1. [Maximum mark: 5]
On any day, Kino travels to school by bus, by car or on foot with probabilities 0.2, 0.1 and 0.7 respectively. The probability that he is late when he travels by bus is \( x \). The probability that he is late when he travels by car is \( 2x \) and the probability that he is late when he travels on foot is 0.25. The probability that, on a randomly chosen day, Kino is late is 0.235.
(a) Find the value of \( x \).
(b) Find the probability that Kino travelled by car given that he was not late.
▶️Answer/Explanation
(a)
Using the law of total probability:
\( 0.2x + 0.1(2x) + 0.7(0.25) = 0.235 \)
Simplify: \( 0.2x + 0.2x + 0.175 = 0.235 \)
Combine terms: \( 0.4x = 0.06 \)
Solve: \( x = 0.15 \)
(b)
First, find \( P(\text{not late}) = 1 – 0.235 = 0.765 \)
Then, \( P(\text{car and not late}) = 0.1 \times (1 – 2 \times 0.15) = 0.07 \)
Using conditional probability:
\( P(\text{car} \mid \text{not late}) = \frac{0.07}{0.765} \approx 0.0915 \)
2. [Maximum mark: 7]
The lengths of the rods produced by a company are normally distributed with mean 55.6 mm and standard deviation 1.2 mm.
(a) In a random sample of 400 of these rods, how many would you expect to have length less than 54.8 mm?
(b) Find the probability that a randomly chosen rod produced by this company has a length that is within half a standard deviation of the mean.
▶️Answer/Explanation
(a)
Standardize: \( z = \frac{54.8 – 55.6}{1.2} = -0.6667 \)
Find probability: \( P(Z < -0.6667) = 1 – 0.7477 = 0.2523 \)
Expected number: \( 400 \times 0.2523 = 100.92 \)
Round to nearest integer: 101
(b)
Range: \( 55.6 \pm 0.5 \times 1.2 = [55.0, 56.2] \)
Standardize bounds:
\( z_1 = \frac{55.0 – 55.6}{1.2} = -0.5 \)
\( z_2 = \frac{56.2 – 55.6}{1.2} = 0.5 \)
Find probability: \( P(-0.5 < Z < 0.5) = 2 \times 0.6915 – 1 = 0.383 \)
3. [Maximum mark: 5]
Three fair 6-sided dice are thrown at the same time repeatedly. The score on each throw is the sum of the numbers on the uppermost faces.
(a) Find the probability that a score of 17 or more is first obtained on the 6th throw.
(b) Find the probability that a score of 17 or more is obtained in fewer than 8 throws.
▶️Answer/Explanation
(a)
Probability of scoring 17+ in one throw: \( \frac{4}{216} = \frac{1}{54} \)
Probability of not scoring 17+ in first 5 throws: \( \left(\frac{53}{54}\right)^5 \)
Probability for 6th throw: \( \left(\frac{53}{54}\right)^5 \times \frac{1}{54} \approx 0.0169 \)
(b)
Probability of not scoring 17+ in first 7 throws: \( \left(\frac{53}{54}\right)^7 \)
Probability in fewer than 8 throws: \( 1 – \left(\frac{53}{54}\right)^7 \approx 0.123 \)
4. [Maximum mark: 7]
The times taken, in minutes, to complete a word processing task by 250 employees are summarised in the table:
| Time (t mins) | Frequency |
|---|---|
| \(0 \leq t < 20\) | 32 |
| \(20 \leq t < 40\) | 46 |
| \(40 \leq t < 50\) | 96 |
| \(50 \leq t < 60\) | 52 |
| \(60 \leq t < 100\) | 24 |
(a) Draw a histogram to represent this information.
(b) Given the estimate of the mean time is 43.2 minutes, calculate an estimate for the standard deviation.
▶️Answer/Explanation
(a)
Calculate frequency densities:
\( \frac{32}{20} = 1.6 \), \( \frac{46}{20} = 2.3 \), \( \frac{96}{10} = 9.6 \), \( \frac{52}{10} = 5.2 \), \( \frac{24}{40} = 0.6 \)
Plot histogram with time on x-axis and frequency density on y-axis
(b)
Midpoints: 10, 30, 45, 55, 80
Calculate \( \sum f x^2 = 32 \times 10^2 + 46 \times 30^2 + \cdots + 24 \times 80^2 = 549900 \)
Variance: \( \frac{549900}{250} – 43.2^2 = 333.36 \)
Standard deviation: \( \sqrt{333.36} \approx 18.3 \) minutes
6. [Maximum mark: 9]
At a call center, 90% of callers are connected immediately. A random sample of 12 callers is chosen.
(a) Find the probability that fewer than 10 of these callers are connected immediately.
A random sample of 80 callers is chosen.
(b) Use an approximation to find the probability that more than 69 are connected immediately.
(c) Justify the use of your approximation.
▶️Answer/Explanation
(a)
Binomial distribution: X ~ B(12, 0.9)
P(X < 10) = 1 – P(X ≥ 10) = 1 – [P(10)+P(11)+P(12)]
= 1 – [¹²C₁₀(0.9)¹⁰(0.1)² + ¹²C₁₁(0.9)¹¹(0.1) + (0.9)¹²]
≈ 1 – [0.2301 + 0.3766 + 0.2824] ≈ 0.111
(b)
Normal approximation: X ~ N(72, 7.2)
With continuity correction: P(X > 69.5)
z = (69.5-72)/√7.2 ≈ -0.9317
P(Z > -0.9317) = Φ(0.9317) ≈ 0.824
(c)
Both np = 72 and nq = 8 are greater than 5, so normal approximation is valid.
7. [Maximum mark: 10]
Consider the word ALLIGATOR.
(a) Find the number of different arrangements where the two As are together and the two Ls are together.
(b) Find the probability that in a random arrangement, the two Ls are together and there are exactly 6 letters between the two As.
(c) Find the number of different 5-letter selections containing at least one A and at most one L.
▶️Answer/Explanation
(a)
Treat AA as one unit and LL as one unit: 7 units total
Number of arrangements: 7! = 5040
(b)
Total arrangements: 9!/(2!2!) = 90720
For condition: A _ _ _ _ _ _ A with LL together
LL can be in 5 positions between/beyond As
Remaining 5 letters can be arranged in 5! ways
Total favorable: 5 × 5! × 2 = 1200
Probability: 1200/90720 = 5/378 ≈ 0.0132
(c)
Cases:
1. 1A and 0L: Choose 4 from {G,I,T,O,R} → 5C4 = 5
2. 1A and 1L: Choose 3 from {G,I,T,O,R} → 5C3 = 10
3. 2A and 0L: Choose 3 from {G,I,T,O,R} → 5C3 = 10
4. 2A and 1L: Choose 2 from {G,I,T,O,R} → 5C2 = 10
Total: 5 + 10 + 10 + 10 = 35