1. [Maximum mark: 6]
(a) Find the number of different arrangements of the 8 letters in the word DECEIVED in which all three Es are together and the two Ds are together.
(b) Find the number of different arrangements of the 8 letters in the word DECEIVED in which the three Es are not all together.
▶️Answer/Explanation
(a) Treat the three Es as one unit (E) and the two Ds as one unit (D). The letters to arrange are: [E], [D], C, I, V. This gives 5 units to arrange.
Number of arrangements = 5! = 120.
(b) Total arrangements of DECEIVED (with 3 Es and 2 Ds) = 8! / (3! × 2!) = 3360.
Arrangements with all three Es together = 6! / 2! = 360 (treating the three Es as one unit).
Arrangements with the three Es not all together = Total arrangements – Arrangements with all three Es together = 3360 – 360 = 3000.
2. [Maximum mark: 6]
(a) There are 6 men and 8 women in a Book Club. The committee of the club consists of five of its members. Mr Lan and Mrs Lan are members of the club. In how many different ways can the committee be selected if exactly one of Mr Lan and Mrs Lan must be on the committee?
(b) In how many different ways can the committee be selected if Mrs Lan must be on the committee and there must be more women than men on the committee?
▶️Answer/Explanation
(a) The committee must include either Mr Lan or Mrs Lan but not both.
If Mr Lan is on the committee, the remaining 4 members are chosen from the other 12 members (excluding Mrs Lan): 12C4 = 495.
Similarly, if Mrs Lan is on the committee, the remaining 4 members are chosen from the other 12 members (excluding Mr Lan): 12C4 = 495.
Total ways = 495 + 495 = 990.
(b) Mrs Lan is already on the committee. The remaining 4 members must include more women than men. Possible scenarios:
1. 3 women and 1 man: 7C3 × 6C1 = 35 × 6 = 210.
2. 4 women and 0 men: 7C4 = 35.
Total ways = 210 + 35 = 245.
3. [Maximum mark: 9]
The times taken to travel to college by 2500 students are summarised in the table.
(a) Draw a histogram to represent this information.
(b) From the data, the estimate of the mean value of t is 31.44. Calculate an estimate of the standard deviation of the times taken to travel to college.
(c) In which class interval does the upper quartile lie?
(d) It was later discovered that the times taken to travel to college by two students were incorrectly recorded. One student’s time was recorded as 15 instead of 5 and the other’s time was recorded as 65 instead of 75. Without doing any further calculations, state with a reason whether the estimate of the standard deviation in part (b) would be increased, decreased or stay the same.
▶️Answer/Explanation
(a) Calculate frequency density for each class interval:
Plot these values on a histogram with time (minutes) on the x-axis and frequency density on the y-axis.
(b) Use midpoints and the formula for standard deviation:
Standard deviation = √229.9264 ≈ 15.2 minutes.
(c) The upper quartile (Q3) corresponds to the 75th percentile. Cumulative frequencies:
– 0 ≤ t < 20: 440.
– 20 ≤ t < 30: 440 + 720 = 1160.
– 30 ≤ t < 40: 1160 + 920 = 2080 (75th percentile lies here, as 2080 > 1875 = 0.75 × 2500).
Thus, the upper quartile lies in the interval 30 ≤ t < 40.
(d) The standard deviation would stay the same because the changes in the recorded times do not affect the frequencies or the class intervals. The data remains within the same bounds, so the spread (standard deviation) is unchanged.
4. [Maximum mark: 10]
Jacob has four coins. One of the coins is biased such that when it is thrown the probability of obtaining a head is 7/10. The other three coins are fair. Jacob throws all four coins once. The number of heads that he obtains is denoted by the random variable X. The probability distribution table for X is as follows.
(a) Show that a = 1/5 and find the values of b and c.
(b) Find E(X).
(c) Jacob throws all four coins together 10 times. Find the probability that he obtains exactly one head on fewer than 3 occasions.
(d) Find the probability that Jacob obtains exactly one head for the first time on the 7th or 8th time that he throws the 4 coins.
▶️Answer/Explanation
(a) For X = 1 (a):
P(1 head) = P(biased coin shows head and others show tails) + P(biased coin shows tail and one fair coin shows head and others show tails) = (7/10)(1/2)³ + (3/10)(3)(1/2)³ = 7/80 + 9/80 = 16/80 = 1/5.
For X = 2 (b):
P(2 heads) = (7/10)(3)(1/2)³ + (3/10)(3)(1/2)³ = 21/80 + 9/80 = 30/80 = 3/8.
For X = 3 (c):
P(3 heads) = (7/10)(3)(1/2)³ + (3/10)(1/2)³ = 21/80 + 3/80 =
5. [Maximum mark: 11]
(a) The lengths, in cm, of the leaves of a particular type are modelled by the distribution N(5.2, 1.5²). Find the probability that a randomly chosen leaf of this type has length less than 6 cm.
(b) The lengths of the leaves of another type are also modelled by a normal distribution. A scientist measures the lengths of a random sample of 500 leaves of this type and finds that 46 are less than 3 cm long and 95 are more than 8 cm long. Find estimates for the mean and standard deviation of the lengths of leaves of this type.
(c) In a random sample of 2000 leaves of this second type, how many would the scientist expect to find with lengths more than 1 standard deviation from the mean?
▶️Answer/Explanation
(a) Standardize X = 6: Z = (6 – 5.2)/1.5 ≈ 0.5333.
P(X < 6) = P(Z < 0.5333) ≈ 0.703.
(b) Let μ be the mean and σ the standard deviation.
For P(X < 3) = 46/500 = 0.092: Z ≈ -1.329.
For P(X > 8) = 95/500 = 0.19: Z ≈ 0.878.
Solve:
(3 – μ)/σ = -1.329 and (8 – μ)/σ = 0.878.
Solving gives μ ≈ 6.01 cm and σ ≈ 2.27 cm.
(c) P(|Z| > 1) = 2 × P(Z > 1) = 2 × (1 – 0.8413) = 0.3174.
Expected number of leaves = 2000 × 0.3174 ≈ 635.
6. [Maximum mark: 8]
(a) Janice is playing a computer game. She has to complete level 1 and level 2 to finish the game. She is allowed at most two attempts at any level. For level 1, the probability that Janice completes it at the first attempt is 0.6. If she fails at her first attempt, the probability that she completes it at the second attempt is 0.3. Show that the probability that Janice moves on to level 2 is 0.72.
(b) Find the probability that Janice finishes the game.
(c) Find the probability that Janice fails exactly one attempt, given that she finishes the game.
▶️Answer/Explanation
(a) P(moves to level 2) = P(completes level 1 on first attempt) + P(fails first attempt but completes on second attempt) = 0.6 + (0.4 × 0.3) = 0.6 + 0.12 = 0.72.
(b) P(finishes game) = P(moves to level 2) × P(completes level 2).
P(completes level 2) = P(completes on first attempt) + P(fails first attempt but completes on second attempt) = 0.4 + (0.6 × 0.2) = 0.4 + 0.12 = 0.52.
Thus, P(finishes game) = 0.72 × 0.52 = 0.3744.
(c) P(fails exactly one attempt | finishes game) = P(fails one attempt and finishes game) / P(finishes game).
P(fails one attempt and finishes game) = P(fails level 1 once and completes level 2 without failing) + P(completes level 1 without failing and fails level 2 once) = (0.4 × 0.3 × 0.4) + (0.6 × 0.6 × 0.2) = 0.048 + 0.072 = 0.12.
Thus, P(fails one attempt | finishes game) = 0.12 / 0.3744 ≈ 0.321.