Combinations in probability | CAIE 9709 Probability & Statistics 1 Notes

Probability & Statistics 1

Mathematics9709

Count favourable and total selections in the same world

P(A)=\frac{\text{favourable}}{\text{total}}

“Favourable” means outcomes that satisfy event A. “Total” means every equally likely outcome in the chosen sample space.

If a fixed-size group is chosen at random without replacement, every unordered group of that size is equally likely. Use combinations.

Worked example

One boy and one girl

Choose 2 children at random from 6 boys and 4 girls.

Favourable pairs: ${}^6C_1{}^4C_1=24$ .
All two-person groups: ${}^{10}C_2=45$ .

Answer

P(\text{one of each})=\frac{24}{45}=\frac8{15}

Key idea

The denominator is not “all people” and not all ordered pairs. It is the number of equally likely groups produced by the random selection. Use the category casework from the previous lesson to count “exactly”, “at least” or “none” in the numerator.

Use permutations when positions matter

If distinct objects are placed in a random line, each complete arrangement is an equally likely outcome. Count arrangements on both levels of the fraction.

Worked example

A and B together in a random line

Five different people, including A and B, stand in random order.

Favourable arrangements: treat A and B as a block, then allow AB or BA. This gives $4!\times2!$ .
All arrangements: $5!$ .

Answer

P(\text{together})=\frac{4!\times2!}{5!}=\frac25

Common mistake

Do not count combinations in the numerator and permutations in the denominator. That compares different kinds of outcome.

Given that changes the denominator

In given that B, outcomes outside B are no longer possible. Count the target outcomes in the numerator and all outcomes satisfying B in the denominator.

Worked example

A conditioned team selection

Choose 3 people from 4 girls and 3 boys. Given that at least 2 girls are chosen, find the probability that all 3 are girls.

Target: all 3 girls, so ${}^4C_3=4$ .
Condition: 2 girls and 1 boy, or 3 girls. This gives ${}^4C_2{}^3C_1+{}^4C_3=18+4=22$ .

Answer

P(\text{all 3 girls}\mid\text{at least 2 girls})=\frac4{22}=\frac2{11}

Examiner note

Before dividing, check three things: the outcomes are equally likely, the numerator satisfies the condition, and both counts describe the same kind of outcome.

Your turnIndependent check [4 marks]

A team of 3 is chosen from 5 girls and 4 boys. Given that the team contains at least 2 girls, find the probability that all 3 are girls.

Show worked answer

The condition contains two cases: 2 girls and 1 boy, or 3 girls.

Answer

\frac{{}^5C_3}{{}^5C_2{}^4C_1+{}^5C_3}=\frac{10}{50}=\frac15

Count favourable and total selections in the same world

P(A)=\frac{\text{favourable}}{\text{total}}

“Favourable” means outcomes that satisfy event A. “Total” means every equally likely outcome in the chosen sample space.

If a fixed-size group is chosen at random without replacement, every unordered group of that size is equally likely. Use combinations.

Worked example

One boy and one girl

Choose 2 children at random from 6 boys and 4 girls.

Favourable pairs: ${}^6C_1{}^4C_1=24$ .
All two-person groups: ${}^{10}C_2=45$ .

Answer

P(\text{one of each})=\frac{24}{45}=\frac8{15}

Key idea

Use permutations when positions matter

If distinct objects are placed in a random line, each complete arrangement is an equally likely outcome. Count arrangements on both levels of the fraction.

Worked example

A and B together in a random line

Five different people, including A and B, stand in random order.

Favourable arrangements: treat A and B as a block, then allow AB or BA. This gives $4!\times2!$ .
All arrangements: $5!$ .

Answer

P(\text{together})=\frac{4!\times2!}{5!}=\frac25

Common mistake

Do not count combinations in the numerator and permutations in the denominator. That compares different kinds of outcome.

Given that changes the denominator

In given that B, outcomes outside B are no longer possible. Count the target outcomes in the numerator and all outcomes satisfying B in the denominator.

Worked example

A conditioned team selection

Choose 3 people from 4 girls and 3 boys. Given that at least 2 girls are chosen, find the probability that all 3 are girls.

Target: all 3 girls, so ${}^4C_3=4$ .
Condition: 2 girls and 1 boy, or 3 girls. This gives ${}^4C_2{}^3C_1+{}^4C_3=18+4=22$ .

Answer

P(\text{all 3 girls}\mid\text{at least 2 girls})=\frac4{22}=\frac2{11}

Examiner note

Before dividing, check three things: the outcomes are equally likely, the numerator satisfies the condition, and both counts describe the same kind of outcome.

Your turnIndependent check [4 marks]

A team of 3 is chosen from 5 girls and 4 boys. Given that the team contains at least 2 girls, find the probability that all 3 are girls.

Show worked answer

The condition contains two cases: 2 girls and 1 boy, or 3 girls.

Answer

\frac{{}^5C_3}{{}^5C_2{}^4C_1+{}^5C_3}=\frac{10}{50}=\frac15