Mathematics 9709/51 — October/November 2025

Walkthrough

We are told that $X$ takes the value $x$ with probability $kx^2$, where $x$ is restricted to the set $\{-2, 1, 2, 3\}$. The first step is to substitute each value of $x$ into the formula $kx^2$ to write the four probabilities in terms of the unknown $k$. Because $x$ is squared, $x = -2$ and $x = 2$ both give $4k$, the value $x = 1$ gives $k$, and $x = 3$ gives $9k$.

A probability distribution must assign non-negative probabilities to every possible value, and the total of all these probabilities must equal 1. We therefore set up the equation $4k + k + 4k + 9k = 1$. The left side simplifies to $18k$, which gives $k = \frac{1}{18}$. This is the unique value of $k$ that makes the four probabilities both non-negative and summing to exactly 1.

Finally we substitute $k = \frac{1}{18}$ back into each of $4k$, $k$, $4k$, and $9k$ to obtain $\frac{4}{18}, \frac{1}{18}, \frac{4}{18}, \frac{9}{18}$ respectively. These are the entries of the probability distribution table.

Key Takeaways

• The sum of all probabilities in a probability distribution must equal 1; this is the basic consistency condition used to determine any unknown constant.
• The formula $kx^2$ is applied to each value of $x$ in the support of $X$ to give the corresponding probability.
• The completed table is the source of all other calculations involving $X$, including expectation, variance, and conditional probabilities.

Common Mistakes

• Forgetting to square the values when forming the probabilities; in particular, the value $-2$ gives $4k$, not $-2k$.
• Omitting one of the four values when summing the probabilities; the value $x = 1$ is easy to miss because it has the smallest contribution.
• Setting the sum equal to 0 instead of 1.

Things to Be Careful About

• The distribution includes a negative value ($x = -2$); this is perfectly valid and causes no issues at this stage.
• The probabilities can be left as fractions with denominator 18, although the equivalents $\frac{4}{18} = \frac{2}{9}$ and $\frac{9}{18} = \frac{1}{2}$ are also acceptable.

Walkthrough

The expectation of a discrete random variable is the probability-weighted average of its values. For each $x$ in the support, we multiply $x$ by its probability from the table and then add the four products together. For our distribution the four products are

Adding these gives $\frac{-8 + 1 + 8 + 27}{18} = \frac{28}{18} = \frac{14}{9}$. The two middle terms ($-\frac{8}{18}$ and $+\frac{8}{18}$) cancel, leaving the simple sum $\frac{1 + 27}{18} = \frac{28}{18}$.

Key Takeaways

• $E(X)$ is computed by multiplying each $x$-value by its probability and summing.
• A negative $x$-value contributes a negative term to the sum; this is the only complication that arises when the support includes negative numbers.

Common Mistakes

• Using the wrong value of $x$ (for example, writing $4 \cdot \frac{4}{18}$ instead of $(-2) \cdot \frac{4}{18}$).
• Forgetting to include the $x = 1$ term in the sum, which has a small contribution and is easy to miss.
• Arithmetic slips when combining the four products.

Things to Be Careful About

• The mark scheme allows both $\frac{14}{9}$ and the decimal $1.56$ (or $1.5\dot{6}$); either is acceptable, but the exact fraction $\frac{14}{9}$ is preferred.
• The mark scheme awards a follow-through (FT) mark here, so even an incorrect value of $k$ from part (a) will still allow the method mark to be earned provided the working is otherwise correct.
• The expectation does not have to lie between the smallest and largest $x$ in general, although for this distribution it does.

Walkthrough

The condition $X > 0$ is a filter that restricts $X$ to the values $\{1, 2, 3\}$, excluding the value $-2$. So the conditioning event $B$ has probability

The event $A = \{X \neq 2\}$ by itself would also include $X = -2$, but we only want values of $X$ that lie in both $A$ and $B$ simultaneously. So $A \cap B$ corresponds to $X \in \{1, 3\}$, with probability

The conditional probability formula $\text{P}(A \mid B) = \frac{\text{P}(A \cap B)}{\text{P}(B)}$ then gives

Key Takeaways

• The conditional probability formula $\text{P}(A \mid B) = \frac{\text{P}(A \cap B)}{\text{P}(B)}$ applies to any events, including those described in terms of a discrete random variable.
• For a discrete random variable, both $\text{P}(A \cap B)$ and $\text{P}(B)$ are obtained by adding up the relevant probabilities from the distribution table.
• The conditioning event acts as a filter: only values of $x$ satisfying the condition are allowed to contribute to either the numerator or the denominator.

Common Mistakes

• Computing $\text{P}(X \neq 2)$ on its own (which would be $1 - \frac{4}{18} = \frac{14}{18}$, also the wrong answer for the denominator) and using it as the numerator instead of restricting to values with $X > 0$.
• Swapping the numerator and denominator.
• Forgetting to include $X = 3$ (the largest value) when computing $\text{P}(X > 0)$.

Things to Be Careful About

• The 'and' in $\text{P}(X \neq 2 \text{ and } X > 0)$ is essential; it requires both conditions to hold simultaneously, so the value $X = -2$ is excluded from the numerator even though it satisfies $X \neq 2$.
• The mark scheme accepts either $\frac{10/18}{14/18}$ or $\frac{5/9}{\frac{7}{9}}$ for the method mark, but the final answer must be simplified to $\frac{5}{7}$ to earn the accuracy mark.
• The fraction $\frac{10}{14}$ must be simplified to $\frac{5}{7}$ to earn the final accuracy mark.

Walkthrough

We are looking for the first success (a 6) to happen on the 8th throw. That means throws 1 to 7 must all fail (not be 6), and throw 8 must succeed. Each throw is independent, so we multiply the probability of failure seven times and then the probability of success once. This is exactly the geometric distribution formula $P(X = n) = q^{n-1}p$.

Key Takeaways

• Recognise "first success on the nth trial" as a geometric distribution problem.
• The probability of failure is $q = 1 - p$.
• Use $q^{n-1}p$ for the first success on the nth trial.

Common Mistakes

• Using $p^n$ instead of $q^{n-1}p$.
• Forgetting that the first 7 throws must all be failures.
• Rounding to fewer than 3 significant figures.

Things to Be Careful About

• The exponent on $\frac{5}{6}$ is 7, not 8, because the 8th throw is the success.
• The exact fraction $\frac{78125}{1679616}$ is acceptable, and the decimal should be given to at least 3 significant figures.

Walkthrough

We want the third 6 to occur on the 7th throw. This means that in the first 6 throws there must be exactly two 6s, and then the 7th throw must be a 6. First count the number of ways to choose which two of the first six throws are 6s: $\binom{6}{2}=15$. For any one of these arrangements, the probability is $(1/6)^2(5/6)^4$ for the first six throws, and then $(1/6)$ for the 7th throw. Multiply by 15 to account for all arrangements. This gives the required probability.

Key Takeaways

• A "kth success on the nth trial" problem requires exactly $k-1$ successes in the first $n-1$ trials and a success on the nth trial.
• Use combinations to count the positions of the successes.
• Multiply the probabilities of the independent events.

Common Mistakes

• Choosing positions among all 7 throws instead of only the first 6.
• Using $\binom{7}{2}$ or $\binom{7}{3}$.
• Forgetting the factor $\binom{6}{2}$.
• Treating the 7th throw as one of the variable positions.

Things to Be Careful About

• The 7th throw is fixed as a 6, so only the first 6 throws can vary.
• The exponent of $\frac{1}{6}$ is 3: two 6s in the first six throws plus the 7th throw.
• The exponent of $\frac{5}{6}$ is 4: four non-6s in the first six throws.
• The exact answer is $\frac{3125}{93312}$ or $\frac{9375}{279936}$, and the decimal is $0.0335$ to 3 significant figures.

Walkthrough

The stem-and-leaf diagram is read row by row. For Browns, the leaves are read from right to left (the leaf closest to the stem is the smallest, so reading right to left already gives ascending order within each row). The key tells us the leaves represent the hundreds digit in dollars, so a stem of 32 and a leaf of 4 means $32,400.

We collect all 27 Browns values and order them. With an odd number of observations ($n = 27$), the median sits exactly at the middle position, which is the $\frac{27+1}{2} = 14$th value. Counting through the ordered list, the 14th entry is $32,400.

For the quartiles, the standard convention with an odd $n$ is to remove the median and split the remaining 26 values into two halves of 13 each. The lower quartile is the median of the lower half (the 7th value overall) and the upper quartile is the median of the upper half (the 21st value overall). Reading from the ordered list, these are $31,300 and $33,500. The IQR is the difference.

Key Takeaways

• Back-to-back stem-and-leaf diagrams display each company's distribution on opposite sides of a common stem; leaves on the left (Browns here) are conventionally read right-to-left for ascending order.
• For a data set of size $n$:
  • Median position = $\frac{n+1}{2}$ (if $n$ odd) or mean of the $\frac{n}{2}$ and $\frac{n}{2}+1$ values (if $n$ even).
  • $Q_1$ and $Q_3$ are the medians of the lower and upper halves (with the median removed when $n$ is odd).
• IQR = $Q_3 - Q_1$; it is a measure of spread that ignores the extreme 25% on each end.

Common Mistakes

• Reading the Browns leaves in the wrong order (left-to-right instead of right-to-left), which would scramble the sorted list and give the wrong median.
• Using $\frac{n}{2}$ instead of $\frac{n+1}{2}$ for the median position with odd $n$, leading to the 13th or 15th value.
• Computing $Q_1$ and $Q_3$ from the whole data set (positions 7 and 21 here are correct; some textbooks use different conventions, so always state the method).
• Forgetting the leaf multiplier — the key tells us leaves are in hundreds, so a leaf of 4 with stem 32 is $32,400, not $324.

Things to Be Careful About

• The mark scheme allows $Q_1$ in the range $31,300–$31,500 and $Q_3$ in the range $33,500–$33,700, so the IQR $2,200 (from $33,500 − $31,300) is the CAO answer.
• If you compute using thousands of dollars throughout, convert back to dollars for the final answer (or state the units clearly).

Walkthrough

The 54 employees are pooled into one combined sample. The combined mean is the total of all salaries divided by the total number of employees, $n = 54$. The data are given in thousand dollars, so adding the two sums $878.3 + 896.5 = 1774.8$ and dividing by 54 gives the mean in thousand dollars, which we then convert to dollars.

For the standard deviation we use the algebraic identity

This identity avoids recomputing the mean of the squares by hand. Because the data are pooled, the total of squares is just $\sum x^2 + \sum y^2$, and the mean is the combined mean we already found. Substituting gives a variance of about $1.8511$ in (thousand dollars)$^2$, and taking the square root gives the standard deviation in thousand dollars, which we then convert to dollars.

Key Takeaways

• For two combined data sets of sizes $n_1, n_2$ with sums $S_1, S_2$ and sums of squares $Q_1, Q_2$: $$\bar{x} = \frac{S_1 + S_2}{n_1 + n_2},\qquad \sigma^2 = \frac{Q_1 + Q_2}{n_1 + n_2} - \bar{x}^2.$$
• Keep the data in their given units (here, thousands of dollars) until the very end, then convert.
• Standard deviation is the square root of the variance; do not report $\sigma^2$ as the final answer.

Common Mistakes

• Dividing each company's mean by 27 and then averaging them; the correct combined mean is the overall total divided by 54 (here the two methods happen to give the same value, but they are not equivalent in general).
• Forgetting to square the mean when computing the variance, or subtracting the mean from $\frac{\sum x^2}{n}$ instead of subtracting its square.
• Reporting the variance as the standard deviation.
• Leaving the answer in thousands without stating it (the mark scheme accepts $32.867$ and $1.36$ in thousands but requires the conversion to dollars for the CAO answers).

Things to Be Careful About

• The mark scheme allows $32,866.67 to 3 or more significant figures for the mean, but the B-mark answer is $32,900 to the nearest $100.
• The mark scheme accepts the variance as $1.85$ or as an un-simplified value, but the standard deviation must be explicitly labelled (sd, $\sigma$, etc.) and not just the variance's square root left unattributed.
• The standard deviation is $1,360, not $1,360^2 — units are dollars, not dollars squared.

Walkthrough

The key to this problem is tracking how the contents of each bag change after every draw. We proceed stage by stage.

Stage 1: Bag A starts with 8 Red and 3 Blue (11 total). Drawing Red (probability $\frac{8}{11}$) means it is discarded, leaving Bag A with 7R, 3B. Drawing Blue (probability $\frac{3}{11}$) means it goes into Bag B, so Bag A becomes 8R, 2B and Bag B becomes 4R, 2B.

Stage 2: We now draw from Bag B. If we came from the Red branch, Bag B still has 4R, 1B (5 total), giving probabilities $\frac{4}{5}$ and $\frac{1}{5}$. If we came from the Blue branch, Bag B has 4R, 2B (6 total), giving $\frac{4}{6}$ and $\frac{2}{6}$. Again, Red is discarded and Blue is transferred to Bag A, which changes Bag A's composition for the next stage.

Stage 3: We draw from Bag A again. The composition depends on the path taken:

• Path (R,R): Bag A has 7R, 3B → probabilities $\frac{7}{10}$, $\frac{3}{10}$
• Path (R,B): Bag A has 7R, 4B → probabilities $\frac{7}{11}$, $\frac{4}{11}$
• Path (B,R): Bag A has 8R, 2B → probabilities $\frac{8}{10}$, $\frac{2}{10}$
• Path (B,B): Bag A has 8R, 3B → probabilities $\frac{8}{11}$, $\frac{3}{11}$

Each branch probability is simply the number of marbles of the desired colour divided by the total number of marbles in the relevant bag at that stage.

Key Takeaways

• When objects are transferred between containers, the probabilities at subsequent stages depend on the entire history of previous draws.
• Always update the contents of each container after every action (discard or transfer) before computing the next probability.
• A tree diagram is an effective tool for visualising sequential conditional probabilities.

Common Mistakes

• Forgetting to update the bag contents after a marble is transferred or discarded, leading to incorrect denominators in later stages.
• Assuming the bag contents remain constant throughout the problem.
• Omitting the Red/Blue labels on branches, which costs a mark even if the probabilities are correct.

Things to Be Careful About

• The problem states "If the marble chosen is red, it is discarded. If the marble chosen is blue, it is placed in bag B" (and vice versa for the second draw). This directional transfer matters — do not confuse which bag gains or loses a marble.
• When a marble is placed into a bag, the total count in that bag increases by 1, and the count of the other bag decreases by 1.
• Always verify that branch probabilities from a single node sum to 1 (e.g., $\frac{4}{5} + \frac{1}{5} = 1$ and $\frac{4}{6} + \frac{2}{6} = 1$).

Walkthrough

The question asks for the probability that all three marbles chosen are the same colour. There are exactly two ways this can happen:

1. All three are Red: this follows the path Red (from Bag A) → Red (from Bag B) → Red (from Bag A again).
2. All three are Blue: this follows the path Blue (from Bag A) → Blue (from Bag B) → Blue (from Bag A again).

These two events are mutually exclusive — they cannot both happen — so we use the addition law: $P(\text{RRR or BBB}) = P(\text{RRR}) + P(\text{BBB})$.

For each path, we use the multiplication law, multiplying the branch probabilities along the path using the values computed in part (a):

• $P(\text{RRR}) = \frac{8}{11} \times \frac{4}{5} \times \frac{7}{10} = \frac{224}{550}$
• $P(\text{BBB}) = \frac{3}{11} \times \frac{2}{6} \times \frac{3}{11} = \frac{18}{726}$

Adding these with a common denominator of 3025 gives $\frac{1307}{3025} \approx 0.432$.

Key Takeaways

• When asked for 'all the same' or 'all different' outcomes in sequential draws, identify the specific paths on the tree that satisfy the condition.
• Mutually exclusive outcomes are combined using the addition law; sequential events along a path are combined using the multiplication law.
• Simplifying fractions before adding can make the arithmetic easier, but finding a common denominator is essential.

Common Mistakes

• Including paths like RRB or BBR — the question asks for ALL THREE to be the same colour, not just two.
• Forgetting that RRR and BBB are mutually exclusive and trying to multiply them together instead of adding.
• Arithmetic errors when adding fractions with different denominators.

Things to Be Careful About

• The decimal answer $0.432$ is given to 3 significant figures in the mark scheme; the exact answer $\frac{1307}{3025}$ is preferred.
• Do not confuse this with 'at least two the same colour' — the question is strict about all three being identical.

Walkthrough

First identify the random variable. Since each day is independent and the probability of wearing a blue jumper is always $0.6$, the number of blue-jumper days in a 7-day week follows a binomial distribution, $X \sim \mathrm{B}(7,0.6)$. The phrase “at least 5 days” means exactly 5, 6 or 7 days. Use the binomial formula for each of these values and add the three probabilities. The total is approximately $0.420$.

Key Takeaways

• A binomial model applies when there are a fixed number of independent trials, two outcomes per trial, and a constant probability of success.
• “At least 5” includes 5, 6 and 7, not just values greater than 5.
• The binomial coefficient ${}^{7}C_{x}$ counts the number of ways to choose which $x$ days are blue-jumper days.

Common Mistakes

• Forgetting to include the $X=7$ term.
• Using $P(X>5)$ instead of $P(X \geq 5)$.
• Using $0.4$ as the success probability instead of $0.6$.
• Writing only the final decimal without showing the unsimplified binomial sum, which is required for the method and accuracy marks.

Things to Be Careful About

• Keep enough decimal places in intermediate terms; the final answer should round to $0.420$.
• The mark scheme allows a final probability in the range $0.4198$ to $0.420$.
• Make sure the binomial coefficient and powers are written correctly for each term.

Walkthrough

First find the probability that Cooper does not wear a jumper on a given day. The three outcomes are mutually exclusive and cover all possibilities, so $P(\text{no jumper}) = 1 - 0.6 - 0.3 = 0.1$. Let $Y$ be the number of no-jumper days in 150 days. Then $Y \sim \mathrm{B}(150, 0.1)$. Since $np = 15$ and $n(1-p) = 135$ are both comfortably greater than 5, the normal approximation is suitable. Compute the mean and variance using $\mu = np$ and $\sigma^2 = np(1-p)$. The event “fewer than 22 days” means $Y \leq 21$. Because the normal distribution is continuous, apply a continuity correction and use $21.5$ as the upper boundary. Standardise using $z = \frac{x - \mu}{\sigma}$, then read the cumulative probability $\Phi(1.769)$ from the normal table to obtain $0.962$.

Key Takeaways

• The normal approximation to the binomial is appropriate when $np$ and $n(1-p)$ are both greater than 5.
• The mean and variance of $\mathrm{B}(n,p)$ are $np$ and $np(1-p)$.
• A continuity correction is needed when approximating a discrete distribution by a continuous one.
• For $P(X < k)$, the continuity-corrected boundary is $k - 0.5$.

Common Mistakes

• Using 22 instead of 21.5 in the standardisation formula.
• Using no continuity correction at all.
• Confusing the variance $13.5$ with the standard deviation; the denominator must be $\sqrt{13.5}$.
• Forgetting to check that the normal approximation is valid before using it.
• Reading the wrong tail of the normal distribution; here we need $P(Z < 1.769)$, not $P(Z > 1.769)$.

Things to Be Careful About

• “Fewer than 22” means 21 or fewer, so the continuity-corrected boundary is $21.5$.
• The final answer should be AWRT $0.962$.
• Keep the standard deviation as $\sqrt{13.5} \approx 3.67423$, not $13.5$.
• The normal approximation is suitable here because $n$ is large and $p$ is not too close to 0 or 1.

Walkthrough

We are told the weights are normally distributed with mean 155 g and standard deviation 6 g, and we want the proportion of bars between 148 g and 160 g. Since we cannot look up probabilities for a normal distribution with mean 155 and sd 6 directly, we standardise: convert each boundary into a Z-score using $Z = (X - \mu)/\sigma$. For 148 g we get $Z = (148-155)/6 = -1.1667$, and for 160 g we get $Z = (160-155)/6 = 0.8333$. Then $\text{P}(148 < X < 160) = \text{P}(-1.1667 < Z < 0.8333)$.

To find this probability, we use the symmetry of the normal curve: the area between $-1.1667$ and $0.8333$ equals $\Phi(0.8333) + \Phi(1.1667) - 1$. Why? Because $\Phi(z)$ gives the area to the left of $z$. The area to the left of $0.8333$ is $\Phi(0.8333)$, and the area to the left of $-1.1667$ is $\Phi(-1.1667) = 1 - \Phi(1.1667)$. So the area between them is $\Phi(0.8333) - (1 - \Phi(1.1667)) = \Phi(0.8333) + \Phi(1.1667) - 1$.

From the standard normal table, $\Phi(0.8333) = 0.7975$ and $\Phi(1.1667) = 0.8784$. So the probability is $0.7975 + 0.8784 - 1 = 0.6759$.

Finally, since we have 350 bars, the expected number weighing between 148 g and 160 g is $0.6759 \times 350 = 236.565$, which is approximately 236 or 237 bars.

Key Takeaways

• Standardising a normal variable with $Z = (X - \mu)/\sigma$ lets us use the standard normal table.
• The probability between two values is found by combining cumulative probabilities, using symmetry where needed.
• The expected count in a sample is the probability multiplied by the sample size.

Common Mistakes

• Using $\sigma^2$ instead of $\sigma$ in the standardisation formula.
• Forgetting to handle the negative Z-score correctly (e.g. using $\Phi(-1.1667)$ directly instead of $1 - \Phi(1.1667)$).
• Not multiplying the probability by 350.
• Rounding the probability too early, leading to a wrong expected count.

Things to Be Careful About

• The marking scheme requires the standardisation formula to be shown, with $\sigma$ not $\sigma^2$ or $\sqrt{\sigma}$.
• No continuity correction is used here because we are dealing with a continuous normal distribution directly.
• The final answer must be a single integer; both 236 and 237 are accepted.

Walkthrough

We are given two facts about the second factory: 8% of bars weigh more than 114.0 g, and 20% weigh less than 99.5 g. We need to find $\mu$ and $\sigma$.

First, convert each percentage to a Z-score. For $\text{P}(X > 114) = 0.08$, we need the Z-value such that the area to its right is 0.08, i.e. the area to its left is 0.92. From the standard normal table, $\Phi(1.406) = 0.92$, so $z = 1.406$. For $\text{P}(X < 99.5) = 0.20$, we need the Z-value such that the area to its left is 0.20, which is $z = -0.842$ (since $\Phi(-0.842) = 0.20$).

Now write the standardisation equations:

These give two linear equations in $\mu$ and $\sigma$:

Subtract the second from the first to eliminate $\mu$:

so $\sigma = 14.5/2.248 = 6.45$.

Then substitute back to find $\mu = 114 - 1.406 \times 6.45 = 105$ (approximately). We can check with the other equation: $\mu = 99.5 + 0.842 \times 6.45 = 105$.

Key Takeaways

• Percentages can be converted to Z-scores using the inverse of the standard normal cumulative function.
• Each condition gives one equation in $\mu$ and $\sigma$; two conditions give two equations to solve simultaneously.
• The elimination method is a clean way to remove $\mu$ and solve for $\sigma$ first.

Common Mistakes

• Using the percentage directly (e.g. 0.08) instead of the corresponding Z-score.
• Mixing up which tail the percentage belongs to (e.g. using a positive Z for a lower-tail probability).
• Using $\sigma^2$ or $\sqrt{\sigma}$ in the standardisation formula.
• Arithmetic errors when subtracting the two equations.

Things to Be Careful About

• The Z-values must be read accurately from the table: $1.406$ (not $1.40$ or $1.41$ rounded too coarsely) and $-0.842$.
• The mark scheme accepts $\mu$ between 104.5 and 105.5, and $\sigma$ between 6.44 and 6.475.
• A continuity correction of $\pm 0.05$ is condoned, but not required here.