Year 11: How to solve word problems using combinations in probability

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In Year 11 Mathematics, you will often see probability questions that involve choosing groups of items or people. When the order does not matter, we use combinations. This article will show you a clear, step-by-step “how to solve it” method for tackling word problems that use combinations in probability.

1. What is a combination?

A combination is a way of selecting items from a group where the order of selection does not matter.

• Choosing 3 students out of a class of 10 to form a committee.
• Choosing 6 numbers from 1 to 45 in a lottery game.
• Selecting 2 cards from a deck of 52 cards.

The number of combinations of choosing $r$ objects from $n$ objects is written as:

C(n, r) or sometimes nC_r

and is calculated using the formula:

C(n, r) = n! / (r! × (n − r)!)

where $n!$ (n factorial) means: $n\ne n\times (n-1)\times (n-2)\times \dots \times 2\times 1$.

2. Combinations in probability

Probability is:

Probability = (number of favourable outcomes) / (number of possible outcomes)

In combination problems:

• The number of possible outcomes is usually “all the ways to choose some objects from a group”, so we use a combination like C(n, r).
• The number of favourable outcomes is the number of ways to choose the “good” or “desired” groups, often also using combinations.

3. A “how to solve it” strategy

Use this four-step strategy for word problems:

1. Understand the problem
2. Plan
3. Do (carry out the plan)
4. Check

Step 1: Understand the problem

When you read the question, ask yourself:

• What is being chosen? (students, cards, balls, etc.)
• How many are being chosen each time?
• What is the event whose probability I need? (e.g. “at least 2 girls”)
• Does order matter? (If not, use combinations.)

Step 2: Plan

Planning involves:

• Finding the total number of possible outcomes using combinations: often C(total items, number chosen).
• Finding the number of favourable outcomes using combinations: sometimes you split into cases (e.g. “exactly 2 girls”, “3 girls”, etc.).
• Using the probability formula: favourable / total.

Step 3: Do (carry out the plan)

Use the combination formula to calculate the values. Keep your working neat and show each step clearly.

Step 4: Check

Check that your answer:

• Is between 0 and 1.
• Makes sense (is it reasonable that the probability is small / large?).
• Matches the wording of the question (for example, “at least 2” means 2 or more).

4. Fully worked example

Step 1: Understand the problem

• Total students: 12
• Girls: 7
• Boys: 5
• We choose 4 students in total.
• We want: exactly 2 girls and 2 boys in the group.
• Order does not matter (it is just a group), so we use combinations.

Step 2: Plan

1. Find the total number of ways to choose 4 students from 12.
    This is C(12, 4).

2. Find the number of favourable ways (exactly 2 girls and 2 boys).

• Choose 2 girls from 7 girls: C(7, 2).
• Choose 2 boys from 5 boys: C(5, 2).

Because we need both choices to happen together (2 girls and 2 boys), we multiply:

Favourable outcomes = C(7, 2) × C(5, 2)

3. Use the probability formula:

Probability = [C(7, 2) × C(5, 2)] / C(12, 4)

Step 3: Do (carry out the plan)

First, calculate C(7, 2):

C(7, 2) = 7! / (2! × 5!) = (7 × 6) / (2 × 1) = 21.

Next, calculate C(5, 2):

C(5, 2) = 5! / (2! × 3!) = (5 × 4) / (2 × 1) = 10.

So the number of favourable outcomes is:

C(7, 2) × C(5, 2) = 21 × 10 = 210.

Now calculate the total number of ways to choose 4 students from 12:

C(12, 4) = 12! / (4! × 8!) = (12 × 11 × 10 × 9) / (4 × 3 × 2 × 1)

= (11880) / 24 = 495.

So the probability is:

Probability = favourable / total = 210 / 495

We can simplify the fraction:

210 / 495 = (210 ÷ 15) / (495 ÷ 15) = 14 / 33.

So the probability that the group contains exactly 2 girls and 2 boys is:

14/33 (approximately 0.424 or 42.4%).

Step 4: Check

• The answer 14/33 is between 0 and 1, so that is OK.
• It seems reasonable: it is not extremely small or extremely large, since there are many possible groups and quite a lot of them have 2 girls and 2 boys.
• We correctly used “exactly 2 girls and 2 boys” (not “at least 2”).

5. Another common type: “at least” problems

Sometimes the problem will say “at least 2 girls” or “at least 1 red ball”. “At least 2” means 2 or more (2, 3, 4, …).

The possible girl counts in the group are: 0, 1, 2, 3, or 4 girls.

“At least 2 girls” means: 2 girls, 3 girls, or 4 girls.

A good plan:

• Find the number of ways to get exactly 2 girls (and 2 boys).
• Find the number of ways to get exactly 3 girls (and 1 boy).
• Find the number of ways to get exactly 4 girls (and 0 boys).
• Add them all up to get the total favourable outcomes.
• Divide by C(12, 4) as before.

Each of these is a combination calculation. For example:

• Exactly 3 girls: C(7, 3) × C(5, 1)
• Exactly 4 girls: C(7, 4) × C(5, 0)

6. Common mistakes to avoid

7. Practice problems

Try these on your own. Use the four-step method: Understand, Plan, Do, Check.

1. A club has 9 members. A subcommittee of 3 people is chosen randomly. How many different subcommittees are possible?
2. A bag contains 5 red balls and 6 blue balls. Three balls are drawn at random without replacement. What is the probability that all three balls are red?
3. A school has a debating team of 10 students: 4 are from Year 11 and 6 are from Year 12. A group of 5 students is chosen at random to represent the school. What is the probability that the group contains exactly 2 students from Year 11 and 3 students from Year 12?
4. In a card game, you are dealt 5 cards from a standard deck of 52 cards. What is the probability that you are dealt exactly 3 hearts? (Hint: There are 13 hearts in the deck.)
5. A committee of 6 students is chosen from a group of 8 boys and 7 girls. Find the probability that the committee has at least 4 girls.

Once you are comfortable with these types of problems, you will find that a lot of probability questions in Year 11 are just different versions of the same idea: use combinations to count favourable and total outcomes, then form the probability as a fraction.