O Level Additional Mathematics Statistics Probability Quiz

O-Level Additional Mathematics Quiz - Statistics Probability (Answer Key)

Total Marks: 60

Section A: Permutations and Combinations

1. Number of ways to arrange 5 distinct books:
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
Answer: 120
[1 mark for correct answer]

2. Choosing 3 from 8 (Order does not matter):
$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
Answer: 56
[2 marks: 1 for formula/setup, 1 for answer]

3. 4-digit numbers from {1,2,3,4,5} without repetition:
$P(5, 4) = 5 \times 4 \times 3 \times 2 = 120$
Answer: 120
[2 marks: 1 for method, 1 for answer]

4. Team of 4 with exactly 2 men (from 6) and 2 women (from 4):
$\binom{6}{2} \times \binom{4}{2} = 15 \times 6 = 90$
Answer: 90
[3 marks: 1 for men selection, 1 for women selection, 1 for product]

5. Arrangements of STATISTICS:
Total letters = 10.
Repeats: S (3), T (3), I (2), A (1), C (1).
Number of arrangements = $\frac{10!}{3! \, 3! \, 2! \, 1! \, 1!} = \frac{3,628,800}{6 \times 6 \times 2} = \frac{3,628,800}{72} = 50,400$
Answer: 50,400
[4 marks: 1 for identifying total, 1 for identifying repeats, 1 for formula, 1 for answer]

Section B: Probability Basics and Laws

6. Prime numbers on a die: {2, 3, 5}. Total outcomes: {1, 2, 3, 4, 5, 6}.
$P(\text{Prime}) = \frac{3}{6} = \frac{1}{2}$
Answer: 0.5 or 1/2
[2 marks: 1 for identifying primes, 1 for probability]

7.
(a) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$= 0.4 + 0.5 - 0.2 = 0.7$
Answer (a): 0.7
[2 marks]

(b) Check independence: Is $P(A \cap B) = P(A) \times P(B)$?
$P(A) \times P(B) = 0.4 \times 0.5 = 0.2$.
Since $0.2 = 0.2$, the events are independent.
Answer (b): Yes, they are independent because $P(A \cap B) = P(A)P(B)$.
[2 marks: 1 for calculation, 1 for conclusion]

8. P(Red then Red) without replacement:
$P(R_1) = \frac{5}{8}$.
$P(R_2 | R_1) = \frac{4}{7}$.
$P(RR) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$
Answer: 5/14 (or approx 0.357)
[3 marks: 1 for first prob, 1 for second prob, 1 for final answer]

9. $P(\text{Rain} \cap \text{Late}) = P(\text{Rain}) \times P(\text{Late} | \text{Rain})$
$= 0.3 \times 0.8 = 0.24$
Answer: 0.24
[2 marks]

10. Let $C$ = Chemistry, $P$ = Physics.
$P(C) = 0.6, P(P) = 0.5, P(C \cap P) = 0.3$.
$P(C \cup P) = 0.6 + 0.5 - 0.3 = 0.8$.
$P(\text{Neither}) = 1 - P(C \cup P) = 1 - 0.8 = 0.2$.
Answer: 0.2
[4 marks: 1 for union formula, 1 for union calc, 1 for complement logic, 1 for answer]

Section C: Discrete Random Variables

11. Sum of probabilities must be 1.
$0.1 + k + 0.3 + 0.4 = 1 \Rightarrow k + 0.8 = 1 \Rightarrow k = 0.2$.
Answer: 0.2
[1 mark]

12. $E(X) = \sum x P(X=x)$
$= 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4)$
$= 0.1 + 0.4 + 0.9 + 1.6 = 3.0$
Answer: 3
[3 marks: 1 for substitution, 1 for working, 1 for answer]

13. $Y \sim B(10, 0.4)$.
$P(Y=3) = \binom{10}{3} (0.4)^3 (0.6)^7$
$\binom{10}{3} = 120$.
$120 \times 0.064 \times 0.0279936 \approx 0.21499$
Answer: 0.215
[3 marks: 1 for formula, 1 for substitution, 1 for answer]

14. Variance of $B(n,p) = np(1-p)$.
$Var(Y) = 10 \times 0.4 \times 0.6 = 2.4$.
Answer: 2.4
[2 marks]

15. $X \sim Po(2.5)$.
$P(X=1) = \frac{e^{-2.5} (2.5)^1}{1!} = 2.5 e^{-2.5}$.
$2.5 \times 0.082085 \approx 0.2052$
Answer: 0.205
[3 marks: 1 for formula, 1 for substitution, 1 for answer]

Section D: Continuous Distributions and Normal Approximation

16. Uniform distribution on $[2, 8]$. Total length = $8 - 2 = 6$.
Range $X > 5$ is $(5, 8]$. Length = $8 - 5 = 3$.
$P(X > 5) = \frac{3}{6} = 0.5$.
Answer: 0.5
[2 marks]

17. $H \sim N(170, 10^2)$. Find $P(H > 185)$.
$Z = \frac{185 - 170}{10} = \frac{15}{10} = 1.5$.
$P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668$.
Answer: 0.0668
[3 marks: 1 for standardization, 1 for table lookup, 1 for final prob]

18. Find $h$ such that $P(H < h) = 0.90$.
From tables, $Z \approx 1.282$ for 0.90.
$\frac{h - 170}{10} = 1.282 \Rightarrow h - 170 = 12.82 \Rightarrow h = 182.82$.
Answer: 182.8 cm (or 183 cm)
[3 marks: 1 for Z value, 1 for equation, 1 for answer]

19. $X \sim B(100, 0.5)$. Approximate with $N(\mu, \sigma^2)$.
$\mu = np = 50$. $\sigma^2 = npq = 100(0.5)(0.5) = 25 \Rightarrow \sigma = 5$.
Find $P(X \ge 55)$. With continuity correction: $P(X_{normal} > 54.5)$.
$Z = \frac{54.5 - 50}{5} = \frac{4.5}{5} = 0.9$.
$P(Z > 0.9) = 1 - 0.8159 = 0.1841$.
Answer: 0.184
[4 marks: 1 for params, 1 for continuity correction, 1 for Z, 1 for final prob]

20. $T \sim N(45, \sigma^2)$. $P(T > 50) = 0.10$.
This implies $P(T < 50) = 0.90$.
$Z$ for 0.90 is approx $1.282$.
$\frac{50 - 45}{\sigma} = 1.282 \Rightarrow \frac{5}{\sigma} = 1.282 \Rightarrow \sigma = \frac{5}{1.282} \approx 3.90$.
Answer: 3.90
[3 marks: 1 for Z value, 1 for setup, 1 for answer]