Worked examples
The question
Question: A bag has 4 red and 6 blue marbles.
2 marbles are picked at random.
What is the probability that both are red?
The probability is the number of ways we want, over all the ways.
P(all red) = ways we wantall ways
P stands for probability.
P = probability
1 Order: yes or no
Does order matter?
No.
Picking Ann, then Bo, is the same group as Bo, then Ann.
So we count groups: combinations.
= a group: order does not matter → nCr
2 Count all the ways
10 C 2: the top 2 numbers from 10 down, over 2 factorial.
all = 10C2 = 10 × 92 × 1
Before we multiply, we make the numbers smaller.
10 on top and 2 at the bottom: 10 divided by 2 is 5.
What is left: 5 times 9.
That is 45 ways.
all = 45
5 × 9 = 45
3 Count the ways we want
4 C 2: the top 2 numbers from 4 down, over 2 factorial.
red = 4C2 = 4 × 32 × 1
Before we multiply, we make the numbers smaller.
4 on top and 2 at the bottom: 4 divided by 2 is 2.
What is left: 2 times 3.
That is 6 ways.
red = 6
2 × 3 = 6
4 Write the fraction
6 ways we want, out of 45.
P(all red) = 645
Careful!
The top and the bottom must count the same way: both are groups, with C.
Careful: count groups on top AND at the bottom
5 Simplify
Both 6 and 45 can be divided by 3.
That makes it simpler.
P(all red) = 6 ÷ 345 ÷ 3
So it is 2/15.
P(all red) = 215
6 Check
Let's check another way: pick one at a time.
The first: 4 out of 10.
Then one fewer of each.
Multiplied, that is 2/15.
The same.
So the answer is right!
410 × 39 = 215
Let's look back.
First: order: yes or no.
Then: count all the ways.
Then: count the ways we want.
Then: write the fraction.
Then: simplify.
Last: check.
▶ Watch it solved
The question
Question: A club has 6 boys and 4 girls.
3 of them are picked at random.
What is the probability that all 3 are girls?
The probability is the number of ways we want, over all the ways.
P(all girls) = ways we wantall ways
P stands for probability.
P = probability
1 Order: yes or no
Does order matter?
No.
Picking Ann, then Bo, is the same group as Bo, then Ann.
So we count groups: combinations.
= a group: order does not matter → nCr
2 Count all the ways
10 C 3: the top 3 numbers from 10 down, over 3 factorial.
all = 10C3 = 10 × 9 × 83 × 2 × 1
Before we multiply, we make the numbers smaller.
9 and 3 both divide by 3: 3.
10 and 2 both divide by 2: 5.
What is left is 5 times 3 times 8.
all = 5 × 3 × 8
5 times 3 is 15.
15 times 8 is 120.
5 × 3 = 15
15 × 8 = 120
So there are 120 ways.
all = 120
3 Count the ways we want
4 C 3: the top 3 numbers from 4 down, over 3 factorial.
girls = 4C3 = 4 × 3 × 23 × 2 × 1
Before we multiply, we make the numbers smaller.
3 and 3 both divide by 3: 1.
4 and 2 both divide by 2: 2.
What is left: 2 times 2.
That is 4 ways.
girls = 4
2 × 2 = 4
4 Write the fraction
4 ways we want, out of 120.
P(all girls) = 4120
Careful!
The top and the bottom must count the same way: both are groups, with C.
Careful: count groups on top AND at the bottom
5 Simplify
Both 4 and 120 can be divided by 4.
That makes it simpler.
P(all girls) = 4 ÷ 4120 ÷ 4
So it is 1/30.
P(all girls) = 130
6 Check
Let's check another way: pick one at a time.
The first: 4 out of 10.
Then one fewer of each.
Multiplied, that is 1/30.
The same.
So the answer is right!
410 × 39 × 28 = 130
Let's look back.
First: order: yes or no.
Then: count all the ways.
Then: count the ways we want.
Then: write the fraction.
Then: simplify.
Last: check.
Answer P(all girls) = 130
▶ Watch it solved
The question
Question: A club has 6 boys and 4 girls.
3 of them are picked at random.
What is the probability that 2 are girls and 1 is a boy?
The probability is the number of ways we want, over all the ways.
P(2 girls, 1 boy) = ways we wantall ways
P stands for probability.
P = probability
1 Order: yes or no
Does order matter?
No.
Picking Ann, then Bo, is the same group as Bo, then Ann.
So we count groups: combinations.
= a group: order does not matter → nCr
2 Count all the ways
10 C 3: the top 3 numbers from 10 down, over 3 factorial.
all = 10C3 = 10 × 9 × 83 × 2 × 1
Before we multiply, we make the numbers smaller.
9 and 3 both divide by 3: 3.
10 and 2 both divide by 2: 5.
What is left is 5 times 3 times 8.
all = 5 × 3 × 8
5 times 3 is 15.
15 times 8 is 120.
5 × 3 = 15
15 × 8 = 120
So there are 120 ways.
all = 120
3 Count the ways we want
4 C 2: the top 2 numbers from 4 down, over 2 factorial.
girls = 4C2 = 4 × 32 × 1
Before we multiply, we make the numbers smaller.
4 on top and 2 at the bottom: 4 divided by 2 is 2.
What is left: 2 times 3.
That is 6 ways.
girls = 6
2 × 3 = 6
Picking 1 of 6 boys: 6 ways.
boys = 6C1 = 6
Each choice of girls goes with each choice of boys.
So we multiply: 6 times 6 is 36.
we want = 6 × 6 = 36
Careful!
We need both at the same time: girls and boys.
And means multiply, not add.
Careful: not 6 + 6 ✗
4 Write the fraction
36 ways we want, out of 120.
P(2 girls, 1 boy) = 36120
5 Simplify
Both 36 and 120 can be divided by 12.
That makes it simpler.
P(2 girls, 1 boy) = 36 ÷ 12120 ÷ 12
So it is 3/10.
P(2 girls, 1 boy) = 310
6 Check
Let's check another way: pick one at a time, in the order G, G, B.
Multiplied, that is 1/10.
GGB: 410 × 39 × 68 = 110
But the 2 girls and the 1 boy can come in 3 orders, each just as likely.
3 times that is 3/10.
The same.
So the answer is right!
3 orders: 3 × 110 = 310
Let's look back.
First: order: yes or no.
Then: count all the ways.
Then: count the ways we want.
Then: write the fraction.
Then: simplify.
Last: check.
Answer P(2 girls, 1 boy) = 310
▶ Watch it solved
