Probability · Study Notes · Edexcel IGCSE 4MA1 Higher

Probability scale, AND vs OR rules, with/without replacement. The 'AND = multiply, OR = add' rule is the most-tested fact in 4MA1 stats.

At a glance

$0 \le P \le 1$ . Always.
$P(\text{not } A) = 1 - P(A)$ .
OR (mutually exclusive): $P(A \cup B) = P(A) + P(B)$ .
AND (independent): $P(A \cap B) = P(A) \times P(B)$ .
Without replacement: denominator decreases.
Expected frequency = $n \times p$ .
Relative frequency = successes / trials (experimental probability).
Conditional: $P(B|A)$ — given $A$ happened.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

6.2 — Calculate the probability of single and combined events.
6.2 — Use the OR rule for mutually exclusive events.
6.2 — Use the AND rule for independent events.
6.2 — Solve problems with and without replacement.
6.2 — Calculate expected frequency.
6.2 — Use relative frequency to estimate probability.

Probability scale and complement

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All probabilities live in $[0, 1]$ . Complement is super useful.

Probability scale. $0 \le P \le 1$ .

$P = 0$ : impossible.
$P = 1$ : certain.
$P = 0.5$ : even chance.

A valid probability lies between 0 and 1 inclusive; values like 1.3 or −0.2 signal an error.

If your calculation gives $P = 1.3$ or $P = -0.2$ , something has gone wrong.

Equally likely outcomes. When all outcomes are equally likely: $P(A) = \dfrac{\text{number of favourable outcomes}}{\text{total outcomes}}$

Worked example. Bag has 4 red and 6 blue balls. $P(\text{red}) = 4/10 = 2/5$ .

Complement. $P(\text{not } A) = 1 - P(A)$ .

Worked example. $P(\text{not red}) = 1 - 2/5 = 3/5$ . (Or directly: $6/10 = 3/5$ .)

When complement is useful. "At least one" problems become easy: $P(\text{at least one}) = 1 - P(\text{none})$ .

Worked example. Roll a die three times. $P(\text{at least one 6})$ ?

Direct: messy (could be 1, 2, or 3 sixes).
Complement: $P(\text{no 6}) = (5/6)^3 = 125/216$ .
$P(\text{at least one}) = 1 - 125/216 = 91/216$ .

Edexcel tip. "At least one" → use complement. Saves work.

$P \in [0, 1]$ .
Complement: $1 - P(A)$ .
'At least one' → complement.
Sketch problem; identify favourable / total.

OR vs AND — mutually exclusive vs independent

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OR = add (mutually exclusive). AND = multiply (independent). The most-tested rule.

OR rule (mutually exclusive events).

Mutually exclusive: events that CANNOT happen at the same time.

$P(A \cup B) = P(A) + P(B)$

Worked example. Roll a die. $P(\text{2 or 5}) = 1/6 + 1/6 = 1/3$ .

AND rule (independent events).

Independent: outcome of one doesn't affect the other.

$P(A \cap B) = P(A) \times P(B)$

Worked example. Two coins. $P(\text{HH}) = 0.5 \times 0.5 = 0.25$ .

Worked example. A bag has 4 red and 6 blue. Take a ball, replace it, take another. $P(\text{both red}) = (4/10)(4/10) = 16/100 = 4/25$ .

Memory aid. "AND = times. OR = add."

OR adds the probabilities of non-overlapping events; AND multiplies for events happening together.

Are events independent?

Coins, dice, spinners: yes.
Picking with replacement: yes.
Picking WITHOUT replacement: NO — second probability changes.

Edexcel tip. Look for keywords:

"Both" / "and": multiply.
"Either" / "or": add (if mutually exclusive).
"Replaced" / "with replacement": independent.
"Not replaced" / "without replacement": NOT independent.

OR (mutually exclusive): add.
AND (independent): multiply.
Mutually exclusive: can't both happen.
Independent: one doesn't affect the other.

Without replacement and conditional probability

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Denominator changes for the second pick. Be careful.

Without replacement. When you don't put the first item back:

The TOTAL decreases by 1 for the second pick.
The favourable count may also decrease (depending on what you picked first).

Worked example. Bag: 3 red, 5 blue (total 8). Pick two WITHOUT replacement. $P(\text{both red})$ ?

$P(\text{1st red}) = 3/8$ .
After taking a red: 2 red, 5 blue, total 7. $P(\text{2nd red}) = 2/7$ .
$P(\text{both red}) = (3/8)(2/7) = 6/56 = 3/28$ .

Note both numerator (3 → 2) AND denominator (8 → 7) decreased.

After the first pick the bag has one fewer ball, so every second-branch fraction has denominator 7.

Worked example. Same bag. $P(\text{1st red, 2nd blue})$ ?

$P(\text{1st red}) = 3/8$ .
After red: 5 blue out of 7. $P(\text{2nd blue}) = 5/7$ .
Product: $(3/8)(5/7) = 15/56$ .

Conditional probability $P(B|A)$ . "Probability of $B$ GIVEN $A$ has happened."

In the example above, $P(\text{2nd red} | \text{1st red}) = 2/7$ .

Tree diagrams are the standard tool — see the Tree Diagrams subtopic.

Edexcel tip. When you see "without replacement", IMMEDIATELY note that the second-pick probability differs. The most common error is using $3/8 \times 3/8$ for both red.

Numerator AND denominator may change.
Independent events: with replacement.
Conditional: $P(B|A)$ given $A$ has happened.
Tree diagrams help.

See the full worked example for probability →

Expected frequency and relative frequency

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Theory predicts; experiment estimates.

Expected frequency. Expected number of successes in $n$ trials:

$E = n \times p$

Worked example. Roll a die 60 times. Expected number of sixes: $E = 60 \times 1/6 = 10$ .

This is the THEORETICAL expectation — actual results vary.

Relative frequency (experimental probability). Estimate probability from data:

$\text{Relative freq.} = \dfrac{\text{number of successes}}{\text{total trials}}$

Worked example. Spinner spun 200 times, lands on red 80 times. Relative frequency of red = $80/200 = 0.4$ .

When does relative frequency = theoretical probability? As trials → ∞ (Law of Large Numbers). With small samples, relative frequency is just an estimate.

Worked example. Tossed a coin 4 times, got 3 heads. Relative frequency = 0.75. Doesn't mean coin is biased — small sample.

Edexcel tip. "Estimate the probability" or "use the data" → relative frequency. "Theoretical probability" → use formula or count outcomes.

Expected: $n \times p$ .
Relative frequency: from experiment.
More trials → better estimate.
Don't conclude bias from small samples.

Quick recap

$P \in [0, 1]$ . $P(\text{not } A) = 1 - P(A)$ .
OR (mutually exclusive): add.
AND (independent): multiply.
Without replacement: denominator decreases.
Expected = $n \times p$ .
Relative frequency = successes / trials.
'At least one' → use complement.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

OR: $P(A \cup B) = P(A) + P(B)$ (mutually exclusive).
AND: $P(A \cap B) = P(A) \times P(B)$ (independent).
Complement: $P(\text{not } A) = 1 - P(A)$ .
Expected: $n \times p$ .
Without replacement: reduce both numerator and denominator.

How it’s examined

Probability questions are 4-8 marks across both papers. Tree diagrams + without-replacement is the most common 6-mark question. Examiner reports flag: (1) AND/OR confusion, (2) not reducing denominator without replacement, (3) probabilities outside $[0,1]$ .

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024, 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on probability , written exactly the way a tutor would explain them at the board.

1Basic probability
Foundation• probability
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Question
A bag has 4 red and 6 blue balls. Find the probability of picking a red ball.
Step-by-step solution
1. Step 1
  $P(\text{red}) = \dfrac{\text{number of red}}{\text{total}}$ .
2. Step 2
  $P(\text{red}) = 4/10 = 2/5$ .
Answer
$P(\text{red}) = 2/5 = 0.4$
2OR rule (mutually exclusive)
Higher• OR
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Question
$P(A) = 0.3$ , $P(B) = 0.5$ , $A$ and $B$ mutually exclusive. Find $P(A \text{ or } B)$ .
Step-by-step solution
1. Step 1
  Mutually exclusive: can't both happen. $P(A \cup B) = P(A) + P(B)$ .
2. Step 2
  $P(A \cup B) = 0.3 + 0.5 = 0.8$ .
Answer
$P(A \cup B) = 0.8$
3AND rule (independent events)
Higher• Adapted from 4MA1/1H May/Jun 2024 Q21• AND, independent
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Question
Two fair coins. Find $P(\text{both heads})$ .
Step-by-step solution
1. Step 1
  Independent events: $P(A \text{ and } B) = P(A) \times P(B)$ .
2. Step 2
  $P(\text{HH}) = 0.5 \times 0.5 = 0.25$ .
Answer
$P(\text{HH}) = 1/4 = 0.25$
4Expected frequency
Foundation• expected frequency
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Question
Probability of rolling a 6 = $1/6$ . If you roll 60 times, how many sixes do you expect?
Step-by-step solution
1. Step 1
  Expected frequency = probability × number of trials.
2. Step 2
  $1/6 \times 60 = 10$ .
Answer
$10$ sixes
5Relative frequency (experimental)
Higher• relative frequency
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Question
A spinner is spun 200 times. It lands on red 80 times. Find the relative frequency of red.
Step-by-step solution
1. Step 1
  Relative frequency = number of successes / total trials.
2. Step 2
  $80 / 200 = 0.4$ .
Answer
$0.4$
Examiner tip
Relative frequency estimates probability from experiments. The more trials, the better the estimate.
6Conditional probability without replacement
Higher• Adapted from 4MA1/2H May/Jun 2024 Q19• conditional
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Question
Bag: 3 red, 5 blue. Pick two without replacement. Find $P(\text{both red})$ .
Step-by-step solution
1. Step 1
  $P(\text{1st red}) = 3/8$ .
2. Step 2
  After taking a red, 2 red and 5 blue remain. $P(\text{2nd red}) = 2/7$ .
3. Step 3
  $P(\text{both red}) = (3/8)(2/7) = 6/56 = 3/28$ .
Answer
$3/28$
Examiner tip
Without replacement: numerator AND denominator change. Common slip: forgetting to reduce denominator.
7Why six rolls do not guarantee a six [1 mark]
Getting started• Paper 1H/2H style — 1 mark• probability, probability language
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Question
Rashid has a fair six-sided dice. He says,

"The probability of getting a $6$ is $\frac{1}{6}$ , so if I roll the dice $6$ times I will definitely get at least one $6$ ."

Explain why Rashid is wrong. [1]
Step-by-step solution
1. Step 1
  Each roll is independent of the others, and each roll has a probability of $\frac{5}{6}$ of not being a $6$ .
2. Step 2
  B1 So it is possible for all six rolls to be something other than $6$ . A probability of $\frac{1}{6}$ means he should expect about one $6$ in six rolls on average, not that one is certain.
Answer
It is possible to roll the dice $6$ times without getting a $6$ ; $\frac{1}{6}$ gives an expected (average) number, not a certainty.
Examiner tip
Examiner tip: An explanation must say what can actually happen: here, that no $6$ at all is possible. Expected frequency is a long-run average, not a promise.

Common mistake: Writing "because it is random" or "because probability is not certain" with no reference to the six rolls. Answers that do not engage with the claim do not score.
8Equally likely letters in a word [2 marks]
Getting started• Paper 1H/2H style — 2 marks• probability, mutually exclusive
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Question
Each letter of the word PARALLEL is written on a separate card. The $8$ cards are put in a bag. Sofia takes one card at random.

Work out the probability that the card shows A or L. [2]
Step-by-step solution
1. Step 1
  M1 Count the letters: P, A, R, A, L, L, E, L. There are $2$ cards showing A and $3$ showing L. A card cannot show both letters, so add the probabilities.
  $\frac{2}{8} + \frac{3}{8}$
2. Step 2
  A1 Add.
  $\frac{5}{8}$
Answer
$\frac{5}{8}$ oe
Examiner tip
Examiner tip: Every card is equally likely, so count cards, not different letters. Check by counting the cards that are not A or L: P, R, E, which is $\frac{3}{8}$ , and $\frac{5}{8} + \frac{3}{8} = 1$ .

Common mistake: Counting the different letters (P, A, R, L, E) and answering $\frac{2}{5}$ . The repeated letters are separate cards and each one counts.
9Sample space for the difference of two spinners [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• probability, sample space
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Question
A fair spinner has $5$ equal sections numbered $1$ to $5$ . Another fair spinner has $4$ equal sections numbered $1$ to $4$ .

Zac spins both spinners. His score is the larger number minus the smaller number (or $0$ if they are equal).

Work out the probability that Zac's score is $2$ . [3]
Step-by-step solution
1. Step 1
  M1 Draw a sample-space grid of the differences, with the first spinner across ( $1$ to $5$ ) and the second down ( $1$ to $4$ ). It has $5 \times 4 = 20$ equally likely cells.
  $5 \times 4 = 20$
2. Step 2
  M1 Find the cells with a difference of $2$ . Writing (first, second), they are $(1, 3)$ , $(2, 4)$ , $(3, 1)$ , $(4, 2)$ and $(5, 3)$ .
  $5 \text{ cells}$
3. Step 3
  A1 Write the probability and simplify.
  $\frac{5}{20} = \frac{1}{4}$
Answer
$\frac{5}{20}$ or $\frac{1}{4}$ oe
Examiner tip
Examiner tip: Fill in the whole grid before counting. A difference of $2$ can come with either spinner showing the bigger number, so check both sides of the diagonal.

Common mistake: Finding only the cases where the first spinner is bigger ( $(3, 1)$ , $(4, 2)$ , $(5, 3)$ ) and answering $\frac{3}{20}$ .
10Finding the number of counters when counters are replaced [5 marks]
Stretch• Paper 1H/2H style — 5 marks• probability, independent events, forming equations
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Question
A bag contains only red and green counters. There are $3$ more green counters than red counters.

A counter is taken at random, its colour is recorded and it is put back. A second counter is then taken at random.

The probability that both counters are red is $\frac{4}{25}$ .

Work out the total number of counters in the bag. [5]
Step-by-step solution
1. Step 1
  M1 Let there be $r$ red counters. Then there are $r + 3$ green, so $2r + 3$ counters in total.
  $P(\text{red}) = \frac{r}{2r + 3}$
2. Step 2
  M1 The counter is put back, so the two draws are independent and the probability is the same each time.
  $\left(\frac{r}{2r + 3}\right)^2 = \frac{4}{25}$
3. Step 3
  M1 Take the positive square root (a probability is positive) and clear the fractions.
  $\frac{r}{2r + 3} = \frac{2}{5} \quad\Rightarrow\quad 5r = 4r + 6$
4. Step 4
  A1 Solve.
  $r = 6$
5. Step 5
  A1 There are $6$ red and $9$ green counters.
  $2 \times 6 + 3 = 15$
Answer
$15$
Examiner tip
Examiner tip: Check the answer in the original condition: $\frac{6}{15} = \frac{2}{5}$ , and $\left(\frac{2}{5}\right)^2 = \frac{4}{25}$ . Read the last line of the question again: it asks for the total, not the number of red counters.

Common mistake: Giving $6$ as the answer, or treating the draws as without replacement and writing $\frac{r}{2r + 3} \times \frac{r - 1}{2r + 2}$ , which is the wrong model when the counter is put back.

Question 1
Paper 1H/2H style — 1 mark1 mark
A fair spinner has $8$ equal sections numbered $1$ to $8$ .

Write down the probability that the spinner lands on a number greater than $5$ . [1]
Model answer
Numbers greater than $5$ : $6$ , $7$ , $8$ , which is $3$ out of $8$ .

Answer: $\frac{3}{8}$
Why this scores
Mark scheme:
- B1 for $\frac{3}{8}$ oe
Why this scores full marks: "Greater than $5$ " does not include $5$ , so there are three numbers, not four. The answer is a fraction, not a ratio such as $3 : 8$ , which is not accepted as a probability.
Question 2
Paper 1H/2H style — 1 mark1 mark
A spinner can land on $A$ , $B$ or $C$ . Zara says that the probability of landing on $A$ is $0.6$ and the probability of landing on $B$ is $0.5$ .

Explain why Zara cannot be right. [1]
Model answer
The spinner cannot land on $A$ and $B$ at the same time, so P( $A$ or $B$ ) $= 0.6 + 0.5 = 1.1$ . A probability cannot be more than $1$ .

Answer: the probabilities add up to more than $1$
Why this scores
Mark scheme:
- B1 for a correct explanation, e.g. $0.6 + 0.5 = 1.1$ , which is greater than $1$ oe
Why this scores full marks: The candidate uses the fact that the outcomes are mutually exclusive, so their probabilities add, and shows the total is too big. " $0.6$ is too high" on its own does not explain anything.
Question 3
Paper 1H/2H style — 2 marks2 marks
A factory tests $600$ light bulbs and finds that $18$ are faulty.

The factory makes $4500$ light bulbs in a day.

Work out an estimate for the number of faulty light bulbs made in a day. [2]
Model answer
$\frac{18}{600} = 0.03$

$0.03 \times 4500 = 135$

Answer: $135$
Why this scores
Mark scheme:
- M1 for $\frac{18}{600} \times 4500$ oe
- A1 for $135$ cao
Why this scores full marks: The relative frequency from the test is used as the estimate of the probability, then multiplied by the number of bulbs. Scaling up ( $4500 \div 600 = 7.5$ , and $18 \times 7.5 = 135$ ) is equally good.
Question 4
Paper 1H/2H style — 2 marks2 marks
Spinner P has $5$ equal sections, $2$ of which are red. Spinner Q has $4$ equal sections, and some of them are red. Both spinners are fair.

Both spinners are spun. The probability that both spinners land on red is $\frac{1}{10}$

Work out the number of red sections on spinner Q. [2]
Model answer
$\frac{2}{5} \times P(\text{Q red}) = \frac{1}{10}$

$P(\text{Q red}) = \frac{1}{10} \div \frac{2}{5} = \frac{1}{4}$

$\frac{1}{4}$ of $4$ sections is $1$ section.

Answer: $1$
Why this scores
Mark scheme:
- M1 for $\frac{1}{10} \div \frac{2}{5}$ oe
- A1 for $1$ cao
Why this scores full marks: The spinners do not affect each other, so P(both red) is the product of the two probabilities; working backwards means dividing by $\frac{2}{5}$ . The candidate then answers the question asked, a number of sections, rather than stopping at the probability $\frac{1}{4}$ .
Question 5
Paper 1H/2H style — 3 marks3 marks
A bag contains only blue, red and green counters.

The probability of taking a blue counter is $\frac{1}{3}$ . The probability of taking a red counter is $\frac{1}{4}$ .

There are $10$ green counters in the bag.

Work out the total number of counters in the bag. [3]
Model answer
$P(\text{green}) = 1 - \left(\frac{1}{3} + \frac{1}{4}\right) = 1 - \frac{7}{12} = \frac{5}{12}$

$\frac{5}{12}$ of the counters is $10$ counters, so $\frac{1}{12}$ is $2$ counters.

$2 \times 12 = 24$

Answer: $24$
Why this scores
Mark scheme:
- M1 for $1 - \left(\frac{1}{3} + \frac{1}{4}\right)$ oe $\left(= \frac{5}{12}\right)$
- M1 for $10 \div \frac{5}{12}$ oe
- A1 for $24$ cao
Why this scores full marks: The candidate turns the green probability into a number of counters by working back from $\frac{5}{12}$ . A check: $\frac{1}{3}$ of $24$ is $8$ blue, $\frac{1}{4}$ of $24$ is $6$ red, and $8 + 6 + 10 = 24$ .
Question 6
Paper 1H/2H style — 3 marks3 marks
A fair coin is thrown three times.

(a) List all the possible outcomes. [1]

(b) Work out the probability of getting exactly two heads. [2]
Model answer
(a) HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

(b) Exactly two heads: HHT, HTH, THH, which is $3$ out of $8$ .

Answer: $\frac{3}{8}$
Why this scores
Mark scheme:
- B1 for all $8$ outcomes listed with no repeats
- M1 for identifying HHT, HTH and THH, or $3 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$
- A1 for $\frac{3}{8}$ oe
Why this scores full marks: The list is built systematically (the first coin fixed as H, then as T), so there are $2 \times 2 \times 2 = 8$ outcomes. The two heads can be in $3$ different positions, and all three are counted; $\frac{1}{8}$ is the answer for one order only.
Question 7
Paper 1H/2H style — 4 marks4 marks
Five students, Amy, Ben, Cara, Dev and Esi, want to be in a quiz team. Two of the five students are chosen at random to be in the team.

(a) List all the possible pairs of students that could be chosen. [2]

(b) Work out the probability that Amy is chosen. [2]
Model answer
(a) Amy and Ben, Amy and Cara, Amy and Dev, Amy and Esi

Ben and Cara, Ben and Dev, Ben and Esi

Cara and Dev, Cara and Esi

Dev and Esi

(b) Amy is in $4$ of the $10$ equally likely pairs.

$\frac{4}{10} = \frac{2}{5}$

Answer: $\frac{2}{5}$
Why this scores
Mark scheme:
- B2 for all $10$ pairs with no repeats and no extras (B1 for at least $6$ correct pairs)
- M1 for $\frac{4}{10}$ , or $\frac{4}{\text{their number of pairs}}$
- A1 for $\frac{2}{5}$ oe
Why this scores full marks: The list is systematic: Amy with each of the other four, then Ben with the three after him, and so on, giving $4 + 3 + 2 + 1 = 10$ pairs. "Amy and Ben" and "Ben and Amy" are the same team, so it is listed once; listing both orders gives $20$ and double counts every team. A check for (b): Amy is not chosen in the $6$ pairs without her, and $1 - \frac{6}{10} = \frac{2}{5}$ .
Question 8
Paper 1H/2H style — 4 marks4 marks
A fair coin is thrown and a fair six-sided dice is rolled.

(a) Work out the probability of getting a head and a $6$ . [1]

(b) Work out the probability of getting a head or a $6$ or both. [3]
Model answer
(a) $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$

Answer: $\frac{1}{12}$

(b) The only outcome that fails is a tail and not a $6$ .

$P(\text{tail and not } 6) = \frac{1}{2} \times \frac{5}{6} = \frac{5}{12}$

$1 - \frac{5}{12} = \frac{7}{12}$

Answer: $\frac{7}{12}$
Why this scores
Mark scheme:
- B1 for $\frac{1}{12}$ oe
- M1 for $\frac{1}{2} \times \frac{5}{6}$
- M1 for $1 - \frac{1}{2} \times \frac{5}{6}$ oe
- A1 for $\frac{7}{12}$ oe
Why this scores full marks: A head and a $6$ are not mutually exclusive (they can both happen), so $\frac{1}{2} + \frac{1}{6} = \frac{8}{12}$ counts the outcome "head and $6$ " twice. Using the complement avoids the double count. A check from the $12$ -outcome sample space: $6$ heads plus the tail with a $6$ gives $7$ outcomes.
Question 9
Paper 1H/2H style — 5 marks5 marks
Kim rolls a dice $250$ times. The table shows some of her results.

Score $1$ : $35$ times. Score $2$ : $40$ times. Score $3$ : $38$ times. Score $4$ : $42$ times. Score $5$ : $35$ times. Score $6$ : not recorded.

(a) Work out the relative frequency of a score of $6$ . [2]

Kim rolls the dice twice more.

(b) Use your answer to (a) to estimate the probability that both rolls are a $6$ . [2]

Kim rolls the dice another $800$ times.

(c) Work out an estimate for the number of times she gets a $6$ . [1]
Model answer
(a) $35 + 40 + 38 + 42 + 35 = 190$

$250 - 190 = 60$

$\frac{60}{250} = 0.24$

Answer: $0.24$

(b) $0.24 \times 0.24 = 0.0576$

Answer: $0.0576$

(c) $0.24 \times 800 = 192$

Answer: $192$
Why this scores
Mark scheme:
- M1 for $250 - (35 + 40 + 38 + 42 + 35)$ $(= 60)$
- A1 for $0.24$ oe
- M1 for $0.24 \times 0.24$ ft their (a)
- A1 for $0.0576$ oe
- B1 ft for $192$
Why this scores full marks: The missing frequency comes from the total of $250$ . The two rolls are independent, so the estimated probability is squared, not doubled ( $0.48$ would be the error). In (c) the relative frequency is used as the probability for the expected frequency.
Question 10
Paper 1H/2H style — 5 marks5 marks
At a school fair, a game costs £ $0.50$ to play. A player rolls a fair six-sided dice and spins a spinner. The probability that the spinner lands on red is $0.4$ . The dice and the spinner are independent.

A player wins £ $5$ if the dice shows $6$ and the spinner lands on red.

$60$ people each play the game once.

Work out an estimate for the profit the school makes. [5]
Model answer
$P(\text{win}) = \frac{1}{6} \times 0.4 = \frac{1}{15}$

Expected number of winners: $\frac{1}{15} \times 60 = 4$

Money taken: $60 \times 0.50 =$ £ $30$

Prizes paid: $4 \times 5 =$ £ $20$

Profit: $30 - 20 = 10$

Answer: £ $10$
Why this scores
Mark scheme:
- M1 for $\frac{1}{6} \times 0.4$
- A1 for $\frac{1}{15}$ oe
- M1 for their $\frac{1}{15} \times 60$ $(= 4)$
- M1 for $60 \times 0.5 - 5 \times$ their $4$
- A1 for $10$ cao
Why this scores full marks: The candidate multiplies the two independent probabilities, uses expected frequency to estimate the number of winners, and then compares the money taken with the prizes paid. The answer is an estimate: the actual number of winners could differ from $4$ .

4MA1 Probability

At a glance

What you’ll learn

Probability scale and complement

OR vs AND — mutually exclusive vs independent

Without replacement and conditional probability

Expected frequency and relative frequency

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Basic probability

2OR rule (mutually exclusive)

3AND rule (independent events)

4Expected frequency

5Relative frequency (experimental)

6Conditional probability without replacement

7Why six rolls do not guarantee a six [1 mark]

8Equally likely letters in a word [2 marks]

9Sample space for the difference of two spinners [3 marks]

10Finding the number of counters when counters are replaced [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10