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 $P(X N)$. (b) Find $P(N X)$. (c) Determine whether…

(a) P(X N) = 3/10 (or 0.3) · (b) P(N X) = 6/25 (or 0.24) · (c) X and N are not independent. · (d) 0.895

A fair coin and an ordinary fair six-sided dice 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 $\text{Var}(X)$.

(a) P(X=0)=1/4, P(X=2)=1/6, P(X=4)=1/6, P(X=6)=1/6, P(X=8)=1/12, P(X=10)=1/12, P(X=12)=1/12 · (b) Var(X) = 14.75 = 59/4

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? (b) 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. Find the value of $\sigma$.

(a) 113 or 114 trees · (b) σ = 3.27

9709/52 Oct/Nov 2024 — Mathematics Solved Past Paper

Q1MediumProbabilityDiscrete 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 $P(X | N)$ .

1M Similar

(b)

Find $P(N | X)$ .

1M Similar

(c)

Determine whether or not $X$ and $N$ are independent events.

1M Similar

DifficultyMedium

Worked solution

Approach

Two events are independent if and only if $\text{P}(X \cap N) = \text{P}(X) \times \text{P}(N)$ . Compute each side from the table and compare.

Working

From the table (out of $120$ students total):

\text{P}(X) = \frac{16 + 22 + 12}{120} = \frac{50}{120}

\text{P}(N) = \frac{12 + 28}{120} = \frac{40}{120}

\text{P}(X \cap N) = \frac{12}{120}

Now test independence:

\text{P}(X) \times \text{P}(N) = \frac{50}{120} \times \frac{40}{120} = \frac{5}{36} \approx 0.1389

Since

\text{P}(X \cap N) = \frac{12}{120} = \frac{1}{10} = 0.1 \neq \frac{5}{36}

the events are not independent.

Answer

X \text{ and } N \text{ are not independent.}

Final answer

X and N are not independent.

Detailed explanation

Walkthrough

Two events $X$ and $N$ are independent when knowing one tells you nothing about the other, which algebraically means $\text{P}(X \cap N) = \text{P}(X) \times \text{P}(N)$ . We compute three quantities from the table.

$\text{P}(X)$ : there are $50$ Year 1 students out of $120$ , so $\text{P}(X) = 50/120$ .
$\text{P}(N)$ : there are $40$ netball players out of $120$ , so $\text{P}(N) = 40/120$ .
$\text{P}(X \cap N)$ : there are $12$ students who are both Year 1 and play netball out of $120$ , so $\text{P}(X \cap N) = 12/120 = 0.1$ .

Multiplying the marginals gives $\frac{50}{120} \times \frac{40}{120} = \frac{5}{36} \approx 0.1389$ . This is not equal to $0.1$ , so the equality fails and the events are not independent.

Key Takeaways

Independence is tested by checking whether $\text{P}(A \cap B) = \text{P}(A) \times \text{P}(B)$ .
The direction of the inequality tells you something about how the events relate: here $\text{P}(X \cap N) < \text{P}(X)\text{P}(N)$ , so being a Year 1 student makes netball less likely than under independence.

Common Mistakes

Concluding independence because neither $\text{P}(X|N) = \text{P}(X)$ nor $\text{P}(N|X) = \text{P}(N)$ "looks obviously" true. The algebraic test is required.
Forgetting to state the conclusion clearly ("not independent").
Computing only $\text{P}(N) \times \text{P}(X)$ without comparing it to $\text{P}(N \cap X)$ .

Things to Be Careful About

The mark scheme requires the three probabilities (or their numerators), the multiplication, the comparison, and the explicit conclusion.
Either showing the product does not equal the intersection, or stating clearly that the events are not independent, earns the mark.

Techniques used

compute marginal probabilitiescompute intersection probabilitytest independence condition

(d)

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.

Find the probability that she succeeds on fewer than 3 of these shots.

3M Similar

DifficultyMedium

Worked solution

Approach

Let $Y$ be the number of successful shots. $Y \sim \text{B}(8, 0.15)$ since there are a fixed number of independent trials ( $n = 8$ ), each with the same success probability ( $p = 0.15$ ). We need

\text{P}(Y < 3) = \text{P}(Y = 0) + \text{P}(Y = 1) + \text{P}(Y = 2)

Working

Using the binomial formula $\text{P}(Y = x) = {}^8\text{C}_x (0.15)^x (0.85)^{8-x}$ :

\text{P}(Y = 0) = (0.85)^8 = 0.27249

\text{P}(Y = 1) = {}^8\text{C}_1 (0.15)^1 (0.85)^7 = 8 \times 0.15 \times (0.85)^7 = 0.38469

\text{P}(Y = 2) = {}^8\text{C}_2 (0.15)^2 (0.85)^6 = 28 \times 0.0225 \times (0.85)^6 = 0.23760

Summing:

\text{P}(Y < 3) = 0.27249 + 0.38469 + 0.23760 = 0.89478

Rounded to three significant figures:

\text{P}(Y < 3) \approx 0.895

Answer

\text{P}(Y < 3) = 0.895

Final answer

0.895

Detailed explanation

Walkthrough

The number of successes in a fixed number of independent trials with constant success probability follows a binomial distribution. Here we have $n = 8$ trials, each with $p = 0.15$ , so $Y \sim \text{B}(8, 0.15)$ .

"Fewer than 3" means $Y \in \{0, 1, 2\}$ , so we add three binomial probabilities. The binomial probability formula is

\text{P}(Y = x) = {}^n\text{C}_x p^x (1-p)^{n-x}.

For $x = 0$ the binomial coefficient is $1$ , giving $(0.85)^8 \approx 0.27249$ . For $x = 1$ , ${}^8\text{C}_1 = 8$ , giving $8 \times 0.15 \times (0.85)^7 \approx 0.38469$ . For $x = 2$ , ${}^8\text{C}_2 = 28$ , giving $28 \times (0.15)^2 \times (0.85)^6 \approx 0.23760$ . Summing the three gives approximately $0.89478$ , which rounds to $0.895$ .

Key Takeaways

Recognise the binomial setting: fixed $n$ , two outcomes, independent trials, constant $p$ .
"Fewer than 3" means summing $x = 0, 1, 2$ — a common source of off-by-one errors.
The complementary calculation $\text{P}(Y \geq 3) = 1 - \text{P}(Y < 3)$ is an alternative valid method (Method 2 in the mark scheme).

Common Mistakes

Using $p = 0.85$ instead of $p = 0.15$ (confusing success with failure).
Computing $\text{P}(Y \leq 3)$ instead of $\text{P}(Y < 3)$ , i.e. including the $x = 3$ term.
Omitting the ${}^8\text{C}_x$ coefficient in the formula.

Things to Be Careful About

The mark scheme accepts $0.8945 \leq p \leq 0.895$ as the final answer; the unrounded sum is $0.89478$ which lies in this range.
A correct unsimplified expression of all three terms earns full marks even if the final addition is not carried out, but the final value is required for the third mark.

Techniques used

recognise binomial distributionapply binomial probability formula for x=0,1,2sum probabilities

Q2MediumPermutations and Combinations

(a)

Find the number of different arrangements of the 9 letters in the word ALGEBRAIC.

1M Similar

(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 As.

3M Similar

DifficultyMedium

Worked solution

Approach

We split the count into three mutually exclusive cases by the number $k$ of letters between the two As: $k = 0, 1, 2$ . In each case we place the As in valid positions and arrange the 7 remaining distinct letters.

Working

There are 7 non-A letters: L, G, E, B, R, I, C. They can be arranged in $7!$ ways.n

For exactly $k$ letters between the As, the left-hand A can be placed in positions $1$ up to $8-k$ , so there are $8-k$ possible position pairs for the As.

Case $k=0$ :

7! \times 8

Case $k=1$ :

7! \times 7

Case $k=2$ :

7! \times 6

These cases cannot overlap, so the total number of arrangements is

7!(8+7+6) = 7! \times 21 = 5040 \times 21 = 105840

Answer

105840

Final answer

105840

Detailed explanation

Walkthrough

We only want arrangements with 0, 1 or 2 letters between the two As. The As are identical, so the best way to count is to choose their positions first and then fill the remaining places with the other letters.

For a fixed value of $k$ , if the left A is in position $p$ , the right A must be in position $p + k + 1$ . Therefore $p$ can run from 1 to $8-k$ , giving $8-k$ possible pairs.

For example, when $k=0$ the As are adjacent, with possible pairs $(1,2), (2,3), \dots, (8,9)$ - 8 pairs. When $k=1$ the pairs are $(1,3), (2,4), \dots, (7,9)$ - 7 pairs. When $k=2$ the pairs are $(1,4), (2,5), \dots, (6,9)$ - 6 pairs.

For every chosen pair of As, the remaining 7 positions are filled by the 7 distinct non-A letters in $7!$ ways. Adding the three mutually exclusive cases gives the total.

Key Takeaways

Counting arrangements with restrictions often benefits from splitting into mutually exclusive cases. Repeated letters can be handled by choosing positions for them first; since the As are identical, each chosen pair of positions corresponds to exactly one arrangement of the As.

Common Mistakes

Counting cases with 3 or more letters between the As.
Treating the two As as distinct, which would double each case.
Adding scenario values but forgetting to multiply by $7!$ .
Using $9!$ in each case, which would count arrangements before placing the As and would be incorrect.

Things to Be Careful About

To earn full marks, show at least one correct scenario for the first B1, then add the three correct scenario totals for the M1 and A1. The cases $k=0,1,2$ are mutually exclusive, so no arrangement is counted twice. If using the complement method, subtract cases with 3, 4, 5, 6 or 7 letters between the As from $\frac{9!}{2!}$ .

Techniques used

split into mutually exclusive cases by gap sizecount valid positions for identical repeated lettersmultiply by arrangements of remaining distinct letterssum case totals

Q3Medium-EasyDiscrete Random Variables

A fair coin and an ordinary fair six-sided dice 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 $X$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

A coin and a dice are independent, so each of the 12 outcomes is equally likely, with probability $\frac{1}{12}$ . For each outcome, determine the value of $X$ , then add the probabilities for any outcomes that give the same value of $X$ .

Working

If the coin shows a tail, $X = 2 \times$ the dice score, so $X$ takes values $2, 4, 6, 8, 10, 12$ , each with probability $\frac{1}{12}$ .

If the coin shows a head, $X$ equals the dice score when it is even, and $X = 0$ otherwise. So $X = 2, 4, 6$ each have probability $\frac{1}{12}$ , and $X = 0$ has probability $\frac{3}{12}$ .

Adding probabilities for the same value of $X$ :

$x$	0	2	4	6	8	10	12
$\text{P}(X = x)$	$\frac{3}{12}$	$\frac{2}{12}$	$\frac{2}{12}$	$\frac{2}{12}$	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{12}$

Answer

\text{P}(X = 0) = \frac{1}{4}, \quad \text{P}(X = 2) = \frac{1}{6}, \quad \text{P}(X = 4) = \frac{1}{6}, \quad \text{P}(X = 6) = \frac{1}{6}

\text{P}(X = 8) = \frac{1}{12}, \quad \text{P}(X = 10) = \frac{1}{12}, \quad \text{P}(X = 12) = \frac{1}{12}

Final answer

P(X=0)=1/4, P(X=2)=1/6, P(X=4)=1/6, P(X=6)=1/6, P(X=8)=1/12, P(X=10)=1/12, P(X=12)=1/12

Detailed explanation

Walkthrough

The coin and dice are independent, so there are $2 \times 6 = 12$ equally likely outcomes. Each outcome has probability $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$ .

For a tail, $X$ is twice the dice score, so $X$ can be $2, 4, 6, 8, 10, 12$ , each with probability $\frac{1}{12}$ .

For a head, if the dice score is even, $X$ is that score, so $X = 2, 4, 6$ each with probability $\frac{1}{12}$ . If the dice score is odd, $X = 0$ , and there are three odd scores, so $\text{P}(X = 0) = \frac{3}{12} = \frac{1}{4}$ .

Now collect repeated values. The value $X = 2$ occurs in two ways: tail with dice score 1, and head with dice score 2. So its probability is $\frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$ . The same happens for $X = 4$ and $X = 6$ . The values $8, 10, 12$ occur only from tails, so each has probability $\frac{1}{12}$ .

Key Takeaways

A probability distribution table must list every possible value of the random variable and assign probabilities that sum to 1. When different outcomes give the same value of $X$ , their probabilities must be added together.

Common Mistakes

Forgetting that a head with an odd dice score gives $X = 0$ .
Forgetting that a tail doubles the dice score.
Not adding the probabilities when the same value of $X$ can arise in more than one way.
Giving probabilities that do not sum to 1.

The mark scheme allows unsimplified probabilities, but decimal answers must be correct to at least 3 significant figures.

Things to Be Careful About

The value $X = 0$ is easy to miss. Also, $X = 2$ , $X = 4$ and $X = 6$ each have two contributing outcomes, so their probabilities are $\frac{2}{12}$ , not $\frac{1}{12}$ .

Techniques used

enumerate equally likely outcomescombine probabilities for repeated values of Xconstruct a probability distribution table

(b)

Find $\text{Var}(X)$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the probability distribution from part (a) to compute $\text{E}(X)$ and $\text{E}(X^2)$ , then apply $\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$ .

Working

From the distribution table:

\text{E}(X) = 0\left(\frac{3}{12}\right) + 2\left(\frac{2}{12}\right) + 4\left(\frac{2}{12}\right) + 6\left(\frac{2}{12}\right) + 8\left(\frac{1}{12}\right) + 10\left(\frac{1}{12}\right) + 12\left(\frac{1}{12}\right)

\text{E}(X) = \frac{4 + 8 + 12 + 8 + 10 + 12}{12} = \frac{54}{12} = 4.5

Next,

\text{E}(X^2) = 0^2\left(\frac{3}{12}\right) + 2^2\left(\frac{2}{12}\right) + 4^2\left(\frac{2}{12}\right) + 6^2\left(\frac{2}{12}\right) + 8^2\left(\frac{1}{12}\right) + 10^2\left(\frac{1}{12}\right) + 12^2\left(\frac{1}{12}\right)

\text{E}(X^2) = \frac{8 + 32 + 72 + 64 + 100 + 144}{12} = \frac{420}{12} = 35

Therefore,

\text{Var}(X) = 35 - (4.5)^2 = 35 - 20.25 = 14.75

Answer

\text{Var}(X) = 14.75 = \frac{59}{4}

Final answer

Var(X) = 14.75 = 59/4

Detailed explanation

Walkthrough

Use the distribution table from part (a). The expected value is found by multiplying each value of $X$ by its probability and adding the results.

The zero term contributes nothing to $\text{E}(X)$ , but it is included for completeness. The nonzero terms give $\frac{4}{12} + \frac{8}{12} + \frac{12}{12} + \frac{8}{12} + \frac{10}{12} + \frac{12}{12} = \frac{54}{12} = 4.5$ .

For variance, use $\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$ . First compute $\text{E}(X^2)$ by multiplying each squared value of $X$ by its probability. The sum is $\frac{8 + 32 + 72 + 64 + 100 + 144}{12} = \frac{420}{12} = 35$ .

Then subtract $[\text{E}(X)]^2 = (4.5)^2 = 20.25$ , giving $35 - 20.25 = 14.75$ , which is $\frac{59}{4}$ .

Key Takeaways

The variance of a discrete random variable is calculated using $\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2$ . This requires two separate sums: one for the expected value and one for the expected value of the squares.

Common Mistakes

Using $\text{Var}(X) = \text{E}(X^2) - \text{E}(X)$ instead of subtracting $[\text{E}(X)]^2$ .

Forgetting to square each value of $X$ when computing $\text{E}(X^2)$ .

Using probabilities from the table incorrectly, especially if the table from part (a) has unsimplified fractions.

The mark scheme allows follow-through from a table whose probabilities sum to 1, but the final answer must be correct.

Things to Be Careful About

Always use the same probabilities from the distribution table. The zero term is harmless but must not be counted as a nonzero contribution.

The final answer can be written as $14.75$ , $14\frac{3}{4}$ , or $\frac{59}{4}$ .

Decimals in intermediate working should be kept accurate enough that the final answer is not affected by rounding.

Techniques used

compute expected value from a distribution tablecompute expected value of X squaredapply the variance formula

Q4MediumThe 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?

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Standardise the height $18.2$ using the normal distribution formula, find the normal probability from the standard normal table, then multiply by the total number of trees to get the expected count.

Working

Let $X$ be the height of a white pine tree, so $X \sim \mathrm{N}(19.8, 2.4^2)$ .

Standardise:

z = \frac{18.2 - 19.8}{2.4} = \frac{-1.6}{2.4} = -0.6667

Find the probability:

\mathrm{P}(X < 18.2) = \mathrm{P}(Z < -0.6667) = 1 - \Phi(0.6667)

From the standard normal table, $\Phi(0.6667) = 0.7477$ , so

\mathrm{P}(X < 18.2) = 1 - 0.7477 = 0.2523

Expected number of trees:

0.2523 \times 450 = 113.5

So we expect approximately 113 or 114 trees.

Answer

113 or 114 trees

Final answer

113 or 114 trees

Detailed explanation

Walkthrough

We are told the heights of white pine trees follow a normal distribution with mean 19.8 m and standard deviation 2.4 m. We need the expected number of trees out of 450 that are shorter than 18.2 m.

First, standardise the value 18.2:

z = \frac{18.2 - 19.8}{2.4} = -0.6667

We want $\mathrm{P}(Z < -0.6667)$ . The standard normal table gives $\mathrm{P}(Z < z)$ for positive $z$ , so using the symmetry of the normal curve:

\mathrm{P}(Z < -0.6667) = 1 - \mathrm{P}(Z < 0.6667) = 1 - \Phi(0.6667)

The table gives $\Phi(0.6667) = 0.7477$ , so the probability is $1 - 0.7477 = 0.2523$ .

Finally, multiply by the total number of trees:

0.2523 \times 450 = 113.5

So we expect about 113 or 114 trees.

Key Takeaways

Standardising a normal variable converts it to the standard normal $Z$ .
The symmetry $\Phi(-z) = 1 - \Phi(z)$ lets you use the table for negative $z$ -values.
The expected count of an event is the probability times the number of trials.

Common Mistakes

Using $\Phi(-0.6667)$ directly without converting to $1 - \Phi(0.6667)$ .
Confusing the z-value with the probability.
Forgetting to multiply by 450.
Leaving the answer as a non-integer.

Things to Be Careful About

The mark scheme requires the probability to be 0.2523, which is less than 0.5.
No continuity correction is allowed here.
Give a single integer answer: 113 or 114.

Techniques used

standardise a normal variableuse the symmetry of the normal tablelook up probabilities in the standard normal tablemultiply a probability by a population size

(b)

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.

Find the value of $\sigma$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the given tail probability to find the corresponding $z$ -value from the standard normal table, then substitute it into the standardisation formula and solve for the standard deviation $\sigma$ .

Working

Let $Y$ be the height of a red pine tree, so $Y \sim \mathrm{N}(23.4, \sigma^2)$ .

Given $\mathrm{P}(Y > 25.5) = 0.26$ , standardise:

\mathrm{P}\left(Z > \frac{25.5 - 23.4}{\sigma}\right) = 0.26

The upper-tail probability 0.26 corresponds to $z = 0.643$ , since $\Phi(0.643) = 0.74$ .

So

\frac{2.1}{\sigma} = 0.643

Solving for $\sigma$ :

\sigma = \frac{2.1}{0.643} = 3.27

Answer

$\sigma = 3.27$

Final answer

σ = 3.27

Detailed explanation

Walkthrough

We are told that red pine tree heights are normal with mean 23.4 m and unknown standard deviation $\sigma$ . We know that 26% of red pines are taller than 25.5 m.

Standardise 25.5:

z = \frac{25.5 - 23.4}{\sigma} = \frac{2.1}{\sigma}

The condition $\mathrm{P}(Y > 25.5) = 0.26$ becomes

\mathrm{P}\left(Z > \frac{2.1}{\sigma}\right) = 0.26

Since the upper-tail probability is 0.26, we need $z$ such that $\Phi(z) = 1 - 0.26 = 0.74$ . The normal table gives $z = 0.643$ .

Set up the equation:

\frac{2.1}{\sigma} = 0.643 \quad \Rightarrow \quad \sigma = \frac{2.1}{0.643} = 3.27

Key Takeaways

To find a $z$ -value from a tail probability, use $\Phi(z) = 1 - \text{tail probability}$ .
The standardisation formula can be rearranged to solve for an unknown standard deviation.

Common Mistakes

Using the lower-tail probability 0.26, which would give a negative $z$ -value.
Forgetting to equate the standardised expression to the $z$ -value.
Using $\sigma^2$ or $\sqrt{\sigma}$ in the standardisation.

Things to Be Careful About

The mark scheme accepts $0.643 \leq z \leq 0.6435$ .
The final value of $\sigma$ must be between 3.26 and 3.27.
Do not apply a continuity correction here.

Techniques used

standardise a normal variableinvert the normal probability to find the z-valuesolve a linear equation for the standard deviation

Q5MediumPermutations 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.

(a)

In how many ways can the group of 7 be chosen?

2M Similar

(b)

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.

In how many ways can the group of 5 be chosen?

4M Similar

DifficultyMedium

Worked solution

Approach

Let $V$ , $G$ and $P$ be the numbers of violinists, guitarists and pianists chosen. The restrictions are $V+G+P=5$ , $V\ge2$ , $G\ge1$ and $P\le1$ . List all valid scenarios, count each with combinations, then add the counts.

Working

Possible scenarios are:

$(V,G,P)$	Number of ways
$(4,1,0)$	$\binom{10}{4}\binom{6}{1}\binom{5}{0}=210\times6\times1=1260$
$(3,2,0)$	$\binom{10}{3}\binom{6}{2}\binom{5}{0}=120\times15\times1=1800$
$(2,3,0)$	$\binom{10}{2}\binom{6}{3}\binom{5}{0}=45\times20\times1=900$
$(3,1,1)$	$\binom{10}{3}\binom{6}{1}\binom{5}{1}=120\times6\times5=3600$
$(2,2,1)$	$\binom{10}{2}\binom{6}{2}\binom{5}{1}=45\times15\times5=3375$

These scenarios are mutually exclusive, so add the counts:

1260+1800+900+3600+3375=10935

Answer

The number of ways is $10935$ .

Final answer

10935

Detailed explanation

Walkthrough

We need a group of 5 with at least 2 violinists, at least 1 guitarist and at most 1 pianist. Let $V$ , $G$ and $P$ be the numbers chosen from each group. Then $V+G+P=5$ , with $V\ge2$ , $G\ge1$ and $P\le1$ .

If $P=0$ , then $V+G=5$ . With $V\ge2$ and $G\ge1$ , the possibilities are $(V,G)=(4,1)$ , $(3,2)$ and $(2,3)$ .

If $P=1$ , then $V+G=4$ . With $V\ge2$ and $G\ge1$ , the possibilities are $(V,G)=(3,1)$ and $(2,2)$ .

So there are five valid scenarios. For each scenario, multiply the number of ways to choose the required number of violinists, guitarists and pianists. For example, $(4,1,0)$ gives $\binom{10}{4}\binom{6}{1}\binom{5}{0}=1260$ . Finally, because the scenarios are mutually exclusive, add all five counts to get $10935$ .

Key Takeaways

When a selection problem has several restrictions, split the problem into disjoint cases based on the possible numbers chosen from each category. Count each case using combinations, then add the results.

Common Mistakes

Forgetting the cases where no pianist is chosen, $P=0$ .
Including invalid scenarios such as $(4,0,1)$ because the group must contain at least one guitarist.
Using permutations instead of combinations, even though the order of selection does not matter.
Adding the same scenario twice or missing one of the five valid scenarios.

Things to Be Careful About

Check that every scenario satisfies all three conditions: $V\ge2$ , $G\ge1$ and $P\le1$ . The counts in each scenario must sum to 5. The mark scheme awards a special case B1 for the final value 10935 even if the method marks are not awarded, but it is safer to show all the scenarios clearly.

Techniques used

enumerate valid count scenarioscount each scenario with combinationssum mutually exclusive counts

Q6Medium-EasyRepresentation 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	64	69	76
Kites	32	40	40	45	47	48	52	54	58	59	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.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

A back-to-back stem-and-leaf diagram uses a common stem for the tens digits. The Falcons' times are placed on the left-hand side, with leaves ordered so that values increase as you read from the stem outwards. The Kites' times are placed on the right-hand side in increasing order. A key must state what each leaf on either side of the stem means.

Working

Use the tens digits 3, 4, 5, 6 and 7 as stems. The completed diagram is:

Falcons	Stem	Kites
9 8	3	2
8 6 4 2	4	0 0 5 7 8
9 8 6 2 1 0	5	2 4 8 9 9
9 4	6	0 1 3 5
6	7

Key: 1|5|4 means 51 minutes for Falcons and 54 minutes for Kites.

Answer

A back-to-back stem-and-leaf diagram as shown, with Falcons on the left and Kites on the right.

Final answer

Back-to-back stem-and-leaf diagram with Falcons on the left and Kites on the right; key 1|5|4 means 51 minutes for Falcons and 54 minutes for Kites.

Detailed explanation

Walkthrough

Start by separating each time into a tens digit and a units digit. The stems are 3, 4, 5, 6 and 7. For the Falcons, whose leaves appear on the left of the stem, we mirror the usual order: the units digit nearest the stem is the smallest number in that stem, and as we move away from the stem the units digits increase. For example, Falcons times 42, 44, 46 and 48 are shown as 8 6 4 2 to the left of stem 4. For the Kites, on the right of the stem, we write the units digits in increasing order away from the stem, so 40, 40, 45, 47 and 48 become 0 0 5 7 8.

Finally, include a key. The key tells the reader how to read a stem and a leaf, and must name both teams and state that the units are minutes.

Key Takeaways

A back-to-back stem-and-leaf diagram compares two distributions by sharing a single stem.
The left-hand leaves are written in increasing order as you move outwards from the stem, so they appear reversed when read left-to-right.
A key must always be given, including the units and team names.

Common Mistakes

Writing the Falcons' leaves in increasing order left-to-right instead of from the stem outwards. This gives 2 4 6 8 instead of 8 6 4 2 and is incorrect.
Forgetting to include the stem 7 for the Falcons' time of 76 minutes.
Missing one of the two 40s for the Kites, or one of the two 59s.
Using commas or other punctuation between leaves; the mark scheme requires no punctuation.
Omitting a key or giving a key that does not identify both teams and minutes.

Things to Be Careful About

All 15 values for each team must be included.
The stem-and-leaf diagram should be compact: each stem appears once, and any empty row (e.g. stem 7 on the Kites' side) is left blank.
If a separate diagram were drawn for each team, the marking scheme expects both stems correct, Falcons represented to the left of the stem, and keys including 'mins' and the team names.

Techniques used

sort data by tens and unitsdraw a back-to-back stem-and-leaf diagram with mirrored leavesinclude a key identifying both teams and units

(b)

Find the median and the interquartile range of the times for the Falcons.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Falcons have 15 ordered times, so the median is the 8th value. The lower quartile is the median of the first 7 values and the upper quartile is the median of the last 7 values. The interquartile range is upper quartile minus lower quartile.

Working

The Falcons' times in order are:

38, 39, 42, 44, 46, 48, 50, 51, 52, 56, 58, 59, 64, 69, 76

Median: the 8th value is 51.

Lower quartile: median of 38, 39, 42, 44, 46, 48, 50 is the 4th value, 44.

Upper quartile: median of 52, 56, 58, 59, 64, 69, 76 is the 4th value, 59.

\text{IQR} = 59 - 44 = 15

Answer

Median = 51 minutes; IQR = 15 minutes.

Final answer

Median = 51 minutes; IQR = 15 minutes

Detailed explanation

Walkthrough

First, list the Falcons' times in increasing order. With 15 values, the median is the middle value, which is the 8th value. The lower quartile is the median of the lower half: the first 7 values. For 7 values the middle is the 4th value, so LQ = 44. The upper quartile is the median of the upper half: the last 7 values, so UQ = 59. Then IQR = UQ - LQ = 59 - 44 = 15.

Key Takeaways

With an odd number of data values, the median is the middle value.
The lower and upper quartiles are the medians of the two halves created by the median.
IQR measures the middle 50% of the data and is unaffected by extreme values.

Common Mistakes

Confusing the median position: for 15 values it is the 8th value, not the 7th or 9th.
Including the median in both halves when finding quartiles. For this dataset the quartiles are medians of the 7 values on each side, not of 8 values.
Forgetting to subtract: the IQR is Q3 - Q1, not the other way round.

Things to Be Careful About

The marking scheme accepts $58 \leqslant UQ \leqslant 64$ and $42 \leqslant LQ \leqslant 46$ if quartiles are approximated differently, but the exact quartiles are 44 and 59.
State the units (minutes) with the answer.
The median must be identified explicitly; it is not enough to write 51 without saying it is the median.

Techniques used

locate the median as the middle value of an ordered setfind the lower and upper quartiles from the two halvessubtract quartiles to obtain the interquartile range

(c)

Let $x$ and $y$ denote the times, in minutes, of a runner from the Falcons and a runner from the Kites respectively.

It is given that

\sum x = 792, \quad \sum x^2 = 43\,504, \quad \sum y = 783, \quad \sum y^2 = 42\,223.

Find the mean and the standard deviation of the times taken by all 30 runners from the two teams.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since all 30 runners are pooled, add the two totals and the two sums of squares. Then use the mean formula and the variance formula.

Working

\sum x + \sum y = 792 + 783 = 1575

\sum x^2 + \sum y^2 = 43504 + 42223 = 85727

Combined mean:

\frac{1575}{30} = 52.5

Combined variance:

\sigma^2 = \frac{85727}{30} - \left(\frac{1575}{30}\right)^2 = 101.3167

So the standard deviation is:

\sigma = \sqrt{101.3167} = 10.1 \text{ minutes}

Answer

Mean = 52.5 minutes; standard deviation = 10.1 minutes.

Final answer

Mean = 52.5 minutes; standard deviation = 10.1 minutes

Detailed explanation

Walkthrough

We are told the sums for the Falcons ( $x$ ) and for the Kites ( $y$
). Since we want the mean of all 30 runners together, combine the totals:

\sum x + \sum y = 1575

and divide by the total number of runners, 30, giving 52.5.

For the variance we need the sum of squares of all 30 values. Add the two given sums of squares:

43504 + 42223 = 85727

The variance of the 30 pooled values is the mean of the squares minus the square of the mean:

\sigma^2 = \frac{85727}{30} - \left(\frac{1575}{30}\right)^2 = 101.3167

Finally take the square root to obtain the standard deviation, 10.1 minutes.

Key Takeaways

When combining two data sets, totals and sums of squares are added, but the divisor is the combined sample size.n- The variance formula $\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2$ works for the pooled data exactly as for a single set.n- The final answer is the standard deviation, so a square root must be taken.

Common Mistakes

Forgetting to combine $\sum x^2$ and $\sum y^2$ before using the variance formula.
Even if the mean is correct, using a variance formula based on separate means instead of combined sums leads to the wrong answer.
Quoting the variance (101.3) as the standard deviation without taking the square root.
Leaving the answer as an unsimplified fraction such as $\frac{1575}{30}$ for the mean; the mark scheme requires $52.5$ , $52\frac{1}{2}$ or $\frac{105}{2}$ .

Things to Be Careful About

Use $n = 30$ , not 15, because there are two teams of 15 runners.
The mark scheme accepts an unsimplified variance formula for M1, but the final standard deviation must be identified as sd, $s$ , std d or $\sigma$ .
The answer 10.1 is an approximate value (AWRT); 10.065... should be rounded to 1 decimal place.

Techniques used

combine sums of values and sums of squarescalculate the combined meanapply the variance formula \sigma^2 = sum x^2/n - mean^2

Q7Medium-HardDiscrete Random VariablesThe Normal Distribution

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 $P(X = 5)$ .

1M Similar

DifficultyEasy

Worked solution

Approach

The number of attempts $X$ until the first goal follows a geometric distribution with success probability $p=0.4$ and failure probability $q=1-p=0.6$ . The event $X=5$ means the first four attempts are misses and the fifth attempt is a goal.

Working

For a geometric distribution, $\mathrm{P}(X=k)=q^{k-1}p$ . Therefore

\mathrm{P}(X=5)=0.6^{4}\times 0.4

Evaluate:

0.6^{4}=0.1296

0.1296\times 0.4=0.05184

This equals

\frac{162}{3125}=0.05184

Answer

\mathrm{P}(X=5)=0.05184=\frac{162}{3125}

Final answer

0.05184 (162/3125)

Detailed explanation

Walkthrough

We are told that each attempt is independent and has probability $0.4$ of success. The random variable $X$ counts the number of attempts needed until the first goal, so $X$ has a geometric distribution. For $X=5$ , the first four attempts must all fail and the fifth must succeed. Each failure has probability $0.6$ , so the probability is

0.6 \times 0.6 \times 0.6 \times 0.6 \times 0.4 = 0.6^4 \times 0.4

This is exactly the geometric formula $\mathrm{P}(X=k)=q^{k-1}p$ with $k=5$ , $p=0.4$ and $q=0.6$ .

Key Takeaways

The geometric distribution models the number of trials until the first success. The formula is $q^{k-1}p$ , where $q=1-p$ . The exponent on $q$ is one less than the attempt number because the first $k-1$ attempts must fail.

Common Mistakes

Using $0.4^5$ instead of $0.6^4 \times 0.4$ .
Forgetting that the first four attempts must be failures.
Confusing the geometric distribution with the binomial distribution.

Things to Be Careful About

The success probability is $0.4$ and the failure probability is $0.6$ . The exact fraction is $\frac{162}{3125}$ , which equals $0.05184$ .

Techniques used

model the number of attempts until first success as a geometric distributionapply the geometric probability formula

(b)

Find $P(3 \leq X \leq 7)$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

For a geometric distribution, $\mathrm{P}(X\le k)=1-q^k$ . We can find $\mathrm{P}(3\le X\le 7)$ as

\mathrm{P}(3\le X\le 7)=\mathrm{P}(X\le 7)-\mathrm{P}(X\le 2)

Working

\mathrm{P}(X\le 7)=1-0.6^7

\mathrm{P}(X\le 2)=1-0.6^2

Subtract:

\mathrm{P}(3\le X\le 7)=(1-0.6^7)-(1-0.6^2)

=0.6^2-0.6^7

Evaluate:

0.6^2=0.36,\qquad 0.6^7=0.0279936

0.36-0.0279936=0.3320064

So

\frac{25938}{78125}=0.3320064

Answer

\mathrm{P}(3\le X\le 7)=0.3320064=\frac{25938}{78125}

Final answer

0.3320064 (25938/78125)

Detailed explanation

Walkthrough

For a geometric distribution, the probability that the first success occurs on or before attempt $k$ is $1-q^k$ . This is because the only way to take more than $k$ attempts is to fail all of the first $k$ attempts, which has probability $q^k$ . Therefore

\mathrm{P}(X\le 7)=1-0.6^7

and

\mathrm{P}(X\le 2)=1-0.6^2

Subtracting these gives the probability that $X$ lies between 3 and 7 inclusive:

\mathrm{P}(3\le X\le 7)=(1-0.6^7)-(1-0.6^2)=0.6^2-0.6^7

Key Takeaways

Cumulative geometric probabilities are useful for interval questions. The identity $\mathrm{P}(X\le k)=1-q^k$ avoids summing many individual terms.

Common Mistakes

Subtracting $\mathrm{P}(X\le 3)$ instead of $\mathrm{P}(X\le 2)$ .
Forgetting that the interval includes both endpoints.
Using $\mathrm{P}(X=7)-\mathrm{P}(X=3)$ instead of cumulative probabilities.

Things to Be Careful About

The interval $3\le X\le 7$ includes $X=3,4,5,6,7$ . The exact fraction is $\frac{25938}{78125}$ .

Techniques used

use the cumulative geometric distributionsubtract cumulative probabilitiesevaluate powers of the failure probability

(c)

Find the probability that Leno scores his second goal on or before his 5th attempt.

3M Similar

DifficultyMedium

Worked solution

Approach

Scoring the second goal on or before the 5th attempt means that, in the first 5 attempts, Leno scores at least 2 goals. Let $Y$ be the number of goals scored in 5 attempts. Then $Y\sim \mathrm{B}(5,0.4)$ .

Working

We need

\mathrm{P}(Y\ge 2)=\sum_{k=2}^{5}\binom{5}{k}(0.4)^k(0.6)^{5-k}

Term by term:

\binom{5}{2}(0.4)^2(0.6)^3=10(0.16)(0.216)=0.3456

\binom{5}{3}(0.4)^3(0.6)^2=10(0.064)(0.36)=0.2304

\binom{5}{4}(0.4)^4(0.6)=5(0.0256)(0.6)=0.0768

\binom{5}{5}(0.4)^5=0.01024

Add the probabilities:

0.3456+0.2304+0.0768+0.01024=0.66304

Answer

\frac{2072}{3125}=0.66304

Final answer

0.66304 (2072/3125)

Detailed explanation

Walkthrough

"Second goal on or before the 5th attempt" means that by the time Leno has taken 5 attempts, he has scored at least 2 goals. Let $Y$ be the number of goals in 5 attempts. Since each attempt is independent with success probability $0.4$ , $Y$ follows a binomial distribution $\mathrm{B}(5,0.4)$ . We need

\mathrm{P}(Y\ge 2)=\mathrm{P}(Y=2)+\mathrm{P}(Y=3)+\mathrm{P}(Y=4)+\mathrm{P}(Y=5)

Each term uses $\binom{5}{k}(0.4)^k(0.6)^{5-k}$ . Adding the four terms gives $0.66304$ .

Key Takeaways

A binomial distribution counts the number of successes in a fixed number of trials. The event "at least two goals" is the sum of the binomial probabilities for $k=2,3,4,5$ . An alternative is $1-\mathrm{P}(Y=0)-\mathrm{P}(Y=1)$ .

Common Mistakes

Including $k=0$ and $k=1$ in the sum.
Using $n=4$ instead of $n=5$ .
Using the geometric distribution formula for the second success.

Things to Be Careful About

"On or before the 5th attempt" includes the 2nd, 3rd, 4th and 5th attempts. The exact fraction is $\frac{2072}{3125}$ .

Techniques used

recognise the event as at least two successes in five trialsapply the binomial probability formulasum binomial probabilities

(d)

Leno has 75 attempts to score a goal.

Use a suitable approximation to find the probability that Leno scores more than 28 goals but fewer than 35 goals.

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

The number of goals in 75 attempts is $X\sim \mathrm{B}(75,0.4)$ . Since $np=30$ and $n(1-p)=45$ are both greater than 5, the normal approximation is suitable. We approximate $X$ by a normal distribution with mean $\mu=30$ and variance $\sigma^2=18$ . Because $X$ is discrete, we apply a continuity correction to $28<X<35$ .

Working

Mean and variance:

\mu=75\times 0.4=30

\sigma^2=75\times 0.4\times 0.6=18

\sigma=\sqrt{18}\approx 4.2426

With the continuity correction, $28<X<35$ becomes

28.5<Y<34.5

Standardise:

z_1=\frac{28.5-30}{\sqrt{18}}=\frac{-1.5}{4.2426}\approx -0.3536

z_2=\frac{34.5-30}{\sqrt{18}}=\frac{4.5}{4.2426}\approx 1.0607

Therefore

\mathrm{P}(28<X<35)\approx \Phi(1.0607)-\Phi(-0.3536)

=\Phi(1.0607)+\Phi(0.3536)-1

=0.8556+0.6383-1

=0.4939

Answer

\mathrm{P}(28<X<35)\approx 0.494

Final answer

0.494

Detailed explanation

Walkthrough

We have $X\sim \mathrm{B}(75,0.4)$ . The mean is

\mu=np=75\times 0.4=30

and the variance is

\sigma^2=np(1-p)=75\times 0.4\times 0.6=18

so the standard deviation is $\sqrt{18}\approx 4.2426$ . Since both $np$ and $n(1-p)$ are greater than 5, the normal approximation is appropriate. The event $28<X<35$ means $X=29,30,31,32,33,34$ . The continuity correction changes this to the continuous interval $28.5<Y<34.5$ . Standardising gives $z_1=-0.3536$ and $z_2=1.0607$ . Using the standard normal table,

\Phi(1.0607)=0.8556,\qquad \Phi(0.3536)=0.6383

so

\mathrm{P}(28<X<35)\approx 0.8556-(1-0.6383)=0.4939\approx 0.494

Key Takeaways

The normal approximation to the binomial uses $\mu=np$ and $\sigma^2=np(1-p)$ . A continuity correction is essential because the binomial is discrete and the normal distribution is continuous.

Common Mistakes

Not applying a continuity correction.
Using $28$ and $35$ instead of $28.5$ and $34.5$ .
Using the variance $18$ in the denominator instead of the standard deviation $\sqrt{18}$ .
Reading $\Phi(-0.3536)$ incorrectly and subtracting a negative number.

Things to Be Careful About

The variance is $18$ , but the standard deviation is $\sqrt{18}\approx 4.2426$ . The final answer must be a probability between 0 and 1. The mark scheme accepts answers approximately equal to $0.494$ .

Techniques used

identify the binomial distributioncompute the mean and variance of the binomial distributionapply the continuity correctionstandardise to the normal distributioncombine standard normal probabilities

Mathematics 9709/52 — October/November 2024

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