iGCSE (2021 Edition)

# 8.05 Probability using combinations

Lesson

Recall the probability of event $A$A occurring can be calculated as

$P\left(A\right)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$P(A)=Number of favourable outcomesTotal number of outcomes

Combinations give us a new technique for counting the number of favourable and total outcomes.

#### Worked example

##### Example 1

A magazine editor is deciding which $3$3 stories to feature on the front cover of a magazine.

(a) If she has $5$5 excellent articles to choose from and decides to choose randomly, how many different selections are possible?

Think: Is order important and are there any restrictions? Order does not appear to be important and there are no restrictions apparent, so we can proceed with the combination formula.

Do: Using $^nC_r=\frac{n!}{r!\left(n-r\right)!}$nCr=n!r!(nr)!, where $n$n is the total number of articles, $n=5$n=5 and $r$r is the number we are choosing, $r=3$r=3

 $^5C_2$5C2​ $=$= $\frac{5!}{3!2!}$5!3!2!​ $=$= $10$10

So there are $10$10 possible selections.

(b) If you were the writer of one of the stories, what is the probability that your article is chosen for the front page?

Think: We have the total number of outcomes, $10$10. We need to calculate the number of favourable outcomes and then we can calculate the probability.

Do: There are three articles picked for the front cover, favourable outcomes will have your article selected plus two out of the remaining four articles.

Hence,

 $\text{Number of favourable outcomes}$Number of favourable outcomes $=$= $^4C_2$4C2​ $=$= $\frac{4!}{2!2!}$4!2!2!​ $=$= $6$6

Thus the probability becomes,

 $P\left(\text{Article selected}\right)$P(Article selected) $=$= $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$Number of favourable outcomesTotal number of outcomes​ $=$= $\frac{6}{10}$610​ $=$= $\frac{3}{5}$35​ or $0.6$0.6

#### Example 2

For lunch each day, Tom eats a sandwich, a sausage roll or sushi and he drinks a coffee, a tea or an orange juice.

What is the probability that Tom eats a sausage roll for lunch?

Think: Are there any special conditions placed on this question? The answer is no, so we go ahead and calculate the number of favourable events and total number of outcomes using combinations. Notice there is an 'and' statement. So we will find the combinations of lunches and drinks separately and then multiply.

Do: Favourable outcomes are those where the sausage roll is selected for lunch:

 $\text{Total favourable outcomes}$Total favourable outcomes $=$= $^1C_1\times^3C_1$1C1​×3C1​ $=$= $1\times3$1×3 $=$= $3$3
 $\text{Total Number of outcomes}$Total Number of outcomes $=$= $^3C_1\times^3C_1$3C1​×3C1​ $=$= $3\times3$3×3 $=$= $9$9

Hence,

 $P\left(\text{Sausage Roll}\right)$P(Sausage Roll) $=$= $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$Number of favourable outcomesTotal number of outcomes​ $=$= $\frac{3}{9}$39​ $=$= $\frac{1}{3}$13​

#### Example 3

Elina is taking three books with her on holiday.

She has $4$4 Mathematics textbooks to choose from, $5$5 novels and $3$3 biographies.

What is the probability she takes one of each?

 $\text{Number of favourable outcomes}$Number of favourable outcomes $=$= $\nCr{4}{1}\times\nCr{5}{1}\times\nCr{3}{1}$4C1×5C1×3C1 $=$= $60$60
 $\text{Total number of outcomes}$Total number of outcomes $=$= $\nCr{12}{3}$12C3 $=$= $220$220

Thus the probability becomes,

 $P\left(\text{One of each}\right)$P(One of each) $=$= $\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$Number of favourable outcomesTotal number of outcomes​ $=$= $\frac{60}{220}$60220​ $=$= $\frac{3}{11}$311​

#### Practice questions

##### QUESTION 1

A box contains 6 pens of different colours: red, green, blue, yellow, black and white. Two pens are drawn at random without replacement.

1. How many possible selections are there?

2. What is the probability of drawing the green and black pens?

##### QUESTION 2

$5$5 people are to be selected from a larger group of $10$10 candidates. If Amelia is among the candidates, what is the probability that she will be among those selected?

### Outcomes

#### 0606C11.1

Recognise and distinguish between a permutation case and a combination case.