5.08 Applications to probability

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​

We can still use this formula for problems involving combinations but we now have 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!(n−r)!​, 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 combinations.

 

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

 

Combinations and conditional probability

Recall that conditional probability involves conditions or restrictions are placed on the total number of outcomes, thus reducing the size of our sample space.

Let's look at some examples involving combinations.

Worked examples

Example 2

For lunch each day, Tom eats a sandwich, a sausage roll or sushi. And he will drink a coffee, a tea or an orange juice with his lunch.

a)  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​

b) Given that Tom drinks a coffee with his lunch, what is the probability he also ate a sandwich?

Think: Is there a condition placed on this question? Yes - we know Tom drinks a coffee, thus reducing the size of our sample space. 

As we have not calculated any probabilities, let's calculate the probability directly with the numbers in the appropriate events:

$P\left(A|B\right)=\frac{n\left(A\cap B\right)}{n\left(B\right)}$P(A|B)=n(A∩B)n(B)​

$P\left(\text{Sandwich}|\text{Coffee}\right)=\frac{n\left(S\cap C\right)}{n\left(C\right)}$P(Sandwich|Coffee)=n(S∩C)n(C)​

Do: There is only one option that has a sandwich and a coffee. The options that have a coffee are number of different selections we have for a food item $^3C_1=3$3C1​=3.

Hence,

$P\left(\text{Sandwich}|\text{Coffee}\right)$P(Sandwich|Coffee)	$=$=	$\frac{n\left(S\cap C\right)}{n\left(C\right)}$n(S∩C)n(C)​
	$=$=	$\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.

a)  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​

 

b)  Given that she took only one Mathematics textbook, what is the probability she took at least one novel?

Think: Is there a condition placed on this question? Yes - we know that only one Mathematics textbook was taken, thus reducing the size of our sample space. 

Using the conditional probability rule, we have:

$P\left(\text{At least 1 novel}|\text{Only 1 Maths}\right)=\frac{n\left(\text{At least 1 novel}\cap\text{Only 1 Maths}\right)}{n\left(\text{Only 1 Maths}\right)}$P(At least 1 novel|Only 1 Maths)=n(At least 1 novel∩Only 1 Maths)n(Only 1 Maths)​

Do: 

Options with only $1$1 Mathematics Textbook and at least $1$1 novel: $=\nCr{4}{1}\times\nCr{5}{1}\times\nCr{3}{1}+\nCr{4}{1}\times\nCr{5}{2}\times\nCr{3}{0}$=4C1×5C1×3C1+4C1×5C2×3C0

Options with only $1$1  Mathematics Textbook: $\nCr{4}{1}\times\nCr{8}{2}$4C1×8C2

Hence,

$P\left(\text{At least 1 novel}|\text{Only 1 Maths}\right)$P(At least 1 novel|Only 1 Maths)	$=$=	$\frac{60+40}{112}$60+40112​
	$=$=	$\frac{25}{28}$2528​

 

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3

Question 4

Question 5