times 6=36$ total ordered outcomes rather than treating unordered pairs as identical.

How can part (ii) be solved without recounting all outcomes?

Since 'same number' and 'different number' are complementary events covering the entire sample space, $P(\text{different}) = 1 - P(\text{same}) = 1 - \frac{1}{6} = \frac{5}{6}$ , avoiding the need to individually count 30 outcomes.