Let's refresh our memories about conditional probability.
As the name suggests, conditions or restrictions are placed on the total number of outcomes, thus reducing the size of our sample space.
In the case we're using combinations to define our sample space, the number of total choices of combinations are reduced.
When considering whether a question is referring to a conditional probability, remember to look out for the following phrases:
- Given that ....
- If ......., what is the probability that .....
Let's launch in with some examples.
Example 1
For lunch each day, Tom eats a sandwich, a sausage roll or sushi. 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 right ahead and calculate our probability as normal.
Do: $\frac{\nCr{1}{1}\times\nCr{3}{1}}{\nCr{3}{1}\times\nCr{3}{1}}$1C1×3C13C1×3C1
=$\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.
The denominator of our fraction will therefore be about what he chooses to eat, since we know what he has chosen to drink.
The numerator must represent his choice of drinking coffee AND eating a sandwich.
Do: $\frac{\nCr{1}{1}\times\nCr{1}{1}}{\nCr{1}{1}\times\nCr{3}{1}}$1C1×1C11C1×3C1
=$\frac{1}{3}$13
Carefully follow the working out for this next example, especially part (b).
Example 2:
Elina is taking three books with her on holiday.
She has $4$4 mathematics textbooks to choose from, $5$5 novels and $3$3 biographies.
a) What is the probability she takes one of each?
=$\frac{\nCr{4}{1}\times\nCr{5}{1}\times\nCr{3}{1}}{\nCr{12}{3}}$4C1×5C1×3C112C3
=$\frac{60}{220}$60220
b) Given that she took only one mathematics textbook, what is the probability she took at least one novel?
=$\frac{\nCr{4}{1}\times\nCr{5}{1}\times\nCr{3}{1}+\nCr{4}{1}\times\nCr{5}{2}\times\nCr{3}{0}}{\nCr{4}{1}\times\nCr{8}{2}}$4C1×5C1×3C1+4C1×5C2×3C04C1×8C2
=$\frac{60+40}{112}$60+40112
=$\frac{100}{112}$100112
Worked Examples
Question 1
A student is choosing two units to study at university: a language and a science unit. They have $4$4 languages and $7$7 science units to choose from.
If they choose one of each, what is the total number of combinations of choices?
If Italian is one of the languages they can choose from, what is the probability they choose Italian as their language?
French is one of the available languages. What is the probability they choose French as their language given that they choose Chemistry as their science unit?
Question 2
A tour guide is taking groups on holiday and wants to mix up the nationalities of the groups, but she does this at random. Each group has $6$6 people.
On this tour there are $10$10 Australians, $7$7 Americans and $12$12 Chinese people.
In one group, what is the probability that there are $2$2 people of each nationality?
Given that there are exactly $2$2 Chinese people in the group, what is the probability that there are at least $3$3 Americans?
What is the probability that there are no Australians in the group given that there are exactly $2$2 Americans?