Mathematics 9709/53 — October/November 2025

Walkthrough

We are told the probability that a randomly chosen student travels by car is $0.4$. The event that none of the 3 selected students travel by car means each selected student is not a car traveller. For one student, $P(\text{not by car}) = 1 - 0.4 = 0.6$. Because travel choices are independent, the probability that all three are not car travellers is the product of three separate probabilities: $0.6 \times 0.6 \times 0.6 = 0.216$.

Key Takeaways

• The complement rule: $P(A') = 1 - P(A)$.
• For independent events, probabilities multiply: $P(A \text{ and } B) = P(A)P(B)$.

Common Mistakes

• Using $0.4^3$ instead of $0.6^3$. The question asks for none by car, so use the complement probability.
• Forgetting that 'none' means every one of the three students must fail to travel by car.

Things to Be Careful About

• 'Not by car' includes travelling by bus or on foot, so its probability is $0.6$, not $0.25$.
• The independence condition is essential; without it we could not multiply the three probabilities.

Walkthrough

First combine the probabilities for car and bus. Since a student cannot travel by both car and bus, these events are mutually exclusive, so $P(\text{car or bus}) = 0.4 + 0.35 = 0.75$. For 11 independent students, the number $X$ who travel by car or bus follows a binomial distribution $X \sim B(11, 0.75)$.

'Fewer than 9' means $X = 0, 1, 2, \dots, 8$. Instead of adding nine probabilities, use the complement: $P(X < 9) = 1 - P(X \geq 9)$, and $X \geq 9$ means $X = 9, 10$ or $11$.

For a binomial distribution, $P(X = x) = \binom{11}{x}(0.75)^x(0.25)^{11-x}$. Substitute $x = 9, 10, 11$ and add:

So

Key Takeaways

• A fixed number of independent trials with two outcomes leads to the binomial distribution.
• Use the complement when a tail probability has fewer terms than its complement.
• The binomial coefficient $\binom{n}{x}$ counts the number of ways of choosing the $x$ successes.

Common Mistakes

• Using $p = 0.4$ or $p = 0.35$ instead of $p = 0.75$. The event is travelling by car or bus.
• Treating 'fewer than 9' as $X \leq 9$; it actually means $X \leq 8$.
• Forgetting the $X = 11$ term when using the complement.
• Giving only a calculator answer without showing the binomial expression, since the mark scheme requires method.

Things to Be Careful About

• Car and bus are mutually exclusive, so their probabilities add to give $0.75$.
• The probability of failure is $0.25$, representing travelling on foot.
• Round only at the end; the final answer should be in the range $0.5445 < p \leq 0.545$.
• The complement of $X<9$ is $X \geq 9$, not $X > 9$ or $X = 9$ only.

Walkthrough

The word DAFFODILS has 9 letters, with two Ds and two Fs. Part (a) fixes the two Ds at the ends. Once the Ds are fixed, only the middle 7 positions remain. The letters available for those positions are $A, F, F, O, I, L, S$. Since the two Fs are identical, the number of arrangements is $\frac{7!}{2!} = 2520$.

The condition "the two Fs are not next to each other" is easier to handle by counting the unwanted cases and subtracting. If the two Fs are together, treat $FF$ as one block. Then the items to arrange are $FF, A, O, I, L, S$, which is 6 distinct items, so there are $6! = 720$ arrangements. Subtracting gives $2520 - 720 = 1800$.

A useful alternative view is to arrange the five distinct letters $A, O, I, L, S$ in $5!$ ways, then place the two Fs into two of the 6 gaps around or between them, choosing $\binom{6}{2} = 15$ ways. This gives $5! \times 15 = 1800$, the same result.

Key Takeaways

• Repeated letters reduce the number of arrangements by dividing by the factorial of each repeated letter.
• Fixing certain letters in specified positions simplifies the counting problem.
• "Not next to each other" can often be counted by subtracting the adjacent cases from the total.
• The gap method is a powerful alternative for placing identical items so that they are separated.

Common Mistakes

• Forgetting to divide by $2!$ for the repeated Fs.
• Including arrangements where the Ds are not at the ends.
• Treating the two Fs as distinct letters.
• In the complement method, forgetting that the total count must also have Ds fixed at the ends.
• Giving an unsupported final answer; the mark scheme requires the method to be shown.

Things to Be Careful About

• The two Fs are identical, so divide by $2!$ when arranging them among the middle letters.
• The two Ds are fixed at the ends, so they are not arranged.
• When subtracting, both counts must be for arrangements with Ds at the ends.
• If using the gap method, choose two different gaps, so use $\binom{6}{2}$, not $6^2$.

Walkthrough

The total number of arrangements of DAFFODILS is $\frac{9!}{2!2!} = 90720$, because the two Ds are identical and the two Fs are identical.

Now count the favourable arrangements. If there are exactly 4 letters between the two Ds, then the positions of the Ds differ by 5. In a 9-letter word, the possible pairs of positions are $(1,6)$, $(2,7)$, $(3,8)$ and $(4,9)$, so there are 4 choices for where the Ds go.

For each choice of D positions, the remaining 7 positions are filled with $A, F, F, O, I, L, S$. Since the two Fs are identical, this can be done in $\frac{7!}{2!} = 2520$ ways.

Therefore the number of favourable arrangements is $4 \times 2520 = 10080$. The probability is

This matches the mark scheme's accepted answer $\frac{1}{9}$ or $0.111$.

Key Takeaways

• Probabilities can be found by enumerating favourable outcomes and dividing by the total number of outcomes.
• When letters are repeated, divide by the factorial of each repeated letter.
• A condition such as "exactly 4 letters between" translates into a fixed position difference, here 5.
• It is important to be consistent about whether identical letters are treated as distinct or identical.

Common Mistakes

• Forgetting to divide the total by $2!2!$ for the repeated Ds and Fs.
• Using the wrong number of D-position pairs: exactly 4 letters between means the positions differ by 5, giving 4 pairs, not 5.
• Treating the Ds as distinct in one part of the calculation and identical in another.
• Not simplifying the probability fraction to $\frac{1}{9}$.

Things to Be Careful About

• If you treat the Ds as distinct, the total is $9!$ and the favourable count is $7! \times 4 \times 2!$; if you treat them as identical, the total is $\frac{9!}{2!2!}$ and the favourable count is $4 \times \frac{7!}{2!}$. Both give $\frac{1}{9}$.
• The two Fs are identical, so divide by $2!$ when arranging the remaining letters.
• "Exactly 4 letters between" means the Ds are 5 positions apart, not 4 positions apart.
• The final probability should be simplified to $\frac{1}{9}$ or given as $0.111$.

Walkthrough

There are $4$ outcomes for the red spinner and $4$ for the blue spinner, so $16$ equally likely ordered pairs. We define $X$ as the higher score, except when the scores are equal, in which case $X=0$. This special rule matters: the equal pairs $(1,1)$, $(2,2)$, $(3,3)$ all give $X=0$, not $X=1,2,3$. For every other pair, $X$ is the larger of the two numbers. Count how many pairs give each value of $X$: for example, $X=4$ occurs only when the red spinner shows $4$, because the blue spinner can never show $4$; this gives $4$ outcomes. Dividing each count by $16$ gives the probability distribution.

Key Takeaways

A probability distribution table lists every possible value of a discrete random variable with its probability. When outcomes are equally likely, probabilities can be found by counting favourable outcomes. Special definitions in the question (such as equal scores giving $X=0$) must be applied before counting.

Common Mistakes

• Forgetting that equal scores give $X=0$, so $(1,1)$, $(2,2)$, $(3,3)$ are not counted as $X=1,2,3$.
• Counting ordered pairs incorrectly, e.g. treating $(2,1)$ and $(1,2)$ as the same outcome when they are different spins.
• Not checking that the probabilities sum to $1$.

Things to Be Careful About

The two spinners have different sets of numbers: red has $1$ to $4$ and blue has $0$ to $3$. This asymmetry means $X=1$ has only one outcome, while $X=4$ has four. Always distinguish the red and blue scores when counting.

Walkthrough

Use the distribution from part (a). The expectation is the weighted average of the values of $X$, using the probabilities as weights: $\text{E}(X)=\sum x\,\text{P}(X=x)$. We compute this first. Then for variance we need $\text{E}(X^2)$, the weighted average of the squares of the values. Finally, $\text{Var}(X)=\text{E}(X^2)-[\text{E}(X)]^2$. The fractions must be converted to a common denominator when subtracting.

Key Takeaways

Variance measures spread around the mean. For a discrete random variable, $\text{Var}(X)=\text{E}(X^2)-[\text{E}(X)]^2$ is often the quickest formula. It is important to square each value before multiplying by its probability when finding $\text{E}(X^2)$.

Common Mistakes

• Forgetting to square the values in $\text{E}(X^2)$ (using $x$ instead of $x^2$).
• Subtracting $[\text{E}(X)]^2$ from $\text{E}(X)$ but using the wrong sign or forgetting the square.
• Using the probabilities from a wrong table; the mark scheme allows follow-through from their table, but the final answer must be consistent.

Things to Be Careful About

Keep fractions exact until the final step. The answer can be written as $\frac{127}{64}$, $1\frac{63}{64}$, or $1.98$. If using a calculator, round only at the end.

Walkthrough

Each spin is independent and the probability of landing on $4$ is $\frac{1}{4}$, so the probability of not landing on $4$ is $\frac{3}{4}$. 'Before the 8th spin' means the first $4$ appears on spin $1,2,\ldots,7$. The complement is that no $4$ appears in the first $7$ spins, which has probability $\left(\frac{3}{4}\right)^7$. Therefore the required probability is $1-\left(\frac{3}{4}\right)^7$. This is the standard geometric distribution idea: the first success occurs within the first $n$ trials.

Key Takeaways

For repeated independent trials with the same probability of success, $\text{P}(\text{first success by trial } n)=1-(1-p)^n$. This is often easier than adding the individual geometric probabilities. The geometric distribution models the number of trials until the first success.

Common Mistakes

• Misreading 'before the 8th spin' as including the 8th spin. It means the first $7$ spins only.
• Using $\left(\frac{1}{4}\right)^7$ instead of $\left(\frac{3}{4}\right)^7$ for the complement.
• Adding only a few terms of the geometric series instead of using the complement.

Things to Be Careful About

The phrase 'before the 8th spin' is a boundary condition. Here it gives $n=7$. If the question said 'by the 8th spin', $n$ would be $8$. The final answer should be given to 3 significant figures: $0.867$.

Walkthrough

We build a probability tree diagram to model the sequential random choices.

Step 1: First choice (Counter from Box A)
Box A contains 3 white and 2 yellow counters. The probability of choosing white is $\frac{3}{5}$ and yellow is $\frac{2}{5}$. These form the first tier of branches.

Step 2: Second choice (First marble)
The box used depends on the counter:

• If white, we use Box W (4 red, 3 blue = 7 marbles). Probability of red is $\frac{4}{7}$, blue is $\frac{3}{7}$.
• If yellow, we use Box Y (5 red, 3 blue = 8 marbles). Probability of red is $\frac{5}{8}$, blue is $\frac{3}{8}$.

Step 3: Third choice (Second marble, without replacement)
Since marbles are not replaced, the total count and composition change:

• Path W-R: Box W now has 3 red, 3 blue (6 total). Probabilities are $\frac{3}{6}$ for red, $\frac{3}{6}$ for blue.
• Path W-B: Box W now has 4 red, 2 blue (6 total). Probabilities are $\frac{4}{6}$ for red, $\frac{2}{6}$ for blue.
• Path Y-R: Box Y now has 4 red, 3 blue (7 total). Probabilities are $\frac{4}{7}$ for red, $\frac{3}{7}$ for blue.
• Path Y-B: Box Y now has 5 red, 2 blue (7 total). Probabilities are $\frac{5}{7}$ for red, $\frac{2}{7}$ for blue.

Key Takeaways

• Tree diagrams are ideal for sequential events where later probabilities depend on earlier outcomes.
• Sampling without replacement changes the denominator and numerator for subsequent branches.
• Always label branches with both the outcome and the conditional probability.

Common Mistakes

• Forgetting to update the number of marbles for the third tier (using 7 or 8 instead of 6 or 7).
• Not labelling the outcomes (W, Y, R, B) clearly at the end of branches.
• Incorrectly calculating conditional probabilities (e.g., using the original box composition for the second draw).

Things to Be Careful About

• Ensure the tree has exactly three tiers as described in the question.
• Verify that probabilities on branches from the same node sum to 1 (e.g., $\frac{3}{6} + \frac{3}{6} = 1$).
• The mark scheme awards marks for the structure, first/second tier probabilities, and third tier probabilities separately.

Walkthrough

We need the probability of getting exactly one red and one blue marble. Since the order of drawing matters for the tree diagram paths, we must consider all sequences that yield one of each.

Step 1: Identify valid paths
From Box W (chosen if counter is white), we can get Red then Blue (WRB) or Blue then Red (WBR).
From Box Y (chosen if counter is yellow), we can get Red then Blue (YRB) or Blue then Red (YBR).

Step 2: Calculate path probabilities
Using the multiplication rule (multiply probabilities along the branches):

• $P(\text{WRB}) = \frac{3}{5} \times \frac{4}{7} \times \frac{3}{6} = \frac{6}{35}$
• $P(\text{WBR}) = \frac{3}{5} \times \frac{3}{7} \times \frac{4}{6} = \frac{6}{35}$
• $P(\text{YRB}) = \frac{2}{5} \times \frac{5}{8} \times \frac{3}{7} = \frac{3}{28}$
• $P(\text{YBR}) = \frac{2}{5} \times \frac{3}{8} \times \frac{5}{7} = \frac{3}{28}$

Step 3: Sum the probabilities
Since these paths are mutually exclusive, we add them:

Key Takeaways

• When asking for a combination of outcomes (e.g., one red and one blue), you must consider all possible orders (sequences) that produce that combination.
• The multiplication rule is applied along each branch of the tree.
• Simplifying fractions early can make addition easier, but finding a common denominator at the end is also valid.

Common Mistakes

• Missing one of the four paths (e.g., only calculating paths from Box W and ignoring Box Y).
• Arithmetic errors when adding fractions with different denominators (35 and 28).
• Not simplifying fractions correctly before adding.

Things to Be Careful About

• Ensure you include both orders (Red-Blue and Blue-Red) for each box scenario.
• The final answer should be exact (fraction form $\frac{39}{70}$) unless a decimal is requested.

Walkthrough

This part asks for a conditional probability: $P(\text{White} \mid \text{1 red and 1 blue})$. The formula for conditional probability is $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.

Step 1: Identify A and B

• Event A: A white counter is chosen from Box A.
• Event B: One red marble and one blue marble are obtained.

Step 2: Calculate P(A ∩ B)
This is the probability that a white counter is chosen AND one red and one blue marble are obtained. Looking at our work in part (b), these are the paths WRB and WBR:

Step 3: Use P(B) from part (b)

Step 4: Apply the formula

Key Takeaways

• Conditional probability restricts the sample space to the given condition (Event B).
• The numerator is the probability of the intersection of the two events.
• Results from previous parts can be directly used in conditional probability calculations.

Common Mistakes

• Using the wrong denominator (e.g., using $\frac{1}{2}$ or some other value instead of the result from part (b)).
• Forgetting that the numerator must only include paths where the counter was white (excluding YRB and YBR).
• Arithmetic errors when dividing fractions.

Things to Be Careful About

• The question asks for the probability given that one red and one blue are obtained, so the denominator MUST be the total probability from part (b), not just the white paths.
• Ensure fractions are simplified correctly at the end ($\frac{24}{39} \to \frac{8}{13}$).

Walkthrough

First, we identify the stems (tens digits) which are 4, 5, 6, and 7. For each stem, we list the leaves (units digits).

For the Linnets, the data is: 48, 51, 54, 57, 59, 60, 64, 64, 65, 68, 70.

• For stem 4, the leaf is 8.
• For stem 5, the leaves are 1, 4, 7, 9. Since they go on the left side of the stem, we write them in descending order from right to left: 9, 7, 4, 1.
• For stem 6, the leaves are 0, 4, 4, 5, 8. Written right to left: 8, 5, 4, 4, 0.
• For stem 7, the leaf is 0.

For the Puffins, the data is: 45, 49, 51, 55, 55, 58, 59, 62, 64, 64, 74.

• For stem 4, the leaves are 5, 9. Written left to right: 5, 9.
• For stem 5, the leaves are 1, 5, 5, 8, 9.
• For stem 6, the leaves are 2, 4, 4.
• For stem 7, the leaf is 4.

Finally, we add a key. A value like $4|5|5$ uses the left leaf 4, stem 5, and right leaf 5 to represent 54 minutes for the Linnets and 55 minutes for the Puffins.

Key Takeaways

• In a back-to-back stem-and-leaf diagram, leaves on the left are ordered in descending order from right to left, while leaves on the right are ordered in ascending order from left to right.
• A key is essential to explain the notation, including the unit of measurement.

Common Mistakes

• Forgetting to order the leaves correctly (left side should be descending right-to-left).
• Adding commas or spaces between leaves on the same side.
• Omitting the key or not including both team names and units in the key.

Things to Be Careful About

• Ensure the stems are aligned vertically.
• Do not split the stems unless specifically asked to do so; a single stem for each tens digit is sufficient here.
• The key must clearly show how to read a value from both sides.

Walkthrough

We are given 11 ordered times for the Linnets. To find the interquartile range, we first locate the lower quartile (LQ) and upper quartile (UQ).

For $n = 11$ data points:

• The LQ is at position $\frac{11+1}{4} = 3$. The 3rd value in the ordered list is 54.
• The UQ is at position $\frac{3(11+1)}{4} = 9$. The 9th value in the ordered list is 65.

The interquartile range is the difference between these two values:
$\text{IQR} = 65 - 54 = 11$

Key Takeaways

• For $n$ data points, the quartiles can be found using positions $\frac{n+1}{4}$ and $\frac{3(n+1)}{4}$.
• The IQR is a measure of spread that is resistant to outliers, calculated as UQ - LQ.

Common Mistakes

• Using the wrong positions for the quartiles (e.g., using $n/4$ instead of $(n+1)/4$).
• Forgetting to subtract LQ from UQ to get the IQR.

Things to Be Careful About

• Ensure the data is in ascending order before finding the quartiles.
• The mark scheme allows for condoning values like 54 and 65 if they are clearly identified in the stem-and-leaf diagram from part (a).

Walkthrough

First, we determine the five-number summary (minimum, LQ, median, UQ, maximum) for both teams.

For the Linnets (11 values):

• Min = 48
• LQ = 3rd value = 54
• Median = 6th value = 60
• UQ = 9th value = 65
• Max = 70

For the Puffins (11 values):

• Min = 45
• LQ = 3rd value = 51
• Median = 6th value = 58
• UQ = 9th value = 64
• Max = 74

Next, we set up a horizontal axis with a linear scale. The data ranges from 45 to 74, so a scale from 45 to 75 is appropriate. The mark scheme allows a scale such as $2\text{ cm} = 10\text{ minutes}$, with at least 3 equally spaced values marked.

We draw the Linnets plot: a whisker from 48 to the box at 54, the box extends to 65 with a median line at 60, and a whisker extends to 70. We label this "Linnets".

We draw the Puffins plot below it on the same scale: a whisker from 45 to the box at 51, the box extends to 64 with a median line at 58, and a whisker extends to 74. We label this "Puffins".

Important rules for box-and-whisker plots:

• Whiskers should not extend to the top/bottom of the box or pass through it.
• The "daylight rule" applies: there should be a gap (daylight) between the end of the whisker and the edge of the box.

Key Takeaways

• A box-and-whisker plot displays the five-number summary and provides a visual representation of the spread and central tendency of data.
• When drawing multiple plots, use the same linear scale for fair comparison.

Common Mistakes

• Using a non-linear or broken scale.
• Forgetting to label the plots or the axis.
• Violating the daylight rule (whiskers touching or crossing the box).
• Using different scales for the two plots.

Things to Be Careful About

• Ensure both plots are drawn on the same linear scale.
• Label the axis with the variable and its unit (e.g., "time taken / min").
• Condone missing vertical lines on the max/min values at the ends of the whiskers.

Walkthrough

We compare the statistics calculated in the previous parts.

For spread:

• Linnets IQR = 11, Puffins IQR = 13. Since 13 > 11, the Puffins have a larger spread in the middle 50% of their data.
• Linnets range = 22, Puffins range = 29. Since 29 > 22, the Puffins have a larger overall spread.

For central tendency:

• Linnets median = 60, Puffins median = 58. Since 60 > 58, the Linnets generally took longer (slower times).

A valid comparison must refer to spread or central tendency. For example: "The times taken by the Puffins are more spread out than those of the Linnets" or "The Linnets have slower times on average than the Puffins."

Key Takeaways

• Comparisons should be based on measurable statistics like IQR, range, or median.
• A valid comparison must explain what the difference means (e.g., more spread out, slower, more consistent).

Common Mistakes

• Simply stating "the median is different" without explaining the implication.
• Comparing only one value without context (e.g., "Puffins have a higher range" is not a complete comparison statement).

Things to Be Careful About

• The mark scheme explicitly states that a comment comparing only the median or range is not acceptable; it must be a comparison about spread or central tendency (e.g., "more spread out", "less consistent", "slower").
• Extra comments are ignored if they do not contradict the correct statement.

Walkthrough

We are told the walking times are normally distributed with mean 14.8 minutes and standard deviation 1.5 minutes. We want the probability that a randomly chosen time is greater than 15.6 minutes.

First standardise 15.6:

This $z$-value says that 15.6 is 0.5333 standard deviations above the mean. The normal table gives the probability that $Z$ is less than this value, $\Phi(0.5333) = 0.7029$. Since we want the probability that $X$ is greater than 15.6, we take the complement:

So the probability is about 0.297.

Key Takeaways

Standardising a normal variable with $Z = \frac{X - \mu}{\sigma}$ lets us use the standard normal table. For an upper-tail probability, subtract the tabled value from 1.

Common Mistakes

• Forgetting to subtract the tabled probability from 1, since the table gives $P(Z < z)$.
• Using the wrong tail, e.g. finding $P(X < 15.6)$ instead of $P(X > 15.6)$.
• Using $\sigma^2$ or $\sqrt{\sigma}$ in the standardisation formula instead of $\sigma = 1.5$.

Things to Be Careful About

• Use $\sigma = 1.5$, not $\sigma^2 = 2.25$.
• Because 15.6 is above the mean, the answer should be less than 0.5.
• No continuity correction is needed when working directly with the normal distribution.

Walkthrough

We know $P(X > t) = 0.9$. Standardising gives $P\left(Z > \frac{t - 14.8}{1.5}\right) = 0.9$.

Because the probability above $t$ is large (0.9), $t$ is below the mean, so the $z$-value is negative. From the normal table, the value with 0.9 above it is $-1.282$. Set the standardised expression equal to $-1.282$ and solve:

Rounded to one decimal place, $t = 12.9$ minutes.

Key Takeaways

Given a probability, use the inverse normal table to find the corresponding $z$-value. For an upper-tail probability greater than 0.5, the $z$-value is negative.

Common Mistakes

• Using $z = +1.282$ instead of $-1.282$, which would give a time above the mean.
• Using the probability 0.9 directly as a $z$-value.
• Using $\sigma^2$ or $\sqrt{\sigma}$ in the standardisation.

Things to Be Careful About

• The critical value is $-1.282$ (or $1.282$ with the sign handled correctly).
• No continuity correction is used here.
• Give the final answer in minutes, AWRT 12.9.

Walkthrough

First find the interval: one minute below the mean is $14.8 - 1 = 13.8$, and one minute above the mean is $14.8 + 1 = 15.8$.

Standardise both boundaries:

The two $z$-values are opposites, so the interval is symmetric about the mean. The probability is

Using $\Phi(0.6667) = 0.7477$:

There are $5 \times 45 = 225$ walking days in a year. The expected number of days is

Since we are counting days, the expected number is 111 days.

Key Takeaways

For an interval symmetric about the mean, use $P(-a < Z < a) = 2\Phi(a) - 1$. To find an expected count, multiply the probability by the total number of trials.

Common Mistakes

• Using 14.3 and 15.3 instead of 13.8 and 15.8; "within one minute of the mean" means $\mu \pm 1$.
• Forgetting to multiply the probability by 225.
• Rounding the probability too early; keep at least four decimal places before multiplying.
• Giving a non-integer like 111.465 as the final number of days.

Things to Be Careful About

• The boundaries are $\mu \pm 1$ minute, not $\mu \pm 1$ standard deviation.
• No continuity correction is needed.
• The expected value is 111.465, so the sensible integer answer is 111 days.