AFCAT Numerical Ability · Probability

Dice: One Die, the 36 Pairs and Three Dice

One die has 6 equally likely faces, two dice 36 ordered pairs and three dice 216 triples; count the favourable ones from a pattern or a grid, or multiply die by die.

All Probability notes

Why this matters

Two dice carry most of this page: sums, products, differences and doublets, all counted on the 6 by 6 grid of ordered pairs. Marks are lost by treating (2, 5) and (5, 2) as one outcome, by miscounting a sum, and on three dice by forgetting how many ways an 'exactly' can be arranged.

Concept 1 of 5: One die: six equally likely faces

A fair die shows 1 to 6, each with probability

\tfrac16

. Count the faces that fit the question and divide by 6.

Definition

Even faces: 2, 4, 6. Odd faces: 1, 3, 5. Prime faces: 2, 3, 5. Multiples of 3: 3, 6.
'Greater than 4' is 5 and 6; 'at least 4' adds the 4.
'Not' a property: count the faces that fail it, or take 1 minus the probability.

One die

P = \dfrac{\text{faces that fit}}{6}

Worked example

A die is thrown. Find the probability of a number that is odd and at least 3.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionEasy

Example 1 · Probability · Dice

When a single die is thrown, what is the probability of getting a perfect square?

Greater than or at least

'Greater than 3' on a die is 4, 5, 6, so

\tfrac12

. 'At least 3' is 3, 4, 5, 6, so

\tfrac23

2 is prime

The prime faces are 2, 3 and 5. Leaving out 2 gives

\tfrac13

instead of

\tfrac12

Drill 8 more on one die: six equally likely faces

Concept 2 of 5: Sums on two dice

Two dice give

6 \times 6 = 36

ordered pairs. A sum of 7 is the most common, made 6 ways. Each step away from 7 loses one way, down to 1 way each for 2 and 12.

Definition

The number of ways to make a sum s (from 2 to 12) is $6 - |s - 7|$ .
(1, 6) and (6, 1) are different outcomes: the two dice are told apart even when thrown together.
For a set of sums, add the ways for each sum, then divide by 36.
'Given the sum is s' shrinks the sample space to that sum's pairs, as on the coins page.

Ways to make a sum

n(\text{sum } s) = 6 - |s - 7|, \quad 2 \le s \le 12

Worked example

Two dice are thrown. Find the probability that the sum is a prime number.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Probability · Dice

When two dice are thrown, what is the probability of getting a sum that is divisible by

4

Count both orders

A sum of 3 comes from (1, 2) and (2, 1): 2 ways, so

\tfrac{1}{18}

. Counting it once gives

\tfrac{1}{36}

The 11 sums are not equally likely

There are 11 possible sums, 2 to 12, but 7 is six times as likely as 2. Never write

\tfrac{1}{11}

for one sum.

Drill 12 more on sums on two dice

Concept 3 of 5: Products, differences and doublets

For anything other than a sum, walk the grid: list the pairs that fit, each order counted, and divide by 36.

Definition

A doublet is both dice the same, (1, 1) to (6, 6): 6 pairs.
Product n: list the factor pairs with both factors from 1 to 6, each in both orders, except a square such as (3, 3), which is one pair.
Difference d (larger minus smaller): $2(6 - d)$ pairs for d from 1 to 5, and 6 pairs (the doublets) for d = 0.
'The first die shows more than the second' is half of the 30 pairs that are not doublets.

Pairs with difference d

n(|a - b| = d) = 2(6 - d), \quad 1 \le d \le 5

Worked example

Two dice are thrown. Find the probability that the product of the numbers is 4.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Probability · Dice

When two dice are thrown, what is the probability that the absolute difference between the numbers is

2

A square product counts once

Product 9 is only (3, 3), one pair. Product 6 is (1, 6), (6, 1), (2, 3) and (3, 2): four pairs.

Differ by means either order

'The numbers differ by 4' includes (5, 1) as well as (1, 5). Counting one order halves the answer.

Drill 6 more on products, differences and doublets

Concept 4 of 5: Die by die: multiply, and one minus none

The dice do not affect each other, so a condition on each die separately multiplies. Both dice show a number greater than 2 with probability

\tfrac46 \times \tfrac46 = \tfrac49

Definition

Independent dice: P(a condition on die 1 and a condition on die 2) is the product of the two.
At least one six from n dice: $1 - \left(\tfrac56\right)^n$ .
Exactly one of two dice shows a given number: $2 \times \tfrac16 \times \tfrac56 = \tfrac{10}{36}$ , the two orders added.
Use the complement whenever 'at least one' appears: one product instead of several cases.

At least one six from n dice

P(\text{at least one six}) = 1 - \left(\tfrac56\right)^n

Worked example

Two dice are thrown. Find the probability that at least one shows a number greater than 4.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 4 · Probability · Dice

What is the probability that when two dice are rolled, the product is even?

Do not add for at least one

At least one six on two dice is not

\tfrac16 + \tfrac16 = \tfrac13

: that counts (6, 6) twice. It is

1 - \tfrac{25}{36} = \tfrac{11}{36}

Exactly one has two orders

'Exactly one die shows a 4' on two dice is

\tfrac16 \times \tfrac56

for each die, so twice that:

\tfrac{10}{36}

Drill 11 more on die by die: multiply, and one minus none

Concept 5 of 5: Three dice: 216 ordered triples

Three dice give

6^3 = 216

ordered triples. Count favourable triples by multiplying the choices die by die, or by choosing which dice do what.

Definition

One triple such as (2, 2, 2) has probability $\tfrac{1}{216}$ . The smallest and largest sums, 3 and 18, can each be made only one way.
Exactly k dice show a six: $^3C_k \left(\tfrac16\right)^k \left(\tfrac56\right)^{3 - k}$ . The $^3C_k$ picks which dice, as $^nC_k$ picked the coins.
All different values: multiply the choices die by die, one fewer each time.
All the same value: 6 triples, (1, 1, 1) to (6, 6, 6).

Exactly k sixes on three dice

P = {}^3C_k \left(\tfrac16\right)^k \left(\tfrac56\right)^{3 - k}

Worked example

Three dice are thrown. Find the probability that exactly two show a number greater than 4.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionHard

Example 5 · Probability · Dice

When three dice are thrown, what is the probability that all three dice show different numbers?

Choose which dice

Exactly two sixes on three dice is not

\left(\tfrac16\right)^2 \times \tfrac56

alone. The die that is not a six can be any of the three, so multiply by 3:

\tfrac{15}{216}

A sum of 4 has three triples

(1, 1, 2), (1, 2, 1) and (2, 1, 1) are three different outcomes, so a sum of 4 has probability

\tfrac{3}{216}

, not

\tfrac{1}{216}

Drill 5 more on three dice: 216 ordered triples

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

One die: six equally likely faces
One die
$P = \dfrac{\text{faces that fit}}{6}$
Sums on two dice
Ways to make a sum
$n(\text{sum } s) = 6 - |s - 7|, \quad 2 \le s \le 12$
Products, differences and doublets
Pairs with difference d
$n(|a - b| = d) = 2(6 - d), \quad 1 \le d \le 5$
Die by die: multiply, and one minus none
At least one six from n dice
$P(\text{at least one six}) = 1 - \left(\tfrac56\right)^n$
Three dice: 216 ordered triples
Exactly k sixes on three dice
$P = {}^3C_k \left(\tfrac16\right)^k \left(\tfrac56\right)^{3 - k}$

Watch out for (10)

Greater than or at least
→ One die: six equally likely faces
2 is prime
→ One die: six equally likely faces
Count both orders
→ Sums on two dice
The 11 sums are not equally likely
→ Sums on two dice
A square product counts once
→ Products, differences and doublets
Differ by means either order
→ Products, differences and doublets
Do not add for at least one
→ Die by die: multiply, and one minus none
Exactly one has two orders
→ Die by die: multiply, and one minus none
Choose which dice
→ Three dice: 216 ordered triples
A sum of 4 has three triples
→ Three dice: 216 ordered triples

Drill the questions on this topic

47 questions from the bank, with cart and Word export.

Drill all 47 questions