Module Topic Test

Relative Frequency and Probability

Maths Standard Year 12 · All 4 lessons · MC checkpoint plus separate short-answer practice

L1, Probability Fundamentals L2, Multistage Events & Trees L3, Relative & Expected Frequency L4, Venn Diagrams & Two-Way Tables
25 MC 8 SA ~55 min
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Complete the 25 multiple choice questions to unlock a sharper next move. The short-answer section below is separate practice.

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Part A, Multiple Choice (1 mark each, 25 marks total)
1 Which of the following values could NOT be a probability? L1
A $0$
B $0.45$
C $\dfrac{3}{4}$
D $1.2$
D, $1.2$. Every probability must satisfy $0 \leq P \leq 1$. Since $1.2 > 1$, it cannot be a probability. The other values all lie between 0 and 1.
2 A fair six-sided die is rolled once. What is the probability of rolling an even number? L1
A $\dfrac{1}{2}$
B $\dfrac{1}{3}$
C $\dfrac{1}{6}$
D $\dfrac{2}{3}$
A, $\dfrac{1}{2}$. $S = \{1,2,3,4,5,6\}$, so $n(S) = 6$. Even numbers are $\{2,4,6\}$, so $n(E) = 3$. $P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$.
3 The probability that it rains tomorrow is $0.35$. What is the probability that it does NOT rain tomorrow? L1
A $0.35$
B $0.65$
C $0.55$
D $1.35$
B, $0.65$. The complement rule gives $P(E') = 1 - P(E) = 1 - 0.35 = 0.65$.
4 A letter is chosen at random from the word STATISTICS. What is the probability that it is the letter T? L1
A $\dfrac{1}{10}$
B $\dfrac{1}{3}$
C $\dfrac{3}{10}$
D $\dfrac{2}{10}$
C, $\dfrac{3}{10}$. STATISTICS has 10 letters (S-T-A-T-I-S-T-I-C-S). The letter T appears 3 times, so $P(T) = \dfrac{3}{10}$.
5 A card is drawn at random from a standard 52-card deck. What is the probability that it is NOT a king? L1
A $\dfrac{1}{13}$
B $\dfrac{4}{52}$
C $\dfrac{12}{13}$
D $\dfrac{1}{4}$
C, $\dfrac{12}{13}$. There are 4 kings, so $P(\text{king}) = \dfrac{4}{52} = \dfrac{1}{13}$. By the complement rule, $P(\text{not king}) = 1 - \dfrac{1}{13} = \dfrac{12}{13}$.
6 A spinner has 8 equal sections numbered 1 to 8. What is the probability of landing on a prime number? L1
A $\dfrac{3}{8}$
B $\dfrac{1}{2}$
C $\dfrac{5}{8}$
D $\dfrac{1}{8}$
B, $\dfrac{1}{2}$. The primes from 1 to 8 are 2, 3, 5 and 7, so $n(E) = 4$. $P(\text{prime}) = \dfrac{4}{8} = \dfrac{1}{2}$. (Note: 1 is not prime.)
7 A probability is given as the fraction $\dfrac{3}{8}$. What is this value expressed as a percentage? L1
A $38\%$
B $30\%$
C $3.75\%$
D $37.5\%$
D, $37.5\%$. Convert the fraction to a decimal, $3 \div 8 = 0.375$, then multiply by 100: $0.375 \times 100 = 37.5\%$.
8 A fair die is rolled and a fair coin is flipped. How many outcomes are in the sample space? L2
A $8$
B $12$
C $6$
D $36$
B, $12$. By the multiplication principle, $n(S) = 6 \times 2 = 12$. There are 6 die outcomes for each of the 2 coin outcomes.
9 A fair coin is flipped twice. What is the probability of getting two heads? L2
A $\dfrac{1}{4}$
B $\dfrac{1}{2}$
C $\dfrac{3}{4}$
D $1$
A, $\dfrac{1}{4}$. The two flips are independent, so $P(\text{HH}) = P(H) \times P(H) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$.
10 A bag contains 3 blue and 2 red marbles. A marble is drawn, its colour noted, then replaced, and a second marble is drawn. What is $P(\text{blue then red})$? L2
A $\dfrac{6}{25}$
B $\dfrac{5}{25}$
C $\dfrac{1}{5}$
D $\dfrac{6}{20}$
A, $\dfrac{6}{25}$. With replacement the draws are independent. $P(\text{blue}) = \dfrac{3}{5}$ and $P(\text{red}) = \dfrac{2}{5}$, so $P(\text{blue then red}) = \dfrac{3}{5} \times \dfrac{2}{5} = \dfrac{6}{25}$.
11 A bag contains 3 blue and 2 red marbles. Two marbles are drawn WITHOUT replacement. What is the probability that both are blue? L2
A $\dfrac{9}{25}$
B $\dfrac{3}{10}$
C $\dfrac{6}{25}$
D $\dfrac{3}{5}$
B, $\dfrac{3}{10}$. First draw $P(\text{blue}) = \dfrac{3}{5}$. After removing one blue, 4 marbles remain (2 blue), so the second draw $P(\text{blue}) = \dfrac{2}{4}$. Multiply: $\dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}$.
12 A fair coin is flipped 3 times. Using the complement, what is the probability of getting at least one head? L2
A $\dfrac{1}{8}$
B $\dfrac{3}{8}$
C $\dfrac{1}{2}$
D $\dfrac{7}{8}$
D, $\dfrac{7}{8}$. The complement of "at least one head" is "no heads" (all tails). $P(\text{TTT}) = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$, so $P(\text{at least one head}) = 1 - \dfrac{1}{8} = \dfrac{7}{8}$.
13 A die is rolled 60 times and the number 3 appears 14 times. What is the relative frequency of rolling a 3? L3
A $\dfrac{7}{30}$
B $\dfrac{1}{6}$
C $\dfrac{30}{7}$
D $\dfrac{14}{6}$
A, $\dfrac{7}{30}$. Relative frequency $= \dfrac{f}{n} = \dfrac{14}{60} = \dfrac{7}{30}$. (The theoretical probability $\dfrac{1}{6}$ is not what the experiment measured.)
14 A spinner has $P(\text{blue}) = 0.25$. In 200 spins, what is the expected frequency of landing on blue? L3
A $25$
B $100$
C $50$
D $0.25$
C, $50$. Expected frequency $E = n \times p = 200 \times 0.25 = 50$.
15 On a fair die, $P(6) = \dfrac{1}{6}$. If the die is rolled 120 times, what is the expected number of sixes? L3
A $6$
B $12$
C $30$
D $20$
D, $20$. $E = n \times p = 120 \times \dfrac{1}{6} = 20$.
16 A quality-control test of 500 items found 15 defective. Based on this relative frequency, how many defective items are expected in the next run of 2000 items? L3
A $60$
B $30$
C $15$
D $150$
A, $60$. Estimate the probability from relative frequency: $p = \dfrac{15}{500} = 0.03$. Then $E = n \times p = 2000 \times 0.03 = 60$.
17 Which statement best describes the law of large numbers? L3
A After a long run of tails, heads becomes more likely to "balance out".
B As the number of trials increases, relative frequency approaches the theoretical probability.
C Relative frequency always equals theoretical probability after 100 trials.
D Large samples guarantee the exact theoretical result every time.
B. The law of large numbers states that as the number of trials increases, the relative (experimental) frequency gets closer to the theoretical probability. It does not force short-run "balancing" (A), nor guarantee an exact match at any finite $n$ (C, D).
18 For two events, $P(A) = 0.5$, $P(B) = 0.4$ and $P(A \cap B) = 0.2$. Using the addition rule, what is $P(A \cup B)$? L4
A $0.9$
B $1.1$
C $0.7$
D $0.2$
C, $0.7$. $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.4 - 0.2 = 0.7$. Subtracting the intersection avoids double-counting the overlap.
19 Two events A and B are mutually exclusive. Which statement is always true? L4
A $P(A \cap B) = 0$, so $P(A \cup B) = P(A) + P(B)$.
B $P(A \cap B) = P(A) \times P(B)$.
C $P(A \cup B) = 1$ always.
D $P(A) = P(B)$.
A. Mutually exclusive events cannot both occur, so their circles do not overlap and $P(A \cap B) = 0$. The addition rule then simplifies to $P(A \cup B) = P(A) + P(B)$.
20 In a survey of 200 people, 90 like coffee, 80 like tea, and 30 like both. How many like NEITHER coffee nor tea? L4
A $30$
B $60$
C $140$
D $50$
B, $60$. Union $= 90 + 80 - 30 = 140$ like at least one. Neither $= 200 - 140 = 60$.
21 In a class of 30 students, 12 play sport, 10 play a musical instrument, and 5 do both. How many students play ONLY sport (and not music)? L4
A $12$
B $5$
C $7$
D $17$
C, $7$. Sport only $= n(\text{sport}) - n(\text{both}) = 12 - 5 = 7$. The 5 who do both are already inside the sport total.
22 A two-way table records 100 people surveyed. 60 are female and 40 are male; 45 females and 30 males own a smartphone. What is $P(\text{owns a smartphone})$? L4
A $\dfrac{45}{100}$
B $\dfrac{1}{2}$
C $\dfrac{3}{4}$
D $\dfrac{30}{100}$
C, $\dfrac{3}{4}$. Total smartphone owners $= 45 + 30 = 75$. $P(\text{smartphone}) = \dfrac{75}{100} = \dfrac{3}{4}$.
23 In a group of 40 people, 22 like action movies (A), 18 like comedies (C), and 8 like both. What is the probability that a randomly chosen person likes at least one of these genres? L4
A $\dfrac{32}{40}$
B $\dfrac{40}{40}$
C $\dfrac{8}{40}$
D $\dfrac{14}{40}$
A, $\dfrac{32}{40}$. Union $= 22 + 18 - 8 = 32$ like at least one. $P(\text{at least one}) = \dfrac{32}{40} = \dfrac{4}{5}$.
24 A bag contains 4 green and 1 yellow marble. A marble is drawn, replaced, then a second is drawn. Using the complement, what is $P(\text{at least one yellow})$? L2
A $\dfrac{16}{25}$
B $\dfrac{1}{25}$
C $\dfrac{2}{5}$
D $\dfrac{9}{25}$
D, $\dfrac{9}{25}$. $P(\text{green}) = \dfrac{4}{5}$, so with replacement $P(\text{no yellow}) = P(\text{GG}) = \dfrac{4}{5} \times \dfrac{4}{5} = \dfrac{16}{25}$. Then $P(\text{at least one yellow}) = 1 - \dfrac{16}{25} = \dfrac{9}{25}$.
25 A drug trial on 800 patients found it effective for 680. Using this relative frequency as the probability, how many of the next 250 patients are expected to benefit? L3
A $170$
B $200$
C $250$
D $213$
D, $213$. Relative frequency $= \dfrac{680}{800} = 0.85$. Then $E = n \times p = 250 \times 0.85 = 212.5 \approx 213$ patients.
Part B, Short Answer (show all working)
1 L1
A box contains 15 balls: 7 red, 4 white and 4 black. A ball is selected at random.
(a) State $n(S)$ and find $P(\text{red})$ as a fraction.
(b) Find $P(\text{not white})$ using the complement.
(c) Express $P(\text{red})$ as a decimal and as a percentage, to 2 decimal places where needed.
(a) $n(S) = 15$. $P(\text{red}) = \dfrac{7}{15}$
(b) $P(\text{white}) = \dfrac{4}{15}$, so $P(\text{not white}) = 1 - \dfrac{4}{15} = \dfrac{11}{15}$
(c) $\dfrac{7}{15} = 0.4\overline{6} \approx 0.47 = 46.67\%$
2 L1
A letter is chosen at random from the word PROBABILITY.
(a) How many letters are there in total?
(b) Find $P(\text{choosing the letter B})$.
(c) Find $P(\text{choosing a vowel})$. (The vowels are A, E, I, O, U.)
(a) PROBABILITY has 11 letters (P-R-O-B-A-B-I-L-I-T-Y).
(b) B appears 2 times: $P(B) = \dfrac{2}{11}$
(c) Vowels present: O, A, I, I, that is 4 vowels. $P(\text{vowel}) = \dfrac{4}{11}$
3 L2
A fair die is rolled and a fair coin is flipped.
(a) State the total number of outcomes $n(S)$.
(b) Find $P(\text{a number greater than 4 and heads})$.
(c) Find $P(\text{tails})$ and explain whether it matches the expected $\tfrac{1}{2}$.
(a) $n(S) = 6 \times 2 = 12$
(b) Numbers greater than 4 are 5 and 6. Favourable cells: (H,5), (H,6). $P = \dfrac{2}{12} = \dfrac{1}{6}$
(c) Tails appears in 6 of the 12 cells: $P(\text{tails}) = \dfrac{6}{12} = \dfrac{1}{2}$, which matches the expected value, confirming the symmetry of the sample space.
4 L2
A bag contains 3 red and 2 white balls. Two balls are drawn WITHOUT replacement.
(a) Write the stage-1 branch probabilities, and the stage-2 probabilities given a red was drawn first.
(b) Find $P(\text{one red and one white, in any order})$.
(c) Is it more likely to draw two reds or one of each colour? Justify.
(a) Stage 1: $P(R) = \dfrac{3}{5}$, $P(W) = \dfrac{2}{5}$. Stage 2 after red: 4 balls left (2R, 2W) so $P(R) = \dfrac{2}{4}$, $P(W) = \dfrac{2}{4}$.
(b) $P(RW) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}$; $P(WR) = \dfrac{2}{5} \times \dfrac{3}{4} = \dfrac{6}{20} = \dfrac{3}{10}$. $P(\text{one of each}) = \dfrac{3}{10} + \dfrac{3}{10} = \dfrac{3}{5}$
(c) $P(RR) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}$. Since $\dfrac{3}{5} > \dfrac{3}{10}$, one of each colour is more likely.
5 L3
A spinner is spun 80 times and lands on green 28 times. The theoretical probability of green is 0.30.
(a) Calculate the relative frequency of green.
(b) How does the relative frequency compare with the theoretical probability?
(c) How many times would you expect green in 500 spins (use the theoretical probability)?
(a) Relative frequency $= \dfrac{28}{80} = \dfrac{7}{20} = 0.35 = 35\%$
(b) $0.35$ is slightly above the theoretical $0.30$. With only 80 spins this variation is plausible; more spins would give a more reliable estimate (law of large numbers).
(c) $E = n \times p = 500 \times 0.30 = 150$ spins expected on green.
6 L3
A factory tests 400 components and finds 12 are faulty.
(a) Estimate the probability that a component is faulty, using relative frequency.
(b) How many faulty components are expected in a run of 5000?
(c) The manager wants fewer than 100 faulty components in the run of 5000. Is this target met at the current rate? Justify.
(a) Relative frequency $= \dfrac{12}{400} = 0.03 = 3\%$, so estimated $P(\text{faulty}) \approx 0.03$
(b) $E = n \times p = 5000 \times 0.03 = 150$ faulty components expected.
(c) The expected 150 is greater than the target of 100, so at the current 3% rate the target is NOT met. The faulty rate would need to fall below $\dfrac{100}{5000} = 0.02$ (2%).
7 L4
In a class of 28 students, 16 own a laptop, 12 own a tablet, and 6 own both.
(a) Find the four Venn diagram regions: laptop only, tablet only, both, and neither.
(b) Find $P(\text{owns a tablet only})$.
(c) Find $P(\text{owns neither})$.
(a) Both $= 6$; laptop only $= 16 - 6 = 10$; tablet only $= 12 - 6 = 6$; neither $= 28 - 10 - 6 - 6 = 6$. Check: $10 + 6 + 6 + 6 = 28$ ✓
(b) $P(\text{tablet only}) = \dfrac{6}{28} = \dfrac{3}{14}$
(c) $P(\text{neither}) = \dfrac{6}{28} = \dfrac{3}{14}$
8 L4
A survey of 200 students asked about after-school activities. 90 play sport, 70 do music, and 30 do both.
(a) Complete a two-way table (rows sport / no sport, columns music / no music) with all totals.
(b) Find $P(\text{sport only})$.
(c) Find $P(\text{participates in at least one activity})$.
(a) Sport and music $= 30$; sport only $= 60$; music only $= 40$; neither $= 70$. Row totals: sport 90, no sport 110. Column totals: music 70, no music 130. Grand total $= 200$.
(b) $P(\text{sport only}) = \dfrac{60}{200} = \dfrac{3}{10}$
(c) At least one $= 60 + 40 + 30 = 130$. $P = \dfrac{130}{200} = \dfrac{13}{20}$. (Or $1 - \dfrac{70}{200} = \dfrac{13}{20}$.)
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