Probability with Two Dice: 36 Outcomes, Class 10 Maths

Quick answer: Two dice give $6 \times 6 = 36$ equally likely outcomes, written as ordered pairs such as $(2, 5)$ . A doublet (both dice the same) has 6 outcomes, so $P = \frac{1}{6}$ . A sum of 7 is the most likely sum, with 6 outcomes. A sum of at least 10 also has 6 outcomes ( $\frac{1}{6}$ ), a prime sum has 15 ( $\frac{5}{12}$ ), and an odd product needs both numbers odd, giving 9 outcomes ( $\frac{1}{4}$ ).

The 36 outcomes

When two dice are thrown, each of the 6 faces of the first die can pair with each of the 6 faces of the second, so $n(S) = 36$ .

Die 1 vs Die 2	1	2	3	4	5	6
1	(1,1)	(1,2)	(1,3)	(1,4)	(1,5)	(1,6)
2	(2,1)	(2,2)	(2,3)	(2,4)	(2,5)	(2,6)
3	(3,1)	(3,2)	(3,3)	(3,4)	(3,5)	(3,6)
4	(4,1)	(4,2)	(4,3)	(4,4)	(4,5)	(4,6)
5	(5,1)	(5,2)	(5,3)	(5,4)	(5,5)	(5,6)
6	(6,1)	(6,2)	(6,3)	(6,4)	(6,5)	(6,6)

The pairs are ordered: $(1, 2)$ and $(2, 1)$ are different outcomes.

Doublets

A doublet is an outcome where both dice show the same number: $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$ .

$P(\text{doublet}) = \frac{6}{36} = \frac{1}{6}$

These are the diagonal of the grid.

Sums of the two dice

Sum	Outcomes	Count	Probability
2	(1,1)	1	1/36
3	(1,2), (2,1)	2	1/18
4	(1,3), (2,2), (3,1)	3	1/12
5	(1,4), (2,3), (3,2), (4,1)	4	1/9
6	(1,5), (2,4), (3,3), (4,2), (5,1)	5	5/36
7	(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)	6	1/6
8	(2,6), (3,5), (4,4), (5,3), (6,2)	5	5/36
9	(3,6), (4,5), (5,4), (6,3)	4	1/9
10	(4,6), (5,5), (6,4)	3	1/12
11	(5,6), (6,5)	2	1/18
12	(6,6)	1	1/36

The counts rise from 1 to 6 and fall back to 1, and they add up to 36. A sum of 7 is six times as likely as a sum of 2 or 12.

Worked example

Two dice are thrown together. Find the probability of getting (i) a doublet, (ii) a sum of at least 10, (iii) a prime sum, (iv) an odd product.

Step 1: The total number of outcomes is 36.

Step 2: Doublet: 6 outcomes, so $P = \frac{6}{36} = \frac{1}{6}$ .

Step 3: Sum at least 10 means a sum of 10, 11 or 12: $3 + 2 + 1 = 6$ outcomes, so $P = \frac{6}{36} = \frac{1}{6}$ .

Step 4: The prime sums possible are 2, 3, 5, 7 and 11: $1 + 2 + 4 + 6 + 2 = 15$ outcomes, so $P = \frac{15}{36} = \frac{5}{12}$ .

Step 5: A product is odd only when both numbers are odd. Each die has 3 odd faces (1, 3, 5), so there are $3 \times 3 = 9$ outcomes and $P = \frac{9}{36} = \frac{1}{4}$ .

Answer: (i) $\frac{1}{6}$ , (ii) $\frac{1}{6}$ , (iii) $\frac{5}{12}$ , (iv) $\frac{1}{4}$ . It follows that $P(\text{even product}) = 1 - \frac{1}{4} = \frac{3}{4}$ .

Summary table

Event	Favourable outcomes	Probability
Doublet	6	1/6
Sum = 7	6	1/6
Sum at least 10	6	1/6
Sum at most 5	10	5/18
Prime sum	15	5/12
Odd product	9	1/4
Even product	27	3/4

Exam tips

"At least 10" means 10 or more; "at most 5" means 5 or less (so 2, 3, 4 or 5 for two dice).
Use the complement where it is quicker: $P(\text{even product}) = 1 - P(\text{odd product})$ .
Draw or write the 6 × 6 grid in rough work so you do not miss outcomes.

Common mistakes

Treating $(1, 2)$ and $(2, 1)$ as one outcome. They are different.
Counting 1 as a prime number. The smallest prime sum is 2.
Taking the total as 12 or 11 (the number of possible sums) instead of 36 outcomes.
Forgetting to simplify the final fraction.

Frequently Asked Questions

How many outcomes are there when two dice are thrown?

There are 6 × 6 = 36 equally likely outcomes.

What is the probability of getting a doublet?

6/36 = 1/6, since there are six doublets from (1, 1) to (6, 6).

Which sum is most likely with two dice?

A sum of 7, which can happen in 6 ways and has probability 1/6.

What is the probability that the product of the two numbers is odd?

1/4, because both numbers must be odd, giving 3 × 3 = 9 favourable outcomes out of 36.

What does "a sum of at least 10" mean?

A sum of 10, 11 or 12, which together have 6 outcomes and probability 1/6.