Probability Equation Questions

What you'll learn

How to turn “two counters are chosen” information into probability equations.
How to handle with replacement and without replacement questions.
How to use ratios like 3:1 to write algebraic numbers of counters.
How to solve the quadratic equations that often appear in grade 8/9 probability.

The big idea

In these questions, you are not just calculating a probability. You are usually given a probability and asked to work backwards to find how many counters were in the bag.

Definition

Probability equation

A probability equation is an equation formed by setting an algebraic probability equal to a given probability, then solving for the unknown number of items.

An event is the outcome you care about, such as “both counters are blue” or “the first counter is red and the second is red”.

The key skill is translating the words into fractions carefully.

Multiplying probabilities along a path

When two things happen in order, multiply the probabilities along the matching path.

For example, if you want “blue then blue”, multiply:

blue)P(\text{blue then blue}) = P(\text{first blue}) \times P(\text{second blue})

A tree diagram is a useful way to organise the changing probabilities, especially when counters are not replaced.

Tree diagram showing probabilities for two draws without replacement

Definition

Without replacement

Without replacement means the first counter is not put back into the bag before the second counter is chosen. So the total number of counters goes down by 1.

Example

Finding a probability without replacement

A bag contains 5 red counters and 3 blue counters. Two counters are chosen at random without replacement. Find the probability that both counters are blue.

There are 8 counters altogether, and 3 of them are blue, so the probability of the first blue counter is:
$38\frac{3}{8}$
If the first counter was blue, there are now 2 blue counters left out of 7 counters altogether. So the probability of the second blue counter is:
$27\frac{2}{7}$
Multiply the two probabilities:
$38×27\frac{3}{8} \times \frac{2}{7}$
Simplify:
$656=328\frac{6}{56} = \frac{3}{28}$

Common Mistake

Forgetting the bag has changed

For questions without replacement, the denominator for the second pick is usually one less than before. The chosen colour may also have one fewer counter left.

With replacement: the bag resets

Definition

With replacement

With replacement means the first counter is put back into the bag before the second counter is chosen. The probabilities stay the same on both picks.

This often makes the equation simpler because the same fraction is repeated.

Example

Solving a with-replacement equation

A bag contains 4 red counters and $y$ blue counters. A counter is chosen, replaced, then another counter is chosen. The probability of getting two red counters is $425\frac{4}{25}$ . Find $y$ .

The total number of counters is $y + 4$ .
The probability of choosing red on one pick is:
$4y+4\frac{4}{y + 4}$
Because the counter is replaced, the probability of red then red is:
$4y+4×4y+4\frac{4}{y + 4} \times \frac{4}{y + 4}$
Set this equal to the given probability:
$(4y+4)2=425\left(\frac{4}{y + 4}\right)^2 = \frac{4}{25}$
Take the positive square root of both sides:
$4y+4=25\frac{4}{y + 4} = \frac{2}{5}$
Cross-multiply and solve:
$4×5=2(y+4)20=2y+812=2yy=6\begin{aligned} 4 \times 5 &= 2(y + 4) \\ 20 &= 2y + 8 \\ 12 &= 2y \\ y &= 6 \end{aligned}$

Tip

Why the positive square root?

A probability cannot be negative, and the number of counters must be positive, so you only use the positive square root in this type of question.

Using ratios to write the number of counters

A ratio compares quantities. If the ratio of red counters to blue counters is 2:1, then you can write:

red counters as $2 k$
blue counters as $k$
total counters as $3 k$

The letter $k$ represents the multiplier that scales the ratio up to the actual numbers.

Example

Using a ratio to form an equation

The ratio of red counters to blue counters is 2:1. Two counters are chosen at random without replacement. The probability that both counters are red is $37\frac{3}{7}$ . Find the number of blue counters.

Use the ratio 2:1. Let the number of blue counters be $k$ , so the number of red counters is $2 k$ .
The total number of counters is:
$3 k$
The probability of the first red counter is:
$2k3k\frac{2k}{3k}$
If one red counter has been removed, there are $2 k - 1$ red counters left, and $3 k - 1$ counters altogether. So:
$red)=2k3k×2k−13k−1P(\text{red then red}) = \frac{2k}{3k} \times \frac{2k - 1}{3k - 1}$
Set this equal to the given probability:
$2k3k×2k−13k−1=37\frac{2k}{3k} \times \frac{2k - 1}{3k - 1} = \frac{3}{7}$
Simplify $2k3k\frac{2k}{3k}$ to $23\frac{2}{3}$ :
$23×2k−13k−1=37\frac{2}{3} \times \frac{2k - 1}{3k - 1} = \frac{3}{7}$
Cross-multiply:
$14 (2 k - 1) = 9 (3 k - 1)$
Expand and solve:
$28k−14=27k−9k=5\begin{aligned} 28k - 14 &= 27k - 9 \\ k &= 5 \end{aligned}$
Since blue counters were represented by $k$ , there are 5 blue counters.

Key Idea

Ratio questions

Do not try to guess the actual numbers straight away. Turn the ratio into algebra first, such as $3 k$ , $k$ , and $4 k$ .

When the equation becomes quadratic

A quadratic equation is an equation containing a squared term, such as $x^2$ . These appear because you often multiply two expressions involving the same unknown.

For example:

x (x - 1)

gives:

x^2 - x

In probability equation questions, you may need to expand brackets, collect terms, factorise, and reject impossible answers.

Example

Unknown number of blue counters

A bag contains 3 red counters and $x$ blue counters. Two counters are chosen at random without replacement. The probability that both counters are blue is $514\frac{5}{14}$ . Find $x$ .

The total number of counters is $x + 3$ .
The probability of the first blue counter is:
$xx+3\frac{x}{x + 3}$
If the first counter was blue, there are $x - 1$ blue counters left, and $x + 2$ counters altogether. So:
$blue)=xx+3×x−1x+2P(\text{blue then blue}) = \frac{x}{x + 3} \times \frac{x - 1}{x + 2}$
Set up the equation:
$xx+3×x−1x+2=514\frac{x}{x + 3} \times \frac{x - 1}{x + 2} = \frac{5}{14}$
Multiply the algebraic fractions:
$x(x−1)(x+3)(x+2)=514\frac{x(x - 1)}{(x + 3)(x + 2)} = \frac{5}{14}$
Cross-multiply:
$14 x (x - 1) = 5 (x + 3) (x + 2)$
Expand both sides:
$14x2−14x=5(x2+5x+6)14x2−14x=5x2+25x+30\begin{aligned} 14x^2 - 14x &= 5(x^2 + 5x + 6) \\ 14x^2 - 14x &= 5x^2 + 25x + 30 \end{aligned}$
Bring everything to one side:
$9x^2 - 39x - 30 = 0$
Divide by 3 to make it simpler:
$3x^2 - 13x - 10 = 0$
Factorise:
$(3 x + 2) (x - 5) = 0$
Solve each factor:
$-\frac{2}{3} \quad \text{or} \quad x = 5$
A number of counters cannot be negative or a fraction, so $x = 5$ .

Common Mistake

Keeping an impossible answer

If a quadratic gives two answers, check them against the context. The number of counters must be a whole number and cannot be negative.

“Show that” probability equations

Sometimes the question asks you to show a particular quadratic first, then solve it. You still begin with the probability statement.

Example

Showing and solving a quadratic

There are $n$ counters in a bag. 6 are red and the rest are blue. Two counters are chosen at random without replacement. The probability of getting two blue counters is $12\frac{1}{2}$ .

Show that $n^2 - 25n + 84 = 0$ , then find $n$ .

If there are $n$ counters altogether and 6 are red, then the number of blue counters is:
$n - 6$
The probability of the first blue counter is:
$n−6n\frac{n - 6}{n}$
After one blue counter is removed, there are $n - 7$ blue counters left, and $n - 1$ counters altogether. So:
$n−6n×n−7n−1=12\frac{n - 6}{n} \times \frac{n - 7}{n - 1} = \frac{1}{2}$
Cross-multiply:
$2 (n - 6) (n - 7) = n (n - 1)$
Expand both sides:
$2(n2−13n+42)=n2−n2n2−26n+84=n2−n\begin{aligned} 2(n^2 - 13n + 42) &= n^2 - n \\ 2n^2 - 26n + 84 &= n^2 - n \end{aligned}$
Bring everything to one side:
$n^2 - 25n + 84 = 0$
Factorise:
$(n - 21) (n - 4) = 0$
Solve:
$\quad \text{or} \quad n = 4$
Since the bag already contains 6 red counters, the total number of counters cannot be 4. Therefore $n = 21$ .

Exam technique

In the exam

Write the number of each colour first, using algebra if needed.
Decide whether the question is with replacement or without replacement before writing the second fraction.
Multiply along the required path, set it equal to the given probability, then solve and reject impossible values.

Self review

Check yourself

If a bag has $x$ blue counters and 5 red counters, what is the total number of counters?
In a without-replacement question, why does the second denominator decrease by 1?
If a quadratic gives two solutions, what checks should you make before choosing your final answer?

Probability Equation Questions

What you'll learn

How to turn “two counters are chosen” information into probability equations.
How to handle with replacement and without replacement questions.
How to use ratios like 3:1 to write algebraic numbers of counters.
How to solve the quadratic equations that often appear in grade 8/9 probability.

The big idea

In these questions, you are not just calculating a probability. You are usually given a probability and asked to work backwards to find how many counters were in the bag.

Definition

Probability equation

A probability equation is an equation formed by setting an algebraic probability equal to a given probability, then solving for the unknown number of items.

An event is the outcome you care about, such as “both counters are blue” or “the first counter is red and the second is red”.

The key skill is translating the words into fractions carefully.

Multiplying probabilities along a path

When two things happen in order, multiply the probabilities along the matching path.

For example, if you want “blue then blue”, multiply:

blue)P(\text{blue then blue}) = P(\text{first blue}) \times P(\text{second blue})

A tree diagram is a useful way to organise the changing probabilities, especially when counters are not replaced.

Tree diagram showing probabilities for two draws without replacement

Definition

Without replacement

Without replacement means the first counter is not put back into the bag before the second counter is chosen. So the total number of counters goes down by 1.

Example

Finding a probability without replacement

A bag contains 5 red counters and 3 blue counters. Two counters are chosen at random without replacement. Find the probability that both counters are blue.

There are 8 counters altogether, and 3 of them are blue, so the probability of the first blue counter is:
$38\frac{3}{8}$
If the first counter was blue, there are now 2 blue counters left out of 7 counters altogether. So the probability of the second blue counter is:
$27\frac{2}{7}$
Multiply the two probabilities:
$38×27\frac{3}{8} \times \frac{2}{7}$
Simplify:
$656=328\frac{6}{56} = \frac{3}{28}$

Common Mistake

Forgetting the bag has changed

For questions without replacement, the denominator for the second pick is usually one less than before. The chosen colour may also have one fewer counter left.

With replacement: the bag resets

Definition

With replacement

With replacement means the first counter is put back into the bag before the second counter is chosen. The probabilities stay the same on both picks.

This often makes the equation simpler because the same fraction is repeated.

Example

Solving a with-replacement equation

A bag contains 4 red counters and $y$ blue counters. A counter is chosen, replaced, then another counter is chosen. The probability of getting two red counters is $425\frac{4}{25}$ . Find $y$ .

The total number of counters is $y + 4$ .
The probability of choosing red on one pick is:
$4y+4\frac{4}{y + 4}$
Because the counter is replaced, the probability of red then red is:
$4y+4×4y+4\frac{4}{y + 4} \times \frac{4}{y + 4}$
Set this equal to the given probability:
$(4y+4)2=425\left(\frac{4}{y + 4}\right)^2 = \frac{4}{25}$
Take the positive square root of both sides:
$4y+4=25\frac{4}{y + 4} = \frac{2}{5}$
Cross-multiply and solve:
$4×5=2(y+4)20=2y+812=2yy=6\begin{aligned} 4 \times 5 &= 2(y + 4) \\ 20 &= 2y + 8 \\ 12 &= 2y \\ y &= 6 \end{aligned}$

Tip

Why the positive square root?

A probability cannot be negative, and the number of counters must be positive, so you only use the positive square root in this type of question.

Using ratios to write the number of counters

A ratio compares quantities. If the ratio of red counters to blue counters is 2:1, then you can write:

red counters as $2 k$
blue counters as $k$
total counters as $3 k$

The letter $k$ represents the multiplier that scales the ratio up to the actual numbers.

Example

Using a ratio to form an equation

Use the ratio 2:1. Let the number of blue counters be $k$ , so the number of red counters is $2 k$ .
The total number of counters is:
$3 k$
The probability of the first red counter is:
$2k3k\frac{2k}{3k}$
If one red counter has been removed, there are $2 k - 1$ red counters left, and $3 k - 1$ counters altogether. So:
$red)=2k3k×2k−13k−1P(\text{red then red}) = \frac{2k}{3k} \times \frac{2k - 1}{3k - 1}$
Set this equal to the given probability:
$2k3k×2k−13k−1=37\frac{2k}{3k} \times \frac{2k - 1}{3k - 1} = \frac{3}{7}$
Simplify $2k3k\frac{2k}{3k}$ to $23\frac{2}{3}$ :
$23×2k−13k−1=37\frac{2}{3} \times \frac{2k - 1}{3k - 1} = \frac{3}{7}$
Cross-multiply:
$14 (2 k - 1) = 9 (3 k - 1)$
Expand and solve:
$28k−14=27k−9k=5\begin{aligned} 28k - 14 &= 27k - 9 \\ k &= 5 \end{aligned}$
Since blue counters were represented by $k$ , there are 5 blue counters.

Key Idea

Ratio questions

Do not try to guess the actual numbers straight away. Turn the ratio into algebra first, such as $3 k$ , $k$ , and $4 k$ .

When the equation becomes quadratic

A quadratic equation is an equation containing a squared term, such as $x^2$ . These appear because you often multiply two expressions involving the same unknown.

For example:

x (x - 1)

gives:

x^2 - x

In probability equation questions, you may need to expand brackets, collect terms, factorise, and reject impossible answers.

Example

Unknown number of blue counters

A bag contains 3 red counters and $x$ blue counters. Two counters are chosen at random without replacement. The probability that both counters are blue is $514\frac{5}{14}$ . Find $x$ .

The total number of counters is $x + 3$ .
The probability of the first blue counter is:
$xx+3\frac{x}{x + 3}$
If the first counter was blue, there are $x - 1$ blue counters left, and $x + 2$ counters altogether. So:
$blue)=xx+3×x−1x+2P(\text{blue then blue}) = \frac{x}{x + 3} \times \frac{x - 1}{x + 2}$
Set up the equation:
$xx+3×x−1x+2=514\frac{x}{x + 3} \times \frac{x - 1}{x + 2} = \frac{5}{14}$
Multiply the algebraic fractions:
$x(x−1)(x+3)(x+2)=514\frac{x(x - 1)}{(x + 3)(x + 2)} = \frac{5}{14}$
Cross-multiply:
$14 x (x - 1) = 5 (x + 3) (x + 2)$
Expand both sides:
$14x2−14x=5(x2+5x+6)14x2−14x=5x2+25x+30\begin{aligned} 14x^2 - 14x &= 5(x^2 + 5x + 6) \\ 14x^2 - 14x &= 5x^2 + 25x + 30 \end{aligned}$
Bring everything to one side:
$9x^2 - 39x - 30 = 0$
Divide by 3 to make it simpler:
$3x^2 - 13x - 10 = 0$
Factorise:
$(3 x + 2) (x - 5) = 0$
Solve each factor:
$-\frac{2}{3} \quad \text{or} \quad x = 5$
A number of counters cannot be negative or a fraction, so $x = 5$ .

Common Mistake

Keeping an impossible answer

If a quadratic gives two answers, check them against the context. The number of counters must be a whole number and cannot be negative.

“Show that” probability equations

Sometimes the question asks you to show a particular quadratic first, then solve it. You still begin with the probability statement.

Example

Showing and solving a quadratic

There are $n$ counters in a bag. 6 are red and the rest are blue. Two counters are chosen at random without replacement. The probability of getting two blue counters is $12\frac{1}{2}$ .

Show that $n^2 - 25n + 84 = 0$ , then find $n$ .

If there are $n$ counters altogether and 6 are red, then the number of blue counters is:
$n - 6$
The probability of the first blue counter is:
$n−6n\frac{n - 6}{n}$
After one blue counter is removed, there are $n - 7$ blue counters left, and $n - 1$ counters altogether. So:
$n−6n×n−7n−1=12\frac{n - 6}{n} \times \frac{n - 7}{n - 1} = \frac{1}{2}$
Cross-multiply:
$2 (n - 6) (n - 7) = n (n - 1)$
Expand both sides:
$2(n2−13n+42)=n2−n2n2−26n+84=n2−n\begin{aligned} 2(n^2 - 13n + 42) &= n^2 - n \\ 2n^2 - 26n + 84 &= n^2 - n \end{aligned}$
Bring everything to one side:
$n^2 - 25n + 84 = 0$
Factorise:
$(n - 21) (n - 4) = 0$
Solve:
$\quad \text{or} \quad n = 4$
Since the bag already contains 6 red counters, the total number of counters cannot be 4. Therefore $n = 21$ .

Exam technique

In the exam

Write the number of each colour first, using algebra if needed.
Decide whether the question is with replacement or without replacement before writing the second fraction.
Multiply along the required path, set it equal to the given probability, then solve and reject impossible values.

Self review

Check yourself

If a bag has $x$ blue counters and 5 red counters, what is the total number of counters?
In a without-replacement question, why does the second denominator decrease by 1?
If a quadratic gives two solutions, what checks should you make before choosing your final answer?

Probability Equation Questions

What you'll learn

The big idea

Multiplying probabilities along a path

With replacement: the bag resets

Using ratios to write the number of counters

When the equation becomes quadratic

“Show that” probability equations

Recap questions

You've reached the end

Probability Equation Questions

What you'll learn

The big idea

Multiplying probabilities along a path

With replacement: the bag resets

Using ratios to write the number of counters

When the equation becomes quadratic

“Show that” probability equations

Recap questions

You've reached the end

Probability Equation Questions

What you'll learn

The big idea

Multiplying probabilities along a path

With replacement: the bag resets

Using ratios to write the number of counters

When the equation becomes quadratic

“Show that” probability equations

Recap questions

You've reached the end

Probability Equation Questions Revision Guide

Probability Equation Questions

What you'll learn

The big idea

Multiplying probabilities along a path

With replacement: the bag resets

Using ratios to write the number of counters

When the equation becomes quadratic

“Show that” probability equations

Recap questions

You've reached the end

Probability Equation Questions Revision Guide