Question 1
(a) Topic: 5.3 Probability
(b) Topic: 5.3 Probability
(c) Topic: 5.3 Probability
(d) 5.4 Discrete random variables
At a college, the students choose exactly one of tennis, hockey, or netball to play. The table shows the numbers of students in Year 1 and Year 2 at the college playing each of these sports.
| Tennis | Hockey | Netball | |
|---|---|---|---|
| Year 1 | 16 | 22 | 12 |
| Year 2 | 24 | 18 | 28 |
One student is chosen at random from the 120 students. Events \(X\) and \(N\) are defined as follows:
- \(X\): the student is in Year 1
- \(N\): the student plays netball
(a) Find \(P(X \cap N)\).
(b) Find \(P(N \mid X)\).
(c) Determine whether or not \(X\) and \(N\) are independent events.
One of the students who plays netball takes 8 shots at goal. On each shot, the probability that she will succeed is 0.15, independently of all other shots.
(d) Find the probability that she succeeds on fewer than 3 of these shots.
▶️Answer/Explanation
Solution :-
(a) Find \(P(X \cap N)\)
\[ P(X \cap N) = \frac{12}{120} = \frac{1}{10} = 0.1 \]
B1: \(0.1\), \(\frac{1}{10}\), 30% OE (Or Equivalent)
(b) Find \(P(N \mid X)\)
\[ P(N \mid X) = \frac{12}{50} = \frac{6}{25} = 0.24 \]
B1: \(0.24\), \(\frac{6}{25}\) OE
(c) Determine whether or not \(X\) and \(N\) are independent events.
\[ P(X \cap N) = \frac{12}{120} \]
\[ P(N) = \frac{40}{120} \]
\[ P(X) = \frac{50}{120} \]
\[ P(N) \times P(X) = \left(\frac{40}{120}\right) \times \left(\frac{50}{120}\right) = \frac{5}{36} \approx 0.1388… \]
\[ 0.1 \neq 0.1388…, \text{ therefore not independent.} \]
B1: \(P(N \cap X)\), \(P(N)\) and \(P(X)\) or \(P(N)\) and \(P(X’)\) notation seen and equated to the values for \(P(N)\), \(P(X)\) and \(P(N \cap X)\) or \(P(N)\) and \(P(X’)\). Calculation stated and evaluated. Not independent clearly stated. \(\frac{5}{36} = \frac{12}{120}\) does not need to be stated. All values OE. Condone consistent use of \(A, B,\) etc. If values for \(P(N), P(X)\) stated, accept \(P(N) \times P(X) = \frac{5}{36}\).
(d) Find the probability that she succeeds on fewer than 3 of these shots.
Method 1:
\[ P(0, 1, 2) = \binom{8}{0} (0.85)^8 + \binom{8}{1} (0.85)^7 (0.15) + \binom{8}{2} (0.85)^6 (0.15)^2 \]
\[ \approx 0.27249 + 0.38469 + 0.23760 \approx 0.895 \]
M1: One term of form \(\binom{n}{r} p^r (1-p)^{(n-r)}\) with \(0 < p < 1\), \(x = 0\) or \(8\).
A1: Correct unsimplified expression, no terms omitted leading to final answer.
B1: \(0.8945 < p < 0.895\).
Method 2:
\[ P(0, 1, 2) = 1 – \left[ \binom{8}{3} (0.85)^5 (0.15)^3 + \binom{8}{4} (0.85)^4 (0.15)^4 + \binom{8}{5} (0.85)^3 (0.15)^5 + \binom{8}{6} (0.85)^2 (0.15)^6 + \binom{8}{7} (0.85) (0.15)^7 + (0.15)^8 \right] \]
\[ \approx 0.895 \]
M1: One term of form \(\binom{n}{r} p^r (1-p)^{(n-r)}\) with \(0 < p < 1\), \(x = 0\) or \(8\).
A1: Correct unsimplified expression. Condone omission of final bracket ‘ ] ‘. If other brackets omitted, allow recovery if \(1 – 0.1052…\) seen.
B1: \(0.8945 < p < 0.895\).
Question 2
(a) Topic: 5.2 Permutations and Combinations
(b) Topic: 5.2 Permutations and Combinations
(a) Find the number of different arrangements of the 9 letters in the word ALGEBRAIC.
(b) Find the number of different arrangements of the 9 letters in the word ALGEBRAIC in which there are no more than two letters between the two A’s.
▶️Answer/Explanation
Solution :-
(a) Finding the total number of arrangements
The total number of ways to arrange the letters in ALGEBRAIC, considering that there are two identical A’s, is given by:
\[ \frac{9!}{2!} = \frac{362880}{2} = 181440 \]
Final Answer: 181440
(b) Finding the number of arrangements with no more than two letters between the two A’s
We count the valid scenarios where the two A’s have 0, 1, or 2 letters between them.
Method 1:
| Scenario | Calculation | Value |
|---|---|---|
| No letters between A’s | \(\ 7! \ \) | 5040 |
| 1 letter between A’s | \(\ 7! \times 7 \ \) | 35280 |
| 2 letters between A’s | \(\ 7! \times 6 \ \) | 30240 |
| Total | \(\ 7! \times (8 + 7 + 6) \ \) | 105840 |
Method 2:
- Ways to place the two A’s with 0 letters between them: \( \binom{8}{2} \times 7! = 40320 \).
- Ways to place the two A’s with 1 letter between them: \( \binom{7}{2} \times 7! = 35280 \).
- Ways to place the two A’s with 2 letters between them: \( \binom{6}{2} \times 7! = 30240 \).
Total: \( 40320 + 35280 + 30240 = 105840 \)
Method 3:
We subtract cases where the two A’s are more than two spaces apart from the total arrangements.
- Arrangements where A’s have 3 or more letters between them: \( 7! \times (5 + 4 + 3 + 2 + 1) = 75600 \).
- Valid arrangements: \( 181440 – 75600 = 105840 \).
Final Answer: 105840
Question 3
(a) Topic: 5.4 Discrete Random Variables
(b) Topic: 5.4 Discrete Random Variables
A fair coin and an ordinary fair six-sided die are thrown at the same time. The random variable \( X \) is defined as follows:
- If the coin shows a tail, \( X \) is twice the score on the dice.
- If the coin shows a head, \( X \) is the score on the dice if the score is even, and \( X \) is 0 otherwise.
(a) Draw up the probability distribution table for \( X \).
(b) Find \( \text{Var}(X) \).
▶️ Answer/Explanation
Solution :-
(a) Probability Distribution Table:
| \( x \) | 0 | 2 | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|---|---|
| \( P(X = x) \) | \(\frac{2}{12}\) | \(\frac{1}{12}\) | \(\frac{2}{12}\) | \(\frac{1}{12}\) | \(\frac{1}{12}\) | \(\frac{1}{12}\) | \(\frac{1}{12}\) |
| Decimal | 0.25 | 0.167 | 0.167 | 0.0833 | 0.0833 | 0.0833 | 0.0833 |
(b) Expected Value \( E(X) \):
\[ E(X) = [0] \times \frac{2}{12} + 2 \times \frac{4}{12} + 4 \times \frac{2}{12} + 6 \times \frac{1}{12} + 8 \times \frac{1}{12} + 10 \times \frac{1}{12} + 12 \times \frac{1}{12} \]
\[ E(X) = \frac{0 + 8 + 8 + 6 + 8 + 10 + 12}{12} = \frac{54}{12} = 4.5 \]
Variance \( \text{Var}(X) \):
\[ E(X^2) = [0] \times \frac{2}{12} + 2^2 \times \frac{4}{12} + 4^2 \times \frac{2}{12} + 6^2 \times \frac{1}{12} + 8^2 \times \frac{1}{12} + 10^2 \times \frac{1}{12} + 12^2 \times \frac{1}{12} \]
\[ E(X^2) = \frac{0 + 16 + 32 + 36 + 64 + 100 + 144}{12} = \frac{392}{12} = 8.5 \]
\[ \text{Var}(X) = E(X^2) – (E(X))^2 \]
\[ \text{Var}(X) = 8.5 – (4.5)^2 = 8.5 – 20.25 = 14.75 \]
Final Answer: \( \frac{59}{4} \) or \( 14.75 \)
Question 4
(a) Topic: 5.5 The Normal Distribution
(b) Topic: 5.5 The Normal Distribution
The heights, in metres, of white pine trees are normally distributed with mean 19.8 and standard deviation 2.4.
In a certain forest, there are 450 white pine trees.
(a) How many of these trees would you expect to have height less than 18.2 metres?
The heights, in metres, of red pine trees are normally distributed with mean 23.4 and standard deviation \( \sigma \). It is known that 26% of red pine trees have height greater than 25.5 metres.
(b) Find the value of \( \sigma \).
▶️ Answer/Explanation
Solution :-
(a) Finding the expected number of trees with height \( < 18.2 \) m
Using the standardization formula:
\[ P(X < 18.2) = P\left( Z < \frac{18.2 – 19.8}{2.4} \right) \]
\[ = P(Z < -0.6667) \]
\[ = 1 – P(Z < 0.6667) \]
\[ = 1 – 0.7477 \]
\[ = 0.2523 \]
Expected number of trees:
\[ 0.2523 \times 450 = 113.5 \]
Final Answer: 113 or 114 trees
(b) Finding \( \sigma \)
\[ P(X > 25.5) = 0.26 \Rightarrow P\left( Z > \frac{25.5 – 23.4}{\sigma} \right) = 0.26 \]
\[ \frac{25.5 – 23.4}{\sigma} = 0.643 \]
\[ \sigma = \frac{2.1}{0.643} \]
\[ \sigma = 3.27 \]
Final Answer: \( 3.26 \leq \sigma \leq 3.27 \)
Question 5
(a) Topic: 5.2 Permutations and Combinations
(b) Topic: 5.2 Permutations and Combinations
In a class of 21 students, there are 10 violinists, 6 guitarists, and 5 pianists. A group of 7 is to be chosen from these 21 students. The group will consist of 4 violinists, 2 guitarists, and 1 pianist.
In how many ways can the group of 7 be chosen?
(a) On another occasion, a group of 5 will be chosen from the 21 students. The group must contain at least 2 violinists, at least 1 guitarist, and at most 1 pianist.
(b) In how many ways can the group of 5 be chosen?
▶️ Answer/Explanation
Solution :-
(a) Finding the number of ways to choose 7 students
\[ \binom{10}{4} \times \binom{6}{2} \times \binom{5}{1} \]
\[ = 210 \times 15 \times 5 \]
\[ = 15750 \]
Final Answer: 15,750 ways
(b) Finding the number of ways to choose 5 students
We analyze different valid scenarios:
| Scenario | Violinists (V) | Guitarists (G) | Pianists (P) | Calculation | Ways |
|---|---|---|---|---|---|
| VVVVG | 4 | 1 | 0 | \(\binom{10}{4} \times \binom{6}{1} \times \binom{5}{0}\) | 1260 |
| VVVGG | 3 | 2 | 0 | \(\binom{10}{3} \times \binom{6}{2} \times \binom{5}{0}\) | 1800 |
| VVGGG | 2 | 3 | 0 | \(\binom{10}{2} \times \binom{6}{3} \times \binom{5}{0}\) | 900 |
| VVVGP | 3 | 1 | 1 | \(\binom{10}{3} \times \binom{6}{1} \times \binom{5}{1}\) | 3600 |
| VVGGP | 2 | 2 | 1 | \(\binom{10}{2} \times \binom{6}{2} \times \binom{5}{1}\) | 3375 |
| Total | 10,935 | ||||
Final Answer: 10,935 ways
Question 6
(a) Topic: 5.1 Representation of Data
(b) Topic: 5.1 Representation of Data
(c) Topic: 5.1 Representation of Data
Teams of 15 runners took part in a charity run last Saturday. The times taken, in minutes, to complete the course by the runners from the Falcons and the runners from the Kites are shown in the table.
| Falcons | 38 | 39 | 42 | 44 | 46 | 48 | 50 | 51 | 52 | 56 | 58 | 59 | 69 | 76 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Kites | 32 | 40 | 40 | 45 | 47 | 48 | 52 | 54 | 58 | 59 | 60 | 61 | 63 | 65 |
(a) Draw a back-to-back stem-and-leaf diagram to represent this information, with the Falcons on the left-hand side.
(b) Find the median and the interquartile range of the times for the Falcons.
(c) Find the mean and the standard deviation of the times taken by all 30 runners from the two teams.
▶️ Answer/Explanation
Solution :-
(a) Back-to-back Stem-and-Leaf Diagram
| Falcons | Stem | Kites |
|---|---|---|
| 9 8 3 | 3 | 2 |
| 6 2 1 0 5 | 4 | 0 0 7 8 |
| 9 6 2 1 0 5 | 5 | 2 4 8 9 |
| 9 4 6 | 6 | 0 1 3 5 |
Key: 15 | 4 means 51 minutes for Falcons and 54 minutes for Kites.
(b) Median and Interquartile Range
Median Calculation:
\[ \text{Median} = 51 \text{ minutes} \]
Interquartile Range (IQR):
\[ IQR = Q_3 – Q_1 = 59 – 44 = 15 \text{ minutes} \]
(c) Mean and Standard Deviation Calculation
Given Data:
\[ \sum x = 792, \quad \sum x^2 = 43,504, \quad \sum y = 783, \quad \sum y^2 = 42,223 \]
Mean Calculation:
\[ \text{Mean} = \frac{792 + 783}{30} = \frac{1575}{30} = 52.5 \]
Variance Calculation:
\[ \text{Variance} = \frac{43504 + 42223}{30} – \left( \frac{792 + 783}{30} \right)^2 \]
\[ = \frac{85727}{30} – \left( \frac{1575}{30} \right)^2 \]
\[ = 101.3167 \]
Standard Deviation:
\[ \sigma = \sqrt{101.3167} = 10.1 \]
Final Answer: Mean = 52.5, Standard Deviation = 10.1
Question 7
(a) Topic: 5.4 Discrete Random Variables
(b) Topic: 5.4 Discrete Random Variables
(C) Topic: 5.4 Discrete Random Variables
(d) Topic: 5.4 Discrete Random Variables
In a game, players attempt to score a goal by kicking a ball into a net. The probability that Leno scores a goal is 0.4 on any attempt, independently of all other attempts. The random variable \( X \) denotes the number of attempts that it takes Leno to score a goal.
(a) Find \( P(X = 5) \).
(b) Find \( P(3 \leq X < 7) \).
(c) Find the probability that Leno scores his second goal on or before his 5th attempt.
Leno has 75 attempts to score a goal.
(d) Use a suitable approximation to find the probability that Leno scores more than 28 goals but fewer than 35 goals.
▶️ Answer/Explanation
Solution :-
(a) Finding \( P(X = 5) \)
The probability mass function for a geometric distribution is given by:
\[ P(X = k) = (1 – p)^{k-1} p \]
Substituting \( p = 0.4 \) and \( k = 5 \):
\[ P(X = 5) = (0.6)^4 \times 0.4 = 0.0518 \]
(b) Finding \( P(3 \leq X < 7) \)
\[ P(3 \leq X < 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) \]
Using the formula:
\[ P(X = 3) = (0.6)^2 \times 0.4 = 0.144 \]
\[ P(X = 4) = (0.6)^3 \times 0.4 = 0.0864 \]
\[ P(X = 5) = (0.6)^4 \times 0.4 = 0.0518 \]
\[ P(X = 6) = (0.6)^5 \times 0.4 = 0.0311 \]
\[ P(3 \leq X < 7) = 0.144 + 0.0864 + 0.0518 + 0.0311 = 0.3133 \]
(c) Finding the probability that Leno scores his second goal on or before his 5th attempt
Using the negative binomial distribution:
\[ P(X \leq 5) = P(2nd \text{ goal at 2nd attempt}) + P(2nd \text{ goal at 3rd attempt}) + P(2nd \text{ goal at 4th attempt}) + P(2nd \text{ goal at 5th attempt}) \]
Using the formula:
\[ P(X = k) = \binom{k-1}{r-1} p^r (1-p)^{k-r} \]
Computing each term:
\[ P(X = 2) = \binom{1}{1} (0.4)^2 (0.6)^0 = 0.16 \]
\[ P(X = 3) = \binom{2}{1} (0.4)^2 (0.6)^1 = 0.192 \]
\[ P(X = 4) = \binom{3}{1} (0.4)^2 (0.6)^2 = 0.1152 \]
\[ P(X = 5) = \binom{4}{1} (0.4)^2 (0.6)^3 = 0.0691 \]
\[ P(X \leq 5) = 0.16 + 0.192 + 0.1152 + 0.0691 = 0.5363 \]
(d) Using a normal approximation to find \( P(28 < X < 35) \)
Since \( X \sim Bin(75, 0.4) \), we approximate using a normal distribution:
Mean:
\[ \mu = np = 75 \times 0.4 = 30 \]
Variance:
\[ \sigma^2 = np(1-p) = 75 \times 0.4 \times 0.6 = 18 \]
Standard deviation:
\[ \sigma = \sqrt{18} = 4.24 \]
Applying continuity correction:
\[ P(28 < X < 35) \Rightarrow P(28.5 < X < 34.5) \]
Converting to standard normal form:
\[ P\left(\frac{28.5 – 30}{4.24} < Z < \frac{34.5 – 30}{4.24} \right) \]
\[ P(-0.35 < Z < 1.06) \]
Using standard normal table:
\[ P(Z < 1.06) = 0.8554, \quad P(Z < -0.35) = 0.3632 \]
\[ P(-0.35 < Z < 1.06) = 0.8554 – 0.3632 = 0.4922 \]