TMUA: Statistics and probability
Statistics and probability at GCSE level.
- Section 1 Part 2 · M1-M7 GCSE-level knowledge (Part 2) · spec M6-M7
- Papers 1 and 2
- 8 practice questions
- No calculator
What the specification covers
- Averages, spread, histograms, cumulative frequency
- Tree diagrams, Venn diagrams, conditional probability
Key ideas
- Mean \(=\tfrac{\text{total}}{n}\); changing one value changes the total, and the mean with it.
- Median: the middle value when ordered (average of the two middle values for an even count).
- Without replacement, the second probability's numerator and denominator both change.
- \(P(A\text{ then }B)=P(A)\times P(B\mid A)\); for 'different colours', add both orders.
Common mistakes
- 'Unique mode' means one value occurs more often than every other.
- Conditional probability: restrict to the given group first.
Exam tip
For small counting problems, list the cases systematically.
Worked example
Worked example
A bag contains 5 red and 3 blue counters. Two counters are taken at random, without replacement. What is the probability that they are different colours?
- \(\frac{13}{28}\)
- \(\frac{15}{64}\)
- \(\frac{15}{32}\)
- \(\frac{15}{28}\)
- \(\frac{15}{56}\)
Answer: D: \(\frac{15}{28}\)
\(P(\text{RB})+P(\text{BR})=\tfrac58\cdot\tfrac37+\tfrac38\cdot\tfrac57=\tfrac{30}{56}=\tfrac{15}{28}\).
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
Five positive integers have mean 6, median 7 and a unique mode of 8. What is the largest possible range of the five integers?
- \(7\)
- \(4\)
- \(5\)
- \(8\)
- \(6\)
Show the answer and solution
Answer: A: \(7\)
In order \(a\le b\le 7\le d\le e\) with total 30. The mode 8 must appear at least twice, and only \(d,e\) can be 8 (three 8s would move the median), so \(d=e=8\) and \(a+b=7\). Then \(a\ne b\) (7 is odd), so no value other than 8 repeats. The range is \(8-a\), largest when \(a=1\) (\(b=6\)): 7.
Question 2
A bag contains 5 red and 2 blue counters. Two counters are taken at random, without replacement. What is the probability that they are different colours?
- \(\frac{10}{49}\)
- \(\frac{10}{21}\)
- \(\frac{20}{49}\)
- \(\frac{11}{21}\)
- \(\frac{5}{21}\)
Show the answer and solution
Answer: B: \(\frac{10}{21}\)
\(P(\text{RB})+P(\text{BR})=\tfrac{5}{7}\cdot\tfrac{2}{6}+\tfrac{2}{7}\cdot\tfrac{5}{6}=\frac{10}{21}\).
Question 3
A bag contains 4 red and 3 blue counters. Two counters are taken at random, without replacement. What is the probability that they are different colours?
- \(\frac{12}{49}\)
- \(\frac{3}{7}\)
- \(\frac{2}{7}\)
- \(\frac{4}{7}\)
- \(\frac{24}{49}\)
Show the answer and solution
Answer: D: \(\frac{4}{7}\)
\(P(\text{RB})+P(\text{BR})=\tfrac{4}{7}\cdot\tfrac{3}{6}+\tfrac{3}{7}\cdot\tfrac{4}{6}=\frac{4}{7}\).
Question 4
A bag contains 6 red and 4 blue counters. Two counters are taken at random, without replacement. What is the probability that they are different colours?
- \(\frac{6}{25}\)
- \(\frac{7}{15}\)
- \(\frac{4}{15}\)
- \(\frac{12}{25}\)
- \(\frac{8}{15}\)
Show the answer and solution
Answer: E: \(\frac{8}{15}\)
\(P(\text{RB})+P(\text{BR})=\tfrac{6}{10}\cdot\tfrac{4}{9}+\tfrac{4}{10}\cdot\tfrac{6}{9}=\frac{8}{15}\).
Question 5
A bag contains 5 red and 5 blue counters. Two counters are taken at random, without replacement. What is the probability that they are different colours?
- \(\frac{1}{4}\)
- \(\frac{4}{9}\)
- \(\frac{1}{2}\)
- \(\frac{5}{18}\)
- \(\frac{5}{9}\)
Show the answer and solution
Answer: E: \(\frac{5}{9}\)
\(P(\text{RB})+P(\text{BR})=\tfrac{5}{10}\cdot\tfrac{5}{9}+\tfrac{5}{10}\cdot\tfrac{5}{9}=\frac{5}{9}\).
Question 6
A bag contains 3 red and 5 blue counters. Two counters are taken at random, without replacement. What is the probability that they are different colours?
- \(\frac{15}{32}\)
- \(\frac{15}{28}\)
- \(\frac{15}{64}\)
- \(\frac{13}{28}\)
- \(\frac{15}{56}\)
Show the answer and solution
Answer: B: \(\frac{15}{28}\)
\(P(\text{RB})+P(\text{BR})=\tfrac{3}{8}\cdot\tfrac{5}{7}+\tfrac{5}{8}\cdot\tfrac{3}{7}=\frac{15}{28}\).
Question 7
A bag contains 7 red and 3 blue counters. Two counters are taken at random, without replacement. What is the probability that they are different colours?
- \(\frac{7}{30}\)
- \(\frac{21}{50}\)
- \(\frac{8}{15}\)
- \(\frac{21}{100}\)
- \(\frac{7}{15}\)
Show the answer and solution
Answer: E: \(\frac{7}{15}\)
\(P(\text{RB})+P(\text{BR})=\tfrac{7}{10}\cdot\tfrac{3}{9}+\tfrac{3}{10}\cdot\tfrac{7}{9}=\frac{7}{15}\).
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