Mathematics 9709/55 — October/November 2025

Walkthrough

This is a geometric distribution problem: we repeat independent trials until the first success. Here a success is rolling a 6, so $p = \frac{1}{6}$, and a failure is rolling any other number, so $1-p = \frac{5}{6}$. For $X = 7$, the first six throws must all be failures and the seventh must be a success. Because the throws are independent, multiply the probabilities: $\left(\frac{5}{6}\right)^6\left(\frac{1}{6}\right)$.

Key Takeaways

The number of trials until the first success in independent repeated trials follows a geometric distribution. For a success probability $p$, $P(X = k) = (1-p)^{k-1}p$.

Common Mistakes

A common error is using $\left(\frac{5}{6}\right)^7$ or $\left(\frac{1}{6}\right)^7$, which ignore the requirement that the 6 occurs exactly on the 7th throw. Another common error is confusing $X = 7$ with at least 7 throws.

Things to Be Careful About

The exponent on $\frac{5}{6}$ is 6, not 7, because only the first 6 throws must be non-6. The final multiplication by $\frac{1}{6}$ is essential. The answer can be left as a fraction or rounded to 3 significant figures.

Walkthrough

We need the probability that the first 6 appears on the 3rd, 4th or 5th throw. For each value of $k$, the first $k-1$ throws must be non-6 and the $k$th throw must be 6. So $P(X=3) = \left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)$, $P(X=4) = \left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right)$ and $P(X=5) = \left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right)$. These events are mutually exclusive, so add their probabilities.

Key Takeaways

For a geometric distribution, probabilities for a range of values are found by summing the individual geometric probabilities. Alternatively, use the cumulative distribution: $P(a \leq X \leq b) = P(X \leq b) - P(X \leq a-1)$, which here gives $\left(\frac{5}{6}\right)^2 - \left(\frac{5}{6}\right)^5$.

Common Mistakes

A common mistake is using the wrong powers, such as $\left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right)$ for $X=3$ instead of $\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)$. Another is forgetting to include one of the three values $X=3,4,5$.

Things to Be Careful About

The lower limit is inclusive: $3 \leq X$ means $X=3$ is included, so the first term has exponent 2, not 3. When adding fractions, use a common denominator or convert to decimals carefully. The exact answer is $\frac{2275}{7776}$; the decimal $0.293$ is rounded to 3 significant figures.

Walkthrough

Since $X$ must take one of the values 1, 2, 3 or 4, the probabilities in the table must total 1. Write

so $4p+0.4=1$ and $p=0.15$. This gives the full distribution: $P(X=1)=0.15$, $P(X=2)=0.4$, $P(X=3)=0.3$, $P(X=4)=0.15$.

The mean is found by taking the weighted sum of scores with their probabilities:

For the variance, use

First calculate $E(X^2)$:

Then subtract the square of the mean:

The mark scheme allows the un-simplified expectation to be used in the variance formula, so keep the working clear.

Key Takeaways

This question tests the basic machinery of a discrete random variable: the probabilities must sum to 1, the expectation is a weighted average, and the variance is the expected square minus the square of the expectation. The formula $\text{Var}(X)=E(X^2)-\left(E(X)\right)^2$ is usually the quickest.

Common Mistakes

• Forgetting that the probabilities must sum to 1 and therefore not solving for $p$.
• Using $E(X)^2$ instead of $\left(E(X)\right)^2$ when subtracting.
• Rounding the mean to $2.5$ before finding the variance, which changes the final answer.
• Writing $\text{Var}(X)=E(X^2)$ without subtracting the square of the mean.

Things to Be Careful About

The final answer should be exact or to sufficient accuracy: $0.8475$ or $\frac{339}{400}$. Keep intermediate values such as $2.45$ and $6.85$, rather than rounding early. The mark scheme gives follow-through credit for using the candidate's own value of $p$ in the variance formula, so do not stop after finding $p$.

Walkthrough

There are 10 independent spins and on each spin the probability of scoring 2 is $0.4$. Hence the number of 2s, $Y$, has a binomial distribution $Y \sim B(10,0.4)$. The binomial model is correct because there are a fixed number of trials, each trial has two outcomes (score 2 or not), and the trials are independent.

We need $P(Y<8)$, which is the same as $P(Y \le 7)$. Instead of summing eight probabilities, use the complement:

and $P(Y \ge 8)=P(Y=8)+P(Y=9)+P(Y=10)$. Use

for $r=8,9,10$. Substituting gives the three terms $0.0106168$, $0.00157286$ and $0.00010486$, so

which rounds to $0.988$.

Key Takeaways

This question uses the binomial distribution. The key is recognising a fixed number of independent trials with the same probability of success. The complement rule is especially useful when the required probability is of the form $P(Y \le k)$ with $k$ close to $n$, because it reduces the number of terms needed.

Common Mistakes

• Taking fewer than 8 as 8 or fewer; it means 0, 1, 2, 3, 4, 5, 6 or 7.
• Using $0.6$ as the probability of scoring 2, or forgetting the factor $(0.6)^{10-r}$.
• Omitting the binomial coefficient $\binom{10}{r}$.
• Rounding the three intermediate probabilities before subtracting from 1, which can change the final answer.

Things to Be Careful About

Check that $p=0.4$ and $q=0.6$ are used in the correct positions. The final probability should be given as $0.988$, with the mark scheme accepting $0.9875 < p \le 0.988$. Keep intermediate terms to at least 4 or 5 decimal places to avoid rounding errors.

Walkthrough

To represent this problem, we build a probability tree diagram level by level.

Step 1: The Coin Toss
A fair coin is tossed, so the first two branches are Head (H) and Tail (T), each with probability $\frac{1}{2}$.

Step 2: First Disc Selection

• If H occurs, Priti draws from Bag X (8 Red, 7 Blue). The probability of drawing Red is $\frac{8}{15}$ and Blue is $\frac{7}{15}$.
• If T occurs, she draws from Bag Y (6 Red, 9 Blue). The probability of drawing Red is $\frac{6}{15}$ and Blue is $\frac{9}{15}$.

Step 3: Second Disc Selection

• From H $\to$ R (Without replacement): Bag X now has 7 Red and 7 Blue discs (14 total). The probabilities are $\frac{7}{14}$ for Red and $\frac{7}{14}$ for Blue.
• From H $\to$ B (Without replacement): Bag X now has 8 Red and 6 Blue discs (14 total). The probabilities are $\frac{8}{14}$ for Red and $\frac{6}{14}$ for Blue.
• From T $\to$ R (With replacement): Bag Y remains unchanged (6 Red, 9 Blue). The probabilities are $\frac{6}{15}$ for Red and $\frac{9}{15}$ for Blue.
• From T $\to$ B (With replacement): Bag Y remains unchanged. The probabilities are $\frac{6}{15}$ for Red and $\frac{9}{15}$ for Blue.

Key Takeaways

• Tree diagrams are ideal for sequential probability problems with conditional outcomes.
• Always update the total number of items and the specific counts when sampling without replacement.
• When sampling with replacement, the probabilities at each stage remain identical.

Common Mistakes

• Forgetting to reduce the total count in the denominator for the second draw when sampling without replacement (e.g., using 15 instead of 14).
• Using the wrong bag's probabilities for the second draw after a Tail (must keep the same bag Y probabilities since it is with replacement).

Things to Be Careful About

• Ensure all branch probabilities from a single node sum to 1 (e.g., $\frac{7}{14} + \frac{7}{14} = 1$).
• Clearly label the outcomes (R and B) at each stage to avoid mixing up the paths.

Walkthrough

We want the probability that both discs chosen are blue. Looking at the tree diagram, there are exactly two paths that lead to two blue discs:

Path 1: Head $\to$ Blue $\to$ Blue

• Probability of Head: $\frac{1}{2}$
• Probability of Blue from Bag X (first draw): $\frac{7}{15}$
• Probability of Blue from Bag X (second draw, without replacement): $\frac{6}{14}$
• Multiply these: $\frac{1}{2} \times \frac{7}{15} \times \frac{6}{14} = \frac{1}{10}$

Path 2: Tail $\to$ Blue $\to$ Blue

• Probability of Tail: $\frac{1}{2}$
• Probability of Blue from Bag Y (first draw): $\frac{9}{15}$
• Probability of Blue from Bag Y (second draw, with replacement): $\frac{9}{15}$
• Multiply these: $\frac{1}{2} \times \frac{9}{15} \times \frac{9}{15} = \frac{9}{50}$

Since these two paths cannot happen at the same time (the coin cannot be both Head and Tail), we add the probabilities:

Key Takeaways

• The multiplication rule is used to find the probability of a specific sequence of events along a path in a tree diagram.
• The addition rule is used to combine probabilities of mutually exclusive outcomes (different paths leading to the same final result).

Common Mistakes

• Forgetting to account for the changing total in the denominator for the second draw from Bag X (using 15 instead of 14).
• Adding probabilities across different stages of the same path instead of multiplying them.

Things to Be Careful About

• Ensure you simplify fractions where possible to make addition easier (e.g., $\frac{6}{14} = \frac{3}{7}$).
• Double-check that you have found all paths that result in the desired outcome.

Walkthrough

We are asked for the probability that the discs come from Bag X, given that at least one disc is red. This is a conditional probability problem.

Step 1: Understand the condition
The condition is 'at least one red disc'. The complement of 'at least one red' is 'no red discs', which means 'both blue'. From part (b), we know $P(\text{both blue}) = \frac{7}{25}$. Therefore:

Step 2: Find the intersection
We need $P(\text{Bag X} \cap \text{at least one red})$. Bag X is chosen if the coin is Head. So we need $P(\text{Head} \cap \text{at least one red})$.
Using the complement idea again for the Head branch: the probability of Head is $\frac{1}{2}$. The probability of Head AND both blue is $\frac{1}{10}$ (from part b). So:

Step 3: Calculate the conditional probability

Key Takeaways

• Conditional probability is calculated as $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
• The complement rule is a powerful tool for finding 'at least one' probabilities.
• When calculating the intersection for a conditional probability, ensure you only consider the paths that satisfy both the event and the condition.

Common Mistakes

• Trying to find the probability of 'at least one red' by adding up all the individual paths, which is prone to arithmetic errors. Using the complement $1 - P(\text{both blue})$ is much faster.
• Forgetting to divide by the probability of the condition (the denominator in the conditional probability formula).
• Calculating $P(\text{Bag X} \mid \text{at least one red})$ as $P(\text{Bag X} \cap \text{at least one red})$ without dividing by $P(\text{at least one red})$.

Things to Be Careful About

• Ensure that the intersection $P(\text{Bag X} \cap \text{at least one red})$ only includes paths from Bag X (Head), not from Bag Y (Tail).
• Simplify fractions before multiplying to avoid large numbers and reduce calculation errors.

Walkthrough

First, we sort both sets of height data in ascending order to make it easy to extract the stems and leaves. The stem is the tens digit (15, 16, 17, 18, 19) and the leaf is the units digit. For a back-to-back diagram, the left side (Pelicans) must have leaves ordered from right to left (largest to smallest as you move away from the stem), while the right side (Swans) has leaves ordered from left to right (smallest to largest as you move away from the stem). Finally, we add a key that clearly states what the numbers represent, including units (cm) and team names.

Key Takeaways

A back-to-back stem-and-leaf diagram allows for the visual comparison of two distributions. The key rule is that leaves on the left side are read right-to-left (closest to the stem is the largest leaf for that stem), and leaves on the right side are read left-to-right.

Common Mistakes

• Forgetting to order the leaves correctly (e.g., writing them left-to-right on the left side).
• Omitting the key or not including units/team names in the key.
• Adding commas or other punctuation between the leaves.

Things to Be Careful About

Ensure that every data point from the original table is represented exactly once in the diagram. Check that the stems cover the full range of the data (15 to 19 in this case).

Walkthrough

With 15 data points, the median is the 8th value when ordered. The lower quartile is the 4th value, and the upper quartile is the 12th value. We read these directly from the sorted Pelicans list. The IQR is simply the difference between the upper and lower quartiles.

Key Takeaways

For a dataset of size $n$, the positions of the median, lower quartile, and upper quartile can be found using $\frac{n+1}{2}$, $\frac{n+1}{4}$, and $\frac{3(n+1)}{4}$ respectively. The IQR measures the spread of the middle 50% of the data.

Common Mistakes

• Using the wrong position formula for quartiles (e.g., using $n/4$ instead of $(n+1)/4$).
• Forgetting to subtract $Q_1$ from $Q_3$ to find the IQR.

Things to Be Careful About

Ensure you are reading the values from the correct dataset (Pelicans, not Swans). The IQR is a measure of spread, not a single quartile value.

Walkthrough

We extract the five-number summary for both teams. For the Swans, the values are given in the question. For the Pelicans, we use the minimum and maximum from the sorted data, and the quartiles calculated in part (b). We then draw two horizontal box-and-whisker plots on the same linear scale. The box spans from $Q_1$ to $Q_3$, with a line inside at the median. Whiskers extend to the minimum and maximum values. Both plots must be labeled, and the axis must have a scale and units.

Key Takeaways

Box-and-whisker plots provide a visual summary of the distribution of data, highlighting the median, quartiles, and extremes. When comparing two datasets, they should be drawn on the same scale to allow direct comparison.

Common Mistakes

• Drawing whiskers that pass through the box or are drawn at the corners of the box.
• Using a non-linear scale or forgetting to label the axis with units.
• Failing to label which plot belongs to which team.

Things to Be Careful About

Ensure the scale is linear and covers the full range of the data. Check the accuracy of the plotted values using the "daylight rule" (there should be a gap between the whisker end and the box if the whisker value is not equal to the quartile).

Walkthrough

We can compare the two teams using either the median (central tendency) or the IQR/range (spread). Since the Swans have a higher median (174 cm > 173 cm), we can say they are generally taller. Since the Swans have a smaller IQR (16 cm < 19 cm), we can say their heights are more consistent in the middle 50%. Any valid comparison based on the calculated statistics is acceptable.

Key Takeaways

When comparing two datasets, always refer to the specific statistic you are using (median, IQR, range) and state the conclusion in the context of the problem (e.g., 'taller', 'more consistent').

Common Mistakes

• Making a vague comparison without referring to a specific statistic.
• Saying one group is 'more consistent' based on range when the IQR is a better measure (though range is acceptable if explicitly stated).

Things to Be Careful About

Ensure the comparison is mathematically correct based on the values calculated in previous parts. For example, do not say the Pelicans are more consistent if their IQR is larger.

Walkthrough

We need the number of unordered teams of 6 athletes chosen from three distinct groups: 8 sprinters, 5 hurdlers and 12 throwers.

The conditions are:

• at least 3 sprinters,
• at most 2 hurdlers,
• at most 1 thrower.

Because the team size is fixed at 6, list the possible triples $(s,h,t)$:

Each triple satisfies all the limits and sums to 6. For each triple, choose the required number from each category using combinations. For example, for $(4,1,1)$, choose 4 of the 8 sprinters, 1 of the 5 hurdlers and 1 of the 12 throwers, giving $\binom{8}{4}\binom{5}{1}\binom{12}{1}$.

The scenarios are mutually exclusive, so add their counts.

Key Takeaways

• An unordered selection is counted with combinations, not permutations.
• "At least" and "at most" restrictions can be handled by listing the disjoint cases that satisfy the limits.
• When the cases are mutually exclusive, the total is the sum of their individual counts.

Common Mistakes

• Missing one of the six valid compositions, or adding an invalid composition such as $(3,1,2)$ or $(2,2,2)$.
• Counting ordered teams instead of unordered selections.
• Double-counting by multiplying the same category choices in different orders.
• Giving only the final total $12600$ without showing method; under the mark scheme a bare total can only earn at most the special-case mark if the method marks are not awarded.

Things to Be Careful About

• "At least 3 sprinters" includes exactly 3, 4, 5 and 6 sprinters.
• "At most" conditions include the upper limit.
• The six scenarios are disjoint; do not add extra scenarios.
• The combination products may be left un-simplified; identification of a scenario can be implied by an un-simplified product.

Walkthrough

There are 8 distinct athletes, so without any restriction the 8 of them can be arranged in $8!$ ways.

For the restriction "the 3 hurdlers do not stand all together", subtract the arrangements where they do stand together. Treat the 3 hurdlers as one block. This block plus the other 5 athletes gives $6$ objects, which can be arranged in $6!$ ways. The 3 hurdlers inside the block can themselves be arranged in $3!$ ways. Hence the forbidden arrangements number $6! \times 3!$.

Subtract this from the total to get the required count.

Key Takeaways

• "Not all together" can be counted by subtracting the forbidden block arrangements from all arrangements.
• When several items must stay together, treat them as a single block, then multiply by the number of internal arrangements.
• All athletes are distinct, so factorial arrangements are appropriate.

Common Mistakes

• Forgetting the $3!$ arrangements of the hurdlers inside the block.
• Using $7! \times 3!$ instead of $6! \times 3!$ for the block case.
• Counting all three hurdlers as identical; they are distinct athletes.
• Stopping after finding the invalid count instead of subtracting it from $8!$.

Things to Be Careful About

• "Not all together" still allows two hurdlers to be next to each other; only arrangements with all three consecutive are excluded.
• The total arrangement is $8!$, not $\binom{8}{3}\times$ etc., because every athlete is distinct.
• Show the subtraction $8! - 6!3!$ clearly to earn the method marks.

Walkthrough

There are 3 hurdlers and 5 non-hurdlers. The condition is about the two gaps between consecutive hurdlers: each must contain at least 2 non-hurdlers.

Let $x_1$ be the number of non-hurdlers before the first hurdler, $x_2$ between the first and second, $x_3$ between the second and third, and $x_4$ after the third hurdler. Then:

Subtract the minimum from the middle gaps by setting $y_2=x_2-2$ and $y_3=x_3-2$:

There are 4 ways to distribute this one remaining non-hurdler among the four gaps.

For each distribution, the 5 distinct non-hurdlers can be arranged in $5!$ ways and the 3 distinct hurdlers in $3!$ ways. Therefore the total is $5!\times 3!\times 4=2880$.

Key Takeaways

• "At least two athletes between any two hurdlers" translates into a gap-count condition.
• Use a distribution/dividing-line model to count the gap sizes before arranging the athletes.
• Arrange the distinct hurdlers and non-hurdlers separately after fixing the gap sizes.

Common Mistakes

• Forgetting the $3!$ arrangements of the hurdlers or the $5!$ arrangements of the non-hurdlers.
• Allowing an internal gap to contain fewer than 2 non-hurdlers.
• Treating the outside gaps as forbidden; they may be empty.
• Counting the gap sizes by trying to choose 3 positions among 8 without handling the at-least-two condition separately.

Things to Be Careful About

• The two middle gaps must each contain at least 2 non-hurdlers.
• The left and right outside gaps may contain 0 non-hurdlers.
• After using 4 non-hurdlers for the minimum middle gaps, the one remaining non-hurdler may go into any of the 4 gaps.
• Do not divide by any factorial; the 8 athletes are all distinct and the formula $5!\times 3!\times 4$ already counts every distinct arrangement once.

Walkthrough

The completion times are normally distributed with mean $46.3$ seconds and standard deviation $6.4$ seconds. To find the probability that a randomly chosen runner took more than $55.1$ seconds, we first convert $55.1$ into a $Z$-score using

The probability $P(X > 55.1)$ is the same as $P(Z > 1.375)$. Standard normal tables give the probability to the left of a $Z$-value, so we look up $\Phi(1.375) = 0.9155$. Since the total probability is 1, the upper-tail probability is $1 - 0.9155 = 0.0845$.

Key Takeaways

• Standardisation converts a normal distribution to the standard normal distribution using $Z = \frac{X - \mu}{\sigma}$.
• Normal tables give left-tail probabilities, so upper-tail probabilities require the complement.
• The answer should be less than 0.5 because $55.1$ is above the mean.

Common Mistakes

• Forgetting to subtract the table value from 1 and giving $0.9155$.
• Using the wrong tail: the question asks for more than $55.1$, so the upper tail is needed.
• Using $\sigma^2$ or $\sqrt{\sigma}$ in the standardisation formula.

Things to Be Careful About

• The standardisation formula is $Z = \frac{X - \mu}{\sigma}$, not $\frac{X - \mu}{\sigma^2}$.
• The table value $\Phi(1.375) = 0.9155$ is the area to the left; the required area is to the right.
• The final answer is accepted in the range $0.0845$ to $0.0846$.

Walkthrough

For the second run, the mean is $39.8$ seconds and 10% of runners took more than $48.6$ seconds. The upper-tail probability is therefore $0.10$, so the lower-tail probability is $0.90$. We first find the standard normal value that has $0.90$ of the distribution to its left; from tables this is $z = 1.282$.

Now standardise $48.6$ in terms of the unknown $\sigma$:

This $Z$-value must equal $1.282$, so

Key Takeaways

• A tail probability can be converted to a standard normal critical value.
• The standardisation formula can be used to solve for an unknown mean or standard deviation.
• The standard deviation must be positive.

Common Mistakes

• Using $0.10$ directly as the $Z$-value instead of finding the critical value $1.282$.
• Using $\sigma^2$ in the denominator of the standardisation formula.
• Confusing the upper-tail probability with the lower-tail probability.

Things to be Careful About

• The critical value for a right-tail probability of $0.10$ is $z = 1.282$.
• The answer is accepted as $6.86$ or $6.87$.
• No continuity correction is needed because no discrete distribution is being approximated.

Walkthrough

We choose 150 runners from the second run. From part (b), the probability that a runner takes more than $48.6$ seconds is $0.10$. Thus the number of such runners, $X$, follows a binomial distribution $B(150, 0.10)$. Because $np = 15$ and $n(1-p) = 135$ are both greater than 5, the binomial distribution can be approximated by a normal distribution.

The normal mean and variance are

so the standard deviation is $\sqrt{13.5} \approx 3.674$.

Since the binomial distribution is discrete, we apply a continuity correction. The event "fewer than 20" corresponds to values $0, 1, \dots, 19$, so the continuous normal value is $19.5$. Then