1. [Maximum mark: 5]
A fair red spinner has edges numbered 1, 2, 2, 3. A fair blue spinner has edges numbered −3, −2, −1, −1. Each spinner is spun once and the number on the edge on which each spinner lands is noted. The random variable X denotes the sum of the resulting two numbers.
(a) Draw up the probability distribution table for X.
(b) Given that E(X) = 0.25, find the value of Var(X).
▶️Answer/Explanation
(a) The probability distribution table for X is:
| X | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| P(X) | 1/16 | 3/16 | 5/16 | 5/16 | 2/16 |
(b) Using Var(X) = E(X²) – [E(X)]²:
E(X²) = (1/16 × 4) + (3/16 × 1) + (5/16 × 0) + (5/16 × 1) + (2/16 × 4) = 19/16
Var(X) = (19/16) – (0.25)² = 19/16 – 1/16 = 18/16 = 1.1875
2. [Maximum mark: 6]
In a certain country, the probability of more than 10 cm of rain on any particular day is 0.18, independently of the weather on any other day.
(a) Find the probability that in any randomly chosen 7-day period, more than 2 days have more than 10 cm of rain.
(b) For 3 randomly chosen 7-day periods, find the probability that exactly two of these periods have at least one day with more than 10 cm of rain.
▶️Answer/Explanation
(a) X ~ B(7, 0.18)
P(X > 2) = 1 – P(X ≤ 2) = 1 – [P(0) + P(1) + P(2)]
= 1 – [0.82⁷ + 7×0.18×0.82⁶ + 21×0.18²×0.82⁵] ≈ 1 – 0.88458 = 0.115
(b) First find P(at least 1 rainy day in 7 days) = 1 – 0.82⁷ ≈ 0.7507
Then Y ~ B(3, 0.7507)
P(Y = 2) = 3 × (0.7507)² × (1-0.7507) ≈ 0.421
3. [Maximum mark: 6]
At a summer camp an arithmetic test is taken by 250 children. The times taken, to the nearest minute, to complete the test were recorded. The results are summarised in the table.
| Time taken (minutes) | 1-30 | 31-45 | 46-65 | 66-75 | 76-100 |
|---|---|---|---|---|---|
| Frequency | 21 | 30 | 68 | 86 | 45 |
(a) Draw a histogram to represent this information.
(b) State which class interval contains the median.
(c) Given that an estimate of the mean time is 61.05 minutes, state what feature of the distribution accounts for the median and the mean being different.
▶️Answer/Explanation
(a) Calculate frequency densities:
1-30: 21/30 = 0.7
31-45: 30/15 = 2
46-65: 68/20 = 3.4
66-75: 86/10 = 8.6
76-100: 45/25 = 1.8
Plot these with correct class boundaries and labeled axes.
(b) The median is in the 66-75 minute class interval.
(c) The distribution is not symmetrical (or is skewed).
4. [Maximum mark: 11]
The weights of male leopards in a particular region are normally distributed with mean 55 kg and standard deviation 6 kg.
(a) Find the probability that a randomly chosen male leopard from this region weighs between 46 and 62 kg.
(b) The weights of female leopards in this region are normally distributed with mean 42 kg and standard deviation σ kg. It is known that 25% of female leopards in the region weigh less than 36 kg. Find the value of σ.
(c) The distributions of the weights of male and female leopards are independent of each other. A male leopard and a female leopard are each chosen at random. Find the probability that both the weights of these leopards are less than 46 kg.
▶️Answer/Explanation
(a) For X ~ N(55, 6²):
P(46 < X < 62) = P(-1.5 < Z < 1.1667) = Φ(1.1667) – (1 – Φ(1.5)) ≈ 0.8784 + 0.9332 – 1 = 0.812
(b) For Y ~ N(42, σ²):
P(Y < 36) = 0.25 ⇒ z = -0.674
(36-42)/σ = -0.674 ⇒ σ = 8.90
(c) P(X < 46) = P(Z < -1.5) = 0.0668
P(Y < 46) = P(Z < (46-42)/8.90) ≈ P(Z < 0.449) ≈ 0.6732
P(both) = 0.0668 × 0.6732 ≈ 0.0450
5. [Maximum mark: 10]
A group of 12 people consists of 3 boys, 4 girls and 5 adults.
(a) In how many ways can a team of 5 people be chosen from the group if exactly one adult is included?
(b) In how many ways can a team of 5 people be chosen from the group if the team includes at least 2 boys and at least 1 girl?
(c) The same group of 12 people stand in a line. How many different arrangements are there in which the 3 boys stand together and an adult is at each end of the line?
▶️Answer/Explanation
(a) ⁵C₁ × ⁷C₄ = 5 × 35 = 175 ways
(b) Possible combinations:
2B, 1G, 2A: ³C₂ × ⁴C₁ × ⁵C₂ = 3×4×10 = 120
2B, 2G, 1A: ³C₂ × ⁴C₂ × ⁵C₁ = 3×6×5 = 90
2B, 3G: ³C₂ × ⁴C₃ = 3×4 = 12
3B, 1G, 1A: ³C₃ × ⁴C₁ × ⁵C₁ = 1×4×5 = 20
3B, 2G: ³C₃ × ⁴C₂ = 1×6 = 6
Total = 120 + 90 + 12 + 20 + 6 = 248 ways
(c) Treat boys as one unit (B), with 2 adults fixed at ends (A _ _ _ _ _ _ _ _ _ A)
Arrange remaining 7 units (B, 2 adults, 4 girls): 7! ways
Arrange boys within their unit: 3! ways
Choose 2 adults for ends: ⁵P₂ ways
Total arrangements = 7! × 3! × 20 = 5040 × 6 × 20 = 4,838,400
6. [Maximum mark: 12]
A factory produces chocolates in three flavors: lemon, orange and strawberry in the ratio 3:5:7 respectively. Nell checks the chocolates on the production line by choosing chocolates randomly one at a time.
(a) Find the probability that the first chocolate with lemon flavor that Nell chooses is the 7th chocolate that she checks.
(b) Find the probability that the first chocolate with lemon flavor that Nell chooses is after she has checked at least 6 chocolates.
‘Surprise’ boxes of chocolates each contain 15 chocolates: 3 are lemon, 5 are orange and 7 are strawberry. Petra has a box of Surprise chocolates. She chooses 3 chocolates at random from the box. She eats each chocolate before choosing the next one.
(c) Find the probability that none of Petra’s 3 chocolates has orange flavor.
(d) Find the probability that each of Petra’s 3 chocolates has a different flavor.
(e) Find the probability that at least 2 of Petra’s 3 chocolates have strawberry flavor given that none of them has orange flavor.
▶️Answer/Explanation
(a) P(6 non-lemon then lemon) = (12/15)⁶ × (3/15) ≈ 0.0524
(b) P(first lemon after 6 checks) = P(first 6 non-lemon) = (12/15)⁶ ≈ 0.262
(c) P(no orange) = (10/15)×(9/14)×(8/13) = 24/91 ≈ 0.264
(d) P(one of each) = (3/15)×(5/14)×(7/13)×6 ≈ 0.231
(e) P(≥2 strawberry | no orange) = [P(2S1L) + P(3S)]/P(no orange)
= [(7/15×6/14×3/13)×3 + (7/15×6/14×5/13)]/(24/91) ≈ 49/60 ≈ 0.817