Mathematics 02: Counting for Probability

Many probability questions become easy once you can count the possible outcomes.

Question 1 — How many outfits can you make?

You have 3 shirts and 2 pairs of trousers. Choose one shirt and one pair of trousers.

Multiplication rule

number of complete choices = choices for step 1 × choices for step 2 × ...

Answer

3 shirts × 2 trousers = 6 outfits

This works because each shirt can be paired with each pair of trousers.

Question 2 — How many ways can gold, silver, and bronze be assigned?

Five runners finish a race. We care about first, second, and third place.

Order matters: Ada, Ben, Chen is different from Ben, Ada, Chen.

Permutation formula

P(n, r) = n! / (n - r)!

Answer

P(5, 3) = 5! / (5 - 3)!
        = 5! / 2!
        = (5 × 4 × 3 × 2 × 1) / (2 × 1)
        = 5 × 4 × 3
        = 60

There are 60 ordered podiums.

Question 3 — How many groups of 3 can be chosen from 5 students?

Now we choose a committee of 3 students. The order does not matter. A committee containing Ada, Ben, and Chen is the same committee in every order.

Combination formula

C(n, r) = n! / (r! × (n - r)!)

C(n, r) is also written as nCr or n choose r.

Answer

C(5, 3) = 5! / (3! × 2!)
        = (5 × 4 × 3 × 2 × 1) / ((3 × 2 × 1) × (2 × 1))
        = 10

There are 10 possible committees.

Question 4 — What is the chance of exactly 2 heads in 3 fair coin flips?

The possible equally likely results are:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Exactly two heads appears in HHT, HTH, and THH: 3 outcomes out of 8.

We can also use combinations.

Formula

P(exactly r successes in n fair trials) = C(n, r) × p^r × (1 - p)^(n-r)

For a fair coin, p = 1/2.

Answer

P(exactly 2 heads) = C(3, 2) × (1/2)^2 × (1/2)^1
                    = 3 × 1/8
                    = 3/8
                    = 0.375

This formula is called the binomial formula. You will study it in more detail later.

How to choose the right formula

Practice

1. How many 4-digit PINs can be made if every digit may be 0–9?
2. How many ways can 2 class representatives be selected from 10 students if president and vice-president are different jobs?
3. How many ways can 2 class representatives be selected from 10 students if they have the same job?

Answers

1. 10 × 10 × 10 × 10 = 10,000
2. P(10, 2) = 10 × 9 = 90
3. C(10, 2) = 10! / (2! × 8!) = 45