O Level Additional Mathematics Statistics Probability Quiz

O-Level Additional Mathematics Quiz - Statistics Probability

ANSWER KEY AND MARKING SCHEME

Total Marks: 40

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Section A: Short Answer (10 marks)

1. A fair six-sided die is rolled once. Find the probability of obtaining a number greater than 4. [2 marks]

Answer: $\frac{2}{6} = \frac{1}{3}$

Marking:

• M1: Identifying favourable outcomes {5, 6} = 2 outcomes, or equivalent reasoning
• A1: Correct simplified probability $\frac{1}{3}$

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2. A bag contains 5 red balls, 3 blue balls, and 2 green balls. One ball is drawn at random. Find the probability that the ball drawn is NOT blue. [2 marks]

Answer: $\frac{5+2}{10} = \frac{7}{10}$ or $1 - \frac{3}{10} = \frac{7}{10}$

Marking:

• M1: Identifying total balls = 10, non-blue balls = 7, or using complement $1 - P(\text{blue})$
• A1: Correct probability $\frac{7}{10}$

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3. The probability that it rains on any given day in a certain month is 0.3. Find the probability that it does NOT rain on a particular day. [1 mark]

Answer: $1 - 0.3 = 0.7$

Marking:

• A1: Correct answer 0.7

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4. A card is drawn at random from a standard pack of 52 playing cards. Find the probability that the card is either a King or a Heart. [3 marks]

Answer: $P(\text{King or Heart}) = P(\text{King}) + P(\text{Heart}) - P(\text{King of Hearts})$
$= \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$

Marking:

• M1: Identifying $P(\text{King}) = \frac{4}{52}$ and $P(\text{Heart}) = \frac{13}{52}$
• M1: Recognising overlap (King of Hearts) and applying addition rule correctly
• A1: Correct simplified probability $\frac{4}{13}$

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5. Two fair coins are tossed. List the sample space and find the probability of obtaining exactly one head. [2 marks]

Answer: Sample space = {HH, HT, TH, TT}
$P(\text{exactly one head}) = \frac{2}{4} = \frac{1}{2}$

Marking:

• M1: Correct sample space listed (all 4 outcomes)
• A1: Correct probability $\frac{1}{2}$

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Section B: Calculation and Problem Solving (20 marks)

6. A box contains 4 white chocolates and 6 dark chocolates. Two chocolates are drawn at random, one after the other, without replacement.

(a) Draw a tree diagram to represent this situation, showing all probabilities on the branches. [3 marks]

Marking:

• B1: First branch correctly labelled with probabilities: $P(W) = \frac{4}{10} = \frac{2}{5}$, $P(D) = \frac{6}{10} = \frac{3}{5}$
• B1: Second branches correctly labelled with conditional probabilities:
  • After W: $P(W|W) = \frac{3}{9} = \frac{1}{3}$, $P(D|W) = \frac{6}{9} = \frac{2}{3}$
  • After D: $P(W|D) = \frac{4}{9}$, $P(D|D) = \frac{5}{9}$
• B1: All four outcomes shown with final probabilities (optional but good practice)

(b) Find the probability that both chocolates drawn are dark. [2 marks]

Answer: $P(DD) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$

Marking:

• M1: Multiplying correct probabilities $\frac{6}{10} \times \frac{5}{9}$
• A1: Correct answer $\frac{1}{3}$

(c) Find the probability that exactly one of the chocolates drawn is white. [3 marks]

Answer: $P(\text{exactly one white}) = P(WD) + P(DW)$
$= \left(\frac{4}{10} \times \frac{6}{9}\right) + \left(\frac{6}{10} \times \frac{4}{9}\right)$
$= \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}$

Marking:

• M1: Identifying the two paths: WD and DW
• M1: Correctly calculating both path probabilities
• A1: Correct answer $\frac{8}{15}$

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7. The probability that a student passes Additional Mathematics is 0.75. The probability that the same student passes Physics is 0.8. The probability that the student passes both subjects is 0.65.

(a) Find the probability that the student passes at least one of the two subjects. [2 marks]

Answer: $P(A \cup P) = P(A) + P(P) - P(A \cap P)$
$= 0.75 + 0.8 - 0.65 = 0.9$

Marking:

• M1: Applying addition rule $P(A \cup P) = P(A) + P(P) - P(A \cap P)$
• A1: Correct answer 0.9

(b) Find the probability that the student passes Physics but not Additional Mathematics. [2 marks]

Answer: $P(P \cap A') = P(P) - P(A \cap P)$
$= 0.8 - 0.65 = 0.15$

Marking:

• M1: Correct method: $P(P) - P(A \cap P)$
• A1: Correct answer 0.15

(c) Determine whether the events "passes Additional Mathematics" and "passes Physics" are independent. Justify your answer. [2 marks]

Answer: For independence, $P(A \cap P) = P(A) \times P(P)$
$P(A) \times P(P) = 0.75 \times 0.8 = 0.6$
But $P(A \cap P) = 0.65 \neq 0.6$
Therefore, the events are NOT independent.

Marking:

• M1: Calculating $P(A) \times P(P) = 0.6$ and comparing with $P(A \cap P) = 0.65$
• A1: Correct conclusion "not independent" with valid justification

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8. In a game, a player spins a fair spinner with sectors labelled 1, 2, 3, 4, and 5. The player wins if the spinner lands on an even number.

(a) Find the probability that the player wins on a single spin. [1 mark]

Answer: $P(\text{win}) = \frac{2}{5}$ (even numbers: 2 and 4)

Marking:

• A1: Correct probability $\frac{2}{5}$ or 0.4

(b) The player spins the spinner three times. Find the probability that the player wins exactly twice. [3 marks]

Answer: Let $p = \frac{2}{5}$ (probability of win), $q = \frac{3}{5}$ (probability of loss)
$P(\text{exactly 2 wins in 3 spins}) = \binom{3}{2} p^2 q^1$
$= 3 \times \left(\frac{2}{5}\right)^2 \times \left(\frac{3}{5}\right)$
$= 3 \times \frac{4}{25} \times \frac{3}{5} = 3 \times \frac{12}{125} = \frac{36}{125}$

Marking:

• M1: Identifying binomial probability with $n=3$, $p=\frac{2}{5}$, $r=2$
• M1: Correct binomial expression $\binom{3}{2} p^2 q^1$
• A1: Correct answer $\frac{36}{125}$ (accept 0.288)

(c) Find the probability that the player wins at least once in three spins. [2 marks]

Answer: $P(\text{at least one win}) = 1 - P(\text{no wins})$
$= 1 - \left(\frac{3}{5}\right)^3$
$= 1 - \frac{27}{125} = \frac{98}{125}$

Marking:

• M1: Using complement: $1 - P(\text{all losses})$
• A1: Correct answer $\frac{98}{125}$ (accept 0.784)

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9. A fair coin is tossed four times. Find the probability of obtaining exactly three heads. [2 marks]

Answer: $P(\text{exactly 3 heads}) = \binom{4}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^1 = 4 \times \frac{1}{8} \times \frac{1}{2} = \frac{4}{16} = \frac{1}{4}$

Marking:

• M1: Using binomial coefficient $\binom{4}{3}$ and correct probability expression
• A1: Correct answer $\frac{1}{4}$ or 0.25

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10. A number is chosen at random from the integers 1 to 20 inclusive. Find the probability that the number is a multiple of 3 or a multiple of 5. [3 marks]

Answer: Multiples of 3: {3, 6, 9, 12, 15, 18} → 6 numbers
Multiples of 5: {5, 10, 15, 20} → 4 numbers
Multiples of both (15): 1 number
$P(\text{multiple of 3 or 5}) = \frac{6 + 4 - 1}{20} = \frac{9}{20}$

Marking:

• M1: Listing or counting multiples of 3 and multiples of 5
• M1: Applying addition rule and subtracting overlap
• A1: Correct answer $\frac{9}{20}$ or 0.45

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Section C: Data Interpretation and Application (10 marks)

11. A survey was conducted on 200 students about their participation in two co-curricular activities: Choir and Robotics.

(a) Find the probability that the student participates in Robotics. [1 mark]

Answer: $P(\text{Robotics}) = \frac{80}{200} = \frac{2}{5} = 0.4$

Marking:

• A1: Correct probability $\frac{2}{5}$ or 0.4

(b) Find the probability that the student participates in Choir, given that the student participates in Robotics. [2 marks]

Answer: $P(\text{Choir} | \text{Robotics}) = \frac{45}{80} = \frac{9}{16}$

Marking:

• M1: Correct conditional probability setup: $\frac{\text{Choir AND Robotics}}{\text{Robotics}} = \frac{45}{80}$
• A1: Correct simplified answer $\frac{9}{16}$ (accept 0.5625)

(c) Find the probability that the student participates in exactly one of the two activities. [2 marks]

Answer: $P(\text{exactly one}) = P(\text{Robotics only}) + P(\text{Choir only})$
$= \frac{35}{200} + \frac{55}{200} = \frac{90}{200} = \frac{9}{20}$

Marking:

• M1: Identifying the two relevant cells: Robotics only (35) and Choir only (55)
• A1: Correct answer $\frac{9}{20}$ or 0.45

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12. The masses of apples harvested from an orchard are normally distributed with mean 150 g and standard deviation 20 g.

(a) Find the probability that a randomly selected apple has a mass between 140 g and 170 g. [2 marks]

Answer: Let $X \sim N(150, 20^2)$
$P(140 < X < 170) = P\left(\frac{140-150}{20} < Z < \frac{170-150}{20}\right)$
$= P(-0.5 < Z < 1.0)$
$= P(Z < 1.0) - P(Z < -0.5)$
$= 0.8413 - 0.3085 = 0.5328 \approx 0.533$

Marking:

• M1: Correct standardisation to $Z$-scores: $z_1 = -0.5$, $z_2 = 1.0$
• A1: Correct probability 0.533 (3 s.f.)

(b) An apple is classified as "premium" if its mass exceeds 180 g. In a batch of 500 apples, estimate the number of premium apples. [3 marks]

Answer: $P(X > 180) = P\left(Z > \frac{180-150}{20}\right) = P(Z > 1.5)$
$= 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668$
Expected number = $500 \times 0.0668 = 33.4 \approx 33$ apples

Marking:

• M1: Correct standardisation to $z = 1.5$
• M1: Correct probability $P(Z > 1.5) = 0.0668$
• A1: Correct estimate 33 (or 33.4)

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13. A bag contains 3 black pens and 2 red pens. Two pens are drawn at random without replacement. Find the probability that both pens are the same colour. [2 marks]

Answer: $P(\text{same colour}) = P(BB) + P(RR)$
$= \left(\frac{3}{5} \times \frac{2}{4}\right) + \left(\frac{2}{5} \times \frac{1}{4}\right) = \frac{6}{20} + \frac{2}{20} = \frac{8}{20} = \frac{2}{5}$

Marking:

• M1: Identifying both paths (BB and RR) and multiplying probabilities correctly
• A1: Correct answer $\frac{2}{5}$ or 0.4

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14. The probability that a train is late on a given day is 0.15. Find the probability that the train is late on exactly two out of five randomly selected days. [3 marks]

Answer: Let $p = 0.15$, $n = 5$, $r = 2$
$P(X = 2) = \binom{5}{2} (0.15)^2 (0.85)^3$
$= 10 \times 0.0225 \times 0.614125 = 10 \times 0.0138178... = 0.138$ (3 s.f.)

Marking:

• M1: Identifying binomial distribution with correct parameters
• M1: Correct binomial expression $\binom{5}{2} p^2 q^3$
• A1: Correct answer 0.138 (3 s.f.)

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15. A class consists of 12 boys and 18 girls. Two students are selected at random. Find the probability that one is a boy and one is a girl. [2 marks]

Answer: Total students = 30
$P(\text{one boy, one girl}) = \frac{\binom{12}{1} \times \binom{18}{1}}{\binom{30}{2}} = \frac{12 \times 18}{435} = \frac{216}{435} = \frac{72}{145}$

Marking:

• M1: Correct method using combinations or probability tree
• A1: Correct answer $\frac{72}{145}$ (accept 0.497)

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Section D: Advanced Problem Solving (10 marks)

16. Events A and B are such that P(A) = 0.6, P(B) = 0.5, and P(A ∪ B) = 0.8. Find P(A ∩ B). [2 marks]

Answer: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$0.8 = 0.6 + 0.5 - P(A \cap B)$
$P(A \cap B) = 1.1 - 0.8 = 0.3$

Marking:

• M1: Applying addition rule correctly
• A1: Correct answer 0.3

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17. A fair die is rolled twice. Find the probability that the sum of the two numbers obtained is at least 10. [3 marks]

Answer: Total outcomes = $6 \times 6 = 36$
Favourable outcomes: (4,6), (5,5), (5,6), (6,4), (6,5), (6,6) → 6 outcomes
$P(\text{sum} \geq 10) = \frac{6}{36} = \frac{1}{6}$

Marking:

• M1: Identifying total outcomes = 36
• M1: Listing or counting favourable outcomes correctly
• A1: Correct answer $\frac{1}{6}$

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18. A committee of 3 people is to be selected from 5 men and 4 women. Find the probability that the committee contains at least 2 women. [3 marks]

Answer: Total ways = $\binom{9}{3} = 84$
Ways with exactly 2 women: $\binom{4}{2} \times \binom{5}{1} = 6 \times 5 = 30$
Ways with exactly 3 women: $\binom{4}{3} \times \binom{5}{0} = 4 \times 1 = 4$
$P(\text{at least 2 women}) = \frac{30 + 4}{84} = \frac{34}{84} = \frac{17}{42}$

Marking:

• M1: Correct total number of ways $\binom{9}{3}$
• M1: Correctly calculating favourable ways (2 women or 3 women)
• A1: Correct answer $\frac{17}{42}$ (accept 0.405)

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19. The probability that a student solves a particular problem is 0.7. The student attempts the problem three times independently. Find the probability that the student solves the problem at least twice. [2 marks]

Answer: $P(\text{at least twice}) = P(X = 2) + P(X = 3)$
$= \binom{3}{2} (0.7)^2 (0.3)^1 + \binom{3}{3} (0.7)^3 (0.3)^0$
$= 3 \times 0.49 \times 0.3 + 1 \times 0.343 \times 1$
$= 0.441 + 0.343 = 0.784$

Marking:

• M1: Identifying binomial probabilities for $X = 2$ and $X = 3$
• A1: Correct answer 0.784

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20. A box contains 8 light bulbs, of which 3 are defective. Two bulbs are selected at random without replacement. Find the probability that exactly one of the selected bulbs is defective. [2 marks]

Answer: $P(\text{exactly one defective}) = P(DG) + P(GD)$
$= \left(\frac{3}{8} \times \frac{5}{7}\right) + \left(\frac{5}{8} \times \frac{3}{7}\right) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}$

Marking:

• M1: Identifying the two paths (defective then good, good then defective) and multiplying correctly
• A1: Correct answer $\frac{15}{28}$ (accept 0.536)

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END OF ANSWER KEY