Probability Equation Questions

What you'll learn

How to translate counter problems into algebraic probability equations.
How to handle with replacement and without replacement questions.
How to use ratios like red:blue = 3:1 in probability equations.
How to solve the resulting linear or quadratic equations and reject impossible answers.

The big idea: probability as a product

At Grade 8/9, the probability is usually not the hardest part. The challenge is setting up the equation accurately, then solving it cleanly.

Definition

Probability

The probability of an event is the number of favourable outcomes divided by the total number of possible outcomes. For two required events in a row, use the product rule: $A)P(A\text{ and }B)=P(A)\times P(B\text{ after }A)$ .

So if a question says “both counters are blue”, that means:

first counter is blue and
second counter is blue.

That is why you multiply the two probabilities.

Example

Two counters without replacement

A bag contains 4 red counters and 3 blue counters. Two counters are taken at random without replacement. Find the probability that both are red.

Without replacement, the first red counter removed changes the second red probability from 4 out of 7 to 3 out of 6.

There are 7 counters in total.
The probability the first counter is red is:
$47\frac{4}{7}$
If a red counter has been removed, there are now 3 red counters left out of 6 counters.
Multiply the probabilities:
$47×36=27\frac{4}{7}\times \frac{3}{6}=\frac{2}{7}$

Key Idea

Main method

Write the number of each colour before the first pick.
Decide whether the counter is replaced.
Write the two probabilities and multiply them.
Set the product equal to the probability given in the question.

With replacement: the probabilities reset

Definition

With replacement

“With replacement” means the first counter is put back into the bag before the second counter is taken, so the total number of counters does not change.

With replacement means the bag returns to its original contents before the second pick.

This often creates a squared expression, because the same probability is used twice.

Example

Finding an unknown with replacement

A bag contains 6 red counters and $y$ blue counters. A counter is chosen, replaced, and then another counter is chosen. The probability of getting two red counters is $964\frac{9}{64}$ . Find $y$ .

Because the counter is replaced, both picks use the same red probability, 6 out of 6 + y.

The total number of counters is $6 + y$ .
Because the counter is replaced, the probability of red is the same both times:
$66+y×66+y=964\frac{6}{6+y}\times \frac{6}{6+y}=\frac{9}{64}$
Write this as a square:
$(66+y)2=964\left(\frac{6}{6+y}\right)^2=\frac{9}{64}$
Take the positive square root of both sides:
$66+y=38\frac{6}{6+y}=\frac{3}{8}$
Cross-multiply and solve:
$48=18+3y30=3yy=10\begin{aligned} 48&=18+3y\\ 30&=3y\\ y&=10 \end{aligned}$

Tip

Square root shortcut

When the same event happens twice with replacement, you may be able to take the square root. Use the positive root because probabilities and numbers of counters cannot be negative.

Without replacement: reduce the count and the total

Definition

Without replacement

“Without replacement” means the first counter is not put back, so the total number of counters decreases by 1 before the second pick.

Without replacement means the first counter stays out, so both the chosen colour count and the total may decrease.

If the first counter was blue, then both the number of blue counters and the total number of counters go down by 1.

Example

Forming and solving a quadratic

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

After one blue counter is removed, the second blue probability becomes x − 1 out of x + 2.

The total number of counters at the start is $x + 3$ .
The probability the first counter is blue is:
$xx+3\frac{x}{x+3}$
If one blue counter is removed, there are $x - 1$ blue counters left out of $x + 2$ counters.
Set up the equation:
$xx+3×x−1x+2=514\frac{x}{x+3}\times \frac{x-1}{x+2}=\frac{5}{14}$
Clear the fractions by cross-multiplying:
$14 x (x - 1) = 5 (x + 3) (x + 2)$
Expand and collect terms:
$14x2−14x=5x2+25x+309x2−39x−30=03x2−13x−10=0\begin{aligned} 14x^2-14x&=5x^2+25x+30\\ 9x^2-39x-30&=0\\ 3x^2-13x-10&=0 \end{aligned}$
Factorise:
$(3 x + 2) (x - 5) = 0$
The solutions are $x = 5$ or $x=−23x=-\frac{2}{3}$ . A number of counters must be a whole number, so $x = 5$ .

Common Mistake

Forgetting the removal

For “without replacement”, the second fraction must change. If the first counter was blue, the second blue probability uses $x - 1$ blue counters and one fewer counter in total.

Ratios: use a multiplier

A ratio tells you the relative sizes of the groups, not necessarily the exact numbers.

Definition

Ratio multiplier

If red:blue = 3:2, write red counters as $3 k$ and blue counters as $2 k$ . The total number of counters is then $5 k$ .

A ratio multiplier turns the parts of the ratio into algebraic numbers of counters.

This turns a ratio question into an algebra question.

Example

Using a ratio in a probability equation

The ratio of red counters to blue counters is 3:2. Two counters are taken without replacement. The probability that both counters are blue is $1495\frac{14}{95}$ . Find the total number of counters.

The ratio gives 3k red and 2k blue counters, then without replacement changes the second blue count to 2k − 1.

Let the number of red counters be $3 k$ and the number of blue counters be $2 k$ .
The total number of counters is $5 k$ .
The probability the first counter is blue is:
$2k5k=25\frac{2k}{5k}=\frac{2}{5}$
If one blue counter is removed, the second blue probability is:
$2k−15k−1\frac{2k-1}{5k-1}$
Set up the equation:
$25×2k−15k−1=1495\frac{2}{5}\times \frac{2k-1}{5k-1}=\frac{14}{95}$
Cross-multiply and solve:
$95⋅2(2k−1)=14⋅5(5k−1)190(2k−1)=70(5k−1)380k−190=350k−7030k=120k=4\begin{aligned} 95\cdot 2(2k-1)&=14\cdot 5(5k-1)\\ 190(2k-1)&=70(5k-1)\\ 380k-190&=350k-70\\ 30k&=120\\ k&=4 \end{aligned}$
The total number of counters is $5 k$ , so the total is 20.

“Show that” quadratic questions

Sometimes the question asks you to show a particular quadratic equation first, then solve it. Your job is to start from the probability statement and work forward until the given equation appears.

Example

Showing and solving a quadratic

There are $n$ counters in a bag. 6 of them are red and the rest are blue. Two counters are taken without replacement. The probability that both counters are blue is $12\frac{1}{2}$ .

If the total is n and 6 are red, then the blue count is n − 6 before the first pick and n − 7 before the second.

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

Since there are 6 red counters, the number of blue counters is $n - 6$ .
The probability the first counter is blue is:
$n−6n\frac{n-6}{n}$
If one blue counter is removed, the second blue probability is:
$n−7n−1\frac{n-7}{n-1}$
Set up the probability equation:
$n−6n×n−7n−1=12\frac{n-6}{n}\times \frac{n-7}{n-1}=\frac{1}{2}$
Cross-multiply:
$2 (n - 6) (n - 7) = n (n - 1)$
Expand and simplify to get the required equation:
$2(n2−13n+42)=n2−n2n2−26n+84=n2−nn2−25n+84=0\begin{aligned} 2(n^2-13n+42)&=n^2-n\\ 2n^2-26n+84&=n^2-n\\ n^2-25n+84&=0 \end{aligned}$
Factorise the quadratic:
$n^2-25n+84=(n-21)(n-4)$
The possible values are $n = 21$ or $n = 4$ . Since there are already 6 red counters in the bag, $n = 4$ is impossible, so $n = 21$ .

Common Mistake

Valid roots only

A quadratic may give two algebraic solutions, but a probability question needs a sensible counting answer: positive, whole-number, and consistent with the information in the question.

Exam technique

In the exam

Define your variable clearly, especially whether it represents one colour or the total number of counters.
Check replacement carefully before writing the second fraction.
After solving, reject any root that gives a negative, fractional, or impossible number of counters.

Self review

Check yourself

Can you explain why “both blue” means multiply two probabilities?
In a ratio question, can you write red:blue = 4:1 as algebraic expressions?
If a quadratic gives two answers, can you justify which one is valid?

Probability Equation Questions

What you'll learn

How to translate counter problems into algebraic probability equations.
How to handle with replacement and without replacement questions.
How to use ratios like red:blue = 3:1 in probability equations.
How to solve the resulting linear or quadratic equations and reject impossible answers.

The big idea: probability as a product

At Grade 8/9, the probability is usually not the hardest part. The challenge is setting up the equation accurately, then solving it cleanly.

Definition

Probability

So if a question says “both counters are blue”, that means:

first counter is blue and
second counter is blue.

That is why you multiply the two probabilities.

Example

Two counters without replacement

A bag contains 4 red counters and 3 blue counters. Two counters are taken at random without replacement. Find the probability that both are red.

Without replacement, the first red counter removed changes the second red probability from 4 out of 7 to 3 out of 6.

There are 7 counters in total.
The probability the first counter is red is:
$47\frac{4}{7}$
If a red counter has been removed, there are now 3 red counters left out of 6 counters.
Multiply the probabilities:
$47×36=27\frac{4}{7}\times \frac{3}{6}=\frac{2}{7}$

Key Idea

Main method

Write the number of each colour before the first pick.
Decide whether the counter is replaced.
Write the two probabilities and multiply them.
Set the product equal to the probability given in the question.

With replacement: the probabilities reset

Definition

With replacement

“With replacement” means the first counter is put back into the bag before the second counter is taken, so the total number of counters does not change.

With replacement means the bag returns to its original contents before the second pick.

This often creates a squared expression, because the same probability is used twice.

Example

Finding an unknown with replacement

A bag contains 6 red counters and $y$ blue counters. A counter is chosen, replaced, and then another counter is chosen. The probability of getting two red counters is $964\frac{9}{64}$ . Find $y$ .

Because the counter is replaced, both picks use the same red probability, 6 out of 6 + y.

The total number of counters is $6 + y$ .
Because the counter is replaced, the probability of red is the same both times:
$66+y×66+y=964\frac{6}{6+y}\times \frac{6}{6+y}=\frac{9}{64}$
Write this as a square:
$(66+y)2=964\left(\frac{6}{6+y}\right)^2=\frac{9}{64}$
Take the positive square root of both sides:
$66+y=38\frac{6}{6+y}=\frac{3}{8}$
Cross-multiply and solve:
$48=18+3y30=3yy=10\begin{aligned} 48&=18+3y\\ 30&=3y\\ y&=10 \end{aligned}$

Tip

Square root shortcut

When the same event happens twice with replacement, you may be able to take the square root. Use the positive root because probabilities and numbers of counters cannot be negative.

Without replacement: reduce the count and the total

Definition

Without replacement

“Without replacement” means the first counter is not put back, so the total number of counters decreases by 1 before the second pick.

Without replacement means the first counter stays out, so both the chosen colour count and the total may decrease.

If the first counter was blue, then both the number of blue counters and the total number of counters go down by 1.

Example

Forming and solving a quadratic

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

After one blue counter is removed, the second blue probability becomes x − 1 out of x + 2.

The total number of counters at the start is $x + 3$ .
The probability the first counter is blue is:
$xx+3\frac{x}{x+3}$
If one blue counter is removed, there are $x - 1$ blue counters left out of $x + 2$ counters.
Set up the equation:
$xx+3×x−1x+2=514\frac{x}{x+3}\times \frac{x-1}{x+2}=\frac{5}{14}$
Clear the fractions by cross-multiplying:
$14 x (x - 1) = 5 (x + 3) (x + 2)$
Expand and collect terms:
$14x2−14x=5x2+25x+309x2−39x−30=03x2−13x−10=0\begin{aligned} 14x^2-14x&=5x^2+25x+30\\ 9x^2-39x-30&=0\\ 3x^2-13x-10&=0 \end{aligned}$
Factorise:
$(3 x + 2) (x - 5) = 0$
The solutions are $x = 5$ or $x=−23x=-\frac{2}{3}$ . A number of counters must be a whole number, so $x = 5$ .

Common Mistake

Forgetting the removal

For “without replacement”, the second fraction must change. If the first counter was blue, the second blue probability uses $x - 1$ blue counters and one fewer counter in total.

Ratios: use a multiplier

A ratio tells you the relative sizes of the groups, not necessarily the exact numbers.

Definition

Ratio multiplier

If red:blue = 3:2, write red counters as $3 k$ and blue counters as $2 k$ . The total number of counters is then $5 k$ .

A ratio multiplier turns the parts of the ratio into algebraic numbers of counters.

This turns a ratio question into an algebra question.

Example

Using a ratio in a probability equation

The ratio of red counters to blue counters is 3:2. Two counters are taken without replacement. The probability that both counters are blue is $1495\frac{14}{95}$ . Find the total number of counters.

The ratio gives 3k red and 2k blue counters, then without replacement changes the second blue count to 2k − 1.

Let the number of red counters be $3 k$ and the number of blue counters be $2 k$ .
The total number of counters is $5 k$ .
The probability the first counter is blue is:
$2k5k=25\frac{2k}{5k}=\frac{2}{5}$
If one blue counter is removed, the second blue probability is:
$2k−15k−1\frac{2k-1}{5k-1}$
Set up the equation:
$25×2k−15k−1=1495\frac{2}{5}\times \frac{2k-1}{5k-1}=\frac{14}{95}$
Cross-multiply and solve:
$95⋅2(2k−1)=14⋅5(5k−1)190(2k−1)=70(5k−1)380k−190=350k−7030k=120k=4\begin{aligned} 95\cdot 2(2k-1)&=14\cdot 5(5k-1)\\ 190(2k-1)&=70(5k-1)\\ 380k-190&=350k-70\\ 30k&=120\\ k&=4 \end{aligned}$
The total number of counters is $5 k$ , so the total is 20.

“Show that” quadratic questions

Sometimes the question asks you to show a particular quadratic equation first, then solve it. Your job is to start from the probability statement and work forward until the given equation appears.

Example

Showing and solving a quadratic

There are $n$ counters in a bag. 6 of them are red and the rest are blue. Two counters are taken without replacement. The probability that both counters are blue is $12\frac{1}{2}$ .

If the total is n and 6 are red, then the blue count is n − 6 before the first pick and n − 7 before the second.

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

Since there are 6 red counters, the number of blue counters is $n - 6$ .
The probability the first counter is blue is:
$n−6n\frac{n-6}{n}$
If one blue counter is removed, the second blue probability is:
$n−7n−1\frac{n-7}{n-1}$
Set up the probability equation:
$n−6n×n−7n−1=12\frac{n-6}{n}\times \frac{n-7}{n-1}=\frac{1}{2}$
Cross-multiply:
$2 (n - 6) (n - 7) = n (n - 1)$
Expand and simplify to get the required equation:
$2(n2−13n+42)=n2−n2n2−26n+84=n2−nn2−25n+84=0\begin{aligned} 2(n^2-13n+42)&=n^2-n\\ 2n^2-26n+84&=n^2-n\\ n^2-25n+84&=0 \end{aligned}$
Factorise the quadratic:
$n^2-25n+84=(n-21)(n-4)$
The possible values are $n = 21$ or $n = 4$ . Since there are already 6 red counters in the bag, $n = 4$ is impossible, so $n = 21$ .

Common Mistake

Valid roots only

A quadratic may give two algebraic solutions, but a probability question needs a sensible counting answer: positive, whole-number, and consistent with the information in the question.

Exam technique

In the exam

Define your variable clearly, especially whether it represents one colour or the total number of counters.
Check replacement carefully before writing the second fraction.
After solving, reject any root that gives a negative, fractional, or impossible number of counters.

Self review

Check yourself

Can you explain why “both blue” means multiply two probabilities?
In a ratio question, can you write red:blue = 4:1 as algebraic expressions?
If a quadratic gives two answers, can you justify which one is valid?

Probability Equation Questions

What you'll learn

The big idea: probability as a product

With replacement: the probabilities reset

Without replacement: reduce the count and the total

Ratios: use a multiplier

“Show that” quadratic questions

Recap questions

You've reached the end

Probability Equation Questions

What you'll learn

The big idea: probability as a product

With replacement: the probabilities reset

Without replacement: reduce the count and the total

Ratios: use a multiplier

“Show that” quadratic questions

Recap questions

You've reached the end

Probability Equation Questions

What you'll learn

The big idea: probability as a product

With replacement: the probabilities reset

Without replacement: reduce the count and the total

Ratios: use a multiplier

“Show that” quadratic questions

Recap questions

You've reached the end

Probability Equation Questions Revision Guide

Probability Equation Questions

What you'll learn

The big idea: probability as a product

With replacement: the probabilities reset

Without replacement: reduce the count and the total

Ratios: use a multiplier

“Show that” quadratic questions

Recap questions

You've reached the end

Probability Equation Questions Revision Guide