Probability (Single Event)
The Skill
What is Probability?
Probability measures how likely an event is to happen, on a scale from 0 to 1:
- 0 = Impossible
- 1 = Certain
- Values in between show likelihood
$$P(\text{event}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$$
Key Vocabulary
- Event: Something that might happen (e.g., "rolling a 6")
- Outcome: A possible result (e.g., 1, 2, 3, 4, 5, or 6)
- Favourable outcomes: The outcomes you want
- Sample space: All possible outcomes
Probability as Different Forms
Probability can be written as a fraction, decimal, or percentage.
$P = \frac{1}{4} = 0.25 = 25\%$
Complementary Events
If $P(A)$ is the probability of event A happening:
$$P(\text{not } A) = 1 - P(A)$$
The probabilities of all possible outcomes sum to 1.
The Traps
Common misconceptions and how to avoid them.
Assuming outcomes are equally likely when they are not "The Fair Fallacy"
The Mistake in Action
There are 3 options: win, lose, or draw. Find P(win).
Student writes: "P(win) = 1/3"
Why It Happens
Students see three outcomes and assume each has equal probability, when the question hasn't stated this. Real-world outcomes are often not equally likely.
The Fix
Only use $\frac{1}{n}$ if outcomes are EQUALLY LIKELY (fair dice, fair coin, etc.)
If the question doesn't say "fair" or "equally likely", you need more information.
For a football match, win/lose/draw are NOT equally likely — they depend on the teams!
Ask yourself: "Does the question tell me these outcomes are equally likely?" If not, you can't assume they are.
Spot the Mistake
Outcomes: win, lose, draw (3 outcomes)
P(win) = 1/3
Click on the line that contains the error.
Confusing expected frequency with probability "The Expected Error"
The Mistake in Action
A coin is flipped 50 times. The probability of heads is 0.5. How many heads are expected?
Student writes: "Expected heads = 0.5"
Why It Happens
Students give the probability as the answer when asked for expected frequency. They don't multiply by the number of trials.
The Fix
Expected frequency = Probability × Number of trials
Expected heads = $0.5 × 50 = 25$
The probability (0.5) tells you the proportion you expect. Multiply by the number of trials to get the actual count.
Think of it as: "If I flip 50 times and expect half to be heads, that's 50 × 0.5 = 25 heads"
Spot the Mistake
P(heads) = 0.5
Number of flips = 50
Expected heads = 0.5
Click on the line that contains the error.
Giving probability greater than 1 "The Impossible Probability"
The Mistake in Action
A bag contains 3 red and 5 blue balls. Find P(red).
Student writes: "P(red) = 3/5 = 1.6"
Why It Happens
Students sometimes divide the wrong way around, or confuse probability with odds. They might divide favourable by unfavourable instead of favourable by total.
The Fix
Probability is always between 0 and 1 (inclusive).
Check your answer: If you get something greater than 1 or negative, something is wrong!
Correct: $P(\text{red}) = \frac{3}{3+5} = \frac{3}{8}$
The denominator is always the TOTAL number of outcomes, not the other outcomes.
Spot the Mistake
Red balls = 3, Blue balls = 5
P(red) = 3/5
= 0.6
Click on the line that contains the error.
Mixing up percentage and decimal forms "The Percentage Pickle"
The Mistake in Action
The probability of winning a prize is 15%. Write this as a decimal.
Student writes: "15% = 1.5"
Why It Happens
Students know they need to remove the percentage sign but aren't sure how to convert. Some divide by 10 instead of 100, or remove the sign without any calculation.
The Fix
Percentage to decimal: Divide by 100 (move digits 2 places right)
$15\% = 15 ÷ 100 = 0.15$
Decimal to percentage: Multiply by 100 (move digits 2 places left)
$0.15 × 100 = 15\%$
Quick check: 15% is less than half, so the decimal must be less than 0.5. Is 1.5 < 0.5? No! So 1.5 can't be right.
Spot the Mistake
Convert 15% to a decimal
15% = 1.5
Click on the line that contains the error.
Subtracting the wrong value for complementary events "The Complement Confusion"
The Mistake in Action
P(rain) = 0.3. Find P(no rain).
Student writes: "P(no rain) = 0.3 − 1 = −0.7"
Why It Happens
Students know that complementary events involve subtraction but subtract in the wrong order, giving a negative probability (which is impossible).
The Fix
The complement formula is: $$P(\text{not } A) = 1 - P(A)$$
It's always 1 minus the probability, never the other way around.
$P(\text{no rain}) = 1 - 0.3 = 0.7$
Check: Negative probability? Impossible! Go back and fix it.
Spot the Mistake
P(rain) = 0.3
P(no rain) = 0.3 − 1 = −0.7
Click on the line that contains the error.
Using wrong total for denominator "The Total Trouble"
The Mistake in Action
A fair dice is rolled. Find the probability of getting a number greater than 4.
Student writes: "Numbers greater than 4 are 5 and 6. P = 2/4"
Why It Happens
Students correctly identify the favourable outcomes but then use the count of listed outcomes (or some other number) as the denominator instead of the total possible outcomes.
The Fix
Always ask: "How many outcomes are possible in total?"
A dice has 6 faces, so total outcomes = 6.
Favourable outcomes (>4) = 5 and 6 = 2 outcomes
$$P(>4) = \frac{2}{6} = \frac{1}{3}$$
The denominator comes from the original situation, not from your working.
Spot the Mistake
Numbers greater than 4: 5, 6
That's 2 numbers
P = 2/4 = 1/2
Click on the line that contains the error.
The Deep Dive
Apply your knowledge with these exam-style problems.
Level 1: Fully Worked
Complete solutions with commentary on each step.
Question
A bag contains 4 red counters, 6 blue counters, and 5 green counters.
A counter is picked at random. Find: (a) P(red) (b) P(blue) (c) P(not green)
Solution
First: Find the total Total counters = 4 + 6 + 5 = 15
(a) P(red) $$P(\text{red}) = \frac{\text{red counters}}{\text{total}} = \frac{4}{15}$$
(b) P(blue) $$P(\text{blue}) = \frac{6}{15} = \frac{2}{5}$$
(Simplified by dividing top and bottom by 3)
(c) P(not green) Method 1: Use complement $$P(\text{not green}) = 1 - P(\text{green}) = 1 - \frac{5}{15} = 1 - \frac{1}{3} = \frac{2}{3}$$
Method 2: Count non-green Not green = red + blue = 4 + 6 = 10 $$P(\text{not green}) = \frac{10}{15} = \frac{2}{3}$$
Question
A fair six-sided dice is rolled once.
Find the probability of rolling: (a) a 4 (b) an odd number (c) a number less than 5 (d) a 7
Solution
Total possible outcomes = 6 (the numbers 1, 2, 3, 4, 5, 6)
(a) P(4) Only one face shows 4. $$P(4) = \frac{1}{6}$$
(b) P(odd) Odd numbers on a dice: 1, 3, 5 (three outcomes) $$P(\text{odd}) = \frac{3}{6} = \frac{1}{2}$$
(c) P(less than 5) Numbers less than 5: 1, 2, 3, 4 (four outcomes) $$P(<5) = \frac{4}{6} = \frac{2}{3}$$
(d) P(7) There is no 7 on a standard dice — this is impossible. $$P(7) = \frac{0}{6} = 0$$
Question
Amy spins a spinner 200 times. Her results are shown below.
| Colour | Frequency |
|---|---|
| Red | 52 |
| Blue | 88 |
| Yellow | 60 |
(a) Estimate the probability of landing on blue. (b) If Amy spins the spinner 500 more times, how many times would she expect to land on red?
Solution
(a) Estimate P(blue)
With experimental probability, we use: $$P(\text{blue}) = \frac{\text{frequency of blue}}{\text{total spins}} = \frac{88}{200} = \frac{11}{25} = 0.44$$
(b) Expected frequency of red
First, find the experimental probability of red: $$P(\text{red}) = \frac{52}{200} = 0.26$$
Then use: Expected frequency = Probability × Number of trials $$\text{Expected red} = 0.26 × 500 = 130$$
Answer: She would expect red about 130 times.
(Note: This is an estimate based on experimental data. The actual result may differ.)
Question
The table shows the probabilities of picking different flavour sweets from a bag.
| Flavour | Probability |
|---|---|
| Strawberry | 0.25 |
| Lemon | 0.15 |
| Orange | x |
| Lime | 0.3 |
Find the value of x.
Solution
Key fact: All probabilities must sum to 1.
Step 1: Write the equation $$0.25 + 0.15 + x + 0.3 = 1$$
Step 2: Simplify the left side $$0.7 + x = 1$$
Step 3: Solve for x $$x = 1 - 0.7 = 0.3$$
Answer: x = 0.3
Check: 0.25 + 0.15 + 0.3 + 0.3 = 1 ✓
Question
(a) Write $\frac{3}{5}$ as a percentage. (b) Write 72% as a fraction in its simplest form. (c) Write 0.125 as a fraction in its simplest form.
Solution
(a) Fraction to percentage $$\frac{3}{5} = \frac{3}{5} × 100\% = \frac{300}{5}\% = 60\%$$
Or: $3 ÷ 5 = 0.6$, then $0.6 × 100 = 60\%$
(b) Percentage to fraction $$72\% = \frac{72}{100}$$
Simplify by dividing by 4: $$\frac{72}{100} = \frac{18}{25}$$
(c) Decimal to fraction $0.125$ has 3 decimal places, so: $$0.125 = \frac{125}{1000}$$
Simplify (÷125): $$\frac{125}{1000} = \frac{1}{8}$$
Level 2: Scaffolded
Fill in the key steps.
Question
A fair spinner has 8 equal sections numbered 1-8.
Find: (a) P(spinning a 5) (b) P(spinning an even number) (c) P(spinning a number greater than 6)
Level 3: Solo
Try it yourself!
Question
A box contains only red, blue, and green marbles.
The probability of picking a red marble is $\frac{2}{5}$. The probability of picking a blue marble is 0.35.
(a) What is the probability of picking a green marble? (b) There are 40 marbles in the box. How many are red? (c) Is it more likely to pick blue or green? Explain your answer.
Show Solution
(a) P(green) All probabilities sum to 1.
First, convert to same form: $\frac{2}{5} = 0.4$
$P(\text{green}) = 1 - 0.4 - 0.35 = 0.25$ or $\frac{1}{4}$
(b) Number of red marbles $P(\text{red}) = \frac{2}{5}$
Number of red = $\frac{2}{5} × 40 = 16$
Answer: 16 red marbles
(c) Comparison $P(\text{blue}) = 0.35$ $P(\text{green}) = 0.25$
$0.35 > 0.25$
Blue is more likely because it has a higher probability.
Examiner's View
Mark allocation patterns:
- Basic probability from description: 1-2 marks
- Probability from frequency table: 2 marks
- Finding P(not A): 2 marks
- Setting up equations using P = 1: 3 marks
Common errors examiners see:
- Writing probability greater than 1
- Confusing "number of outcomes" with frequency
- Not simplifying fractions when required
- Forgetting that probabilities must sum to 1
What gains full marks:
- Always show the fraction setup: favourable/total
- Simplify fractions unless told otherwise
- For "not" questions, show 1 − P(event)
- Check your answer is between 0 and 1
AQA Notes
AQA likes using probability with frequency tables from experiments. The question will say "estimate the probability" when using experimental data.