Mathematics 9709/52 — May/June 2024

Walkthrough

This is a geometric distribution problem. Each year is an independent trial with success probability $p = 0.3$ and failure probability $1 - p = 0.7$. For the first success to occur on the 7th attempt, Rajesh must fail in each of the first 6 years and then succeed in the 7th year. Because the trials are independent, the probability is the product of the six failure probabilities and the one success probability: $(0.7)^6 \times 0.3$. This is exactly the geometric distribution formula $P(X = k) = (1-p)^{k-1}p$ with $k = 7$.

Key Takeaways

• The geometric distribution models the number of trials until the first success.
• The probability of the first success on the $k$th trial is $(1-p)^{k-1}p$.
• Independent trials allow us to multiply individual probabilities.

Common Mistakes

• Using $0.3^6$ instead of $(0.7)^6 \times 0.3$.
• Forgetting to include the final success probability $0.3$.
• Rounding to fewer than 3 significant figures.

Things to Be Careful About

• The exponent on $0.7$ is 6 (the number of failures before the success), not 7.
• The answer must be given to at least 3 significant figures.

Walkthrough

"Before his 6th attempt" means the first success occurs on attempt 1, 2, 3, 4, or 5. There are two valid approaches.

Method 1 (complement): The probability that the first success occurs on or after the 6th attempt is the probability that the first 5 attempts are all failures, which is $(0.7)^5$. Therefore $P(X < 6) = 1 - (0.7)^5 = 1 - 0.16807 = 0.83193 \approx 0.832$.

Method 2 (direct summation): Sum the geometric probabilities for $X = 1, 2, 3, 4, 5$:

Both methods give the same result.

Key Takeaways

• The complement method is often easier than summing many geometric terms.
• The geometric distribution can be summed directly for a small number of trials.

Common Mistakes

• Using $0.7^5$ directly instead of $1 - 0.7^5$.
• Including the 6th attempt in the range ("before his 6th attempt" excludes attempt 6).
• Rounding to fewer than 3 significant figures.

Things to Be Careful About

• "Before his 6th attempt" means attempts 1 to 5, not 1 to 6.
• The mark scheme accepts both methods and requires the answer to at least 3 significant figures.

Walkthrough

This is a negative binomial problem. For the second success to occur on the 10th attempt, there must be exactly one success in the first 9 attempts and a success on the 10th attempt. The number of ways to place the one success among the first 9 attempts is ${}^9C_1 = 9$. Each such arrangement has probability $(0.7)^8 \times (0.3)^2$ (8 failures and 2 successes). Therefore the total probability is:

Key Takeaways

• The negative binomial distribution gives the probability of the $r$th success on the $n$th trial: $\binom{n-1}{r-1}(1-p)^{n-r}p^r$.
• Combinatorics (binomial coefficients) is used to count the arrangements.

Common Mistakes

• Using $\binom{10}{2}$ instead of $\binom{9}{1}$.
• Forgetting to multiply by the binomial coefficient.
• Using the wrong exponents on $0.7$ and $0.3$.

Things to Be Careful About

• The binomial coefficient must be over the first 9 attempts (since the 10th attempt is fixed as a success).
• The exponents must sum correctly: 8 failures + 2 successes = 10 attempts.
• The answer must be given to at least 3 significant figures.

Walkthrough

Start by identifying the two possible outcomes that give no red marble. If the coin lands heads, Seva takes one marble, so no red means the marble is blue: probability $\frac{1}{3}\times\frac{5}{9}$. If the coin lands tails, he takes two marbles without replacement, so no red means both are blue: first blue $\frac{5}{9}$, then second blue $\frac{4}{8}$, giving $\frac{2}{3}\times\frac{5}{9}\times\frac{4}{8}$. These two cases are mutually exclusive, so add them. This gives $\frac{10}{27}$. Since 'at least one red' is the complement of 'no red', subtract from 1: $1-\frac{10}{27}=\frac{17}{27}$.

Key Takeaways

The key idea is to use the complement when 'at least one' appears. Also, when selections are made without replacement, the denominator changes after the first selection. Probabilities of disjoint cases are added, and probabilities along a branch are multiplied.

Common Mistakes

A common mistake is to forget the tail branch entirely, or to treat the second marble as selected with replacement, using $\frac{5}{9}$ again instead of $\frac{4}{8}$. Another mistake is to compute the probability of at least one red directly but miss one of the red-containing cases.

Things to Be Careful About

Remember that if the coin is tails, two marbles are chosen without replacement, so the second probability is out of 8. Keep the fractions unsimplified until the end to make adding easier. The probability of no red marbles is $\frac{10}{27}$, which is also needed in part (b).

Walkthrough

We need the probability that the coin was a head, given that no red marble was selected. Write this as $\mathrm{P}(\text{head} \mid \text{no red})$. By the conditional probability formula, divide the probability of head and no red by the probability of no red. Head and no red means head and blue: $\frac{1}{3}\times\frac{5}{9}=\frac{5}{27}$. The probability of no red was found in part (a): $\frac{10}{27}$. Dividing gives $\frac{5/27}{10/27}=\frac{5}{10}=\frac{1}{2}$.

Key Takeaways

Conditional probability reverses the conditioning: $\mathrm{P}(A \mid B)$ uses the intersection $A \cap B$ divided by $\mathrm{P}(B)$, not by $\mathrm{P}(A)$. The denominator is the total probability of the condition, which may come from a previous part.

Common Mistakes

A common mistake is to use $\mathrm{P}(\text{no red} \mid \text{head})$ instead of $\mathrm{P}(\text{head} \mid \text{no red})$, or to divide by 1 instead of by $\frac{10}{27}$. Another mistake is to forget that the numerator must be the intersection, not just the probability of a head.

Things to Be Careful About

Make sure the denominator is the total probability of 'no red marbles', including both the head-blue and tail-blue-blue branches. The final answer $\frac{1}{2}$ is exact; do not round unnecessarily.

Walkthrough

The weight of an orange is modelled by a normal distribution with mean $\mu = 131$ and standard deviation $\sigma = 54$. A large orange is one that weighs at least 184 grams, so we need $\text{P}(X \geq 184)$.

First, standardise the boundary value 184:

This tells us that 184 grams is about 0.9815 standard deviations above the mean. The normal table gives the probability that $Z$ is less than 0.9815, which is $\Phi(0.9815) \approx 0.837$. Since we want the probability that $Z$ is greater than 0.9815, we take the complement:

Finally, multiply by 100 to express this as a percentage: $16.3\%$.

Key Takeaways

• The standardisation formula $z = \frac{x - \mu}{\sigma}$ is the bridge between any normal distribution and the standard normal distribution.
• The standard normal table gives left-tail probabilities, so for an upper tail you must subtract from 1.
• A probability can be converted directly to a percentage by multiplying by 100.

Common Mistakes

• Using $\Phi(0.9815) = 0.837$ directly as the answer instead of taking $1 - 0.837$.
• Forgetting to multiply by 100 and giving 0.163 instead of 16.3%.
• Using the variance $54^2$ instead of the standard deviation 54 in the standardisation formula.

Things to Be Careful About

• Since the normal distribution is continuous, $\text{P}(X \geq 184)$ is the same as $\text{P}(X > 184)$; no continuity correction is needed.
• The boundary 184 lies above the mean, so the $z$-value is positive and the upper-tail probability is less than 0.5, which is a good sanity check.
• The mark scheme accepts any reasonable rounding of the final percentage, so 16.3% is the expected answer.

Walkthrough

We want the greatest weight that is still classified as small. Since 20% of oranges are small, the threshold $w$ satisfies $\text{P}(X < w) = 0.20$.

Standardising gives:

The value $\frac{w - 131}{54}$ is the $z$-score such that the area to its left is 0.20. Looking up 0.20 in the standard normal table gives $z \approx -0.842$. The negative sign is essential: 20% is below the mean, so the threshold must be below 131 grams.

Now solve:

So the greatest possible weight of a small orange is approximately 85.5 grams.

Key Takeaways

• The inverse normal problem starts from a probability and works backwards to a $z$-value, then to the original data value.
• A probability below 0.5 corresponds to a $z$-value below the mean, so the $z$-value is negative.
• The standardisation formula can be rearranged to solve for the unknown boundary $x$.

Common Mistakes

• Using a positive $z$-value of 0.842 instead of $-0.842$, giving $w = 176.5$ grams instead of 85.5 grams.
• Using 0.2 directly as the $z$-value rather than finding the $z$-value whose cumulative probability is 0.2.
• Using 0.8 or another probability instead of 0.2 when reading the table.
• Mixing up signs in the equation, such as writing $w = 131 - 54(0.842)$ but then producing a negative answer.

Things to Be Careful About

• The final answer must be positive, so the signs in the equation must be consistent.
• The mark scheme accepts $85.5 \leq w \leq 85.6$, so rounding to 85.5 is fine.
• Use the standard deviation 54, not the variance $54^2$.
• “Greatest possible weight” is the boundary itself; because the normal distribution is continuous, whether the boundary is included or not does not affect the probability.

Walkthrough

The back-to-back stem-and-leaf diagram gives every Petral salary directly: each stem on the left is read in conjunction with each leaf to its left, in the order shown, using the key $2\mid31\mid5 \Rightarrow$ $31,200. The 19 Petral salaries must first be written out in ascending order, because a stem-and-leaf diagram already orders them row-by-row. Once the 19 values are listed, the median of an odd number of observations is the single middle value — the 10th out of 19 — which is the 32000 (the lone "0" leaf on stem 32). The lower quartile is the median of the lower half (the 5th of 19), giving 31200. The upper quartile is the median of the upper half (the 15th of 19), giving 33500. Subtracting gives the IQR.

Key Takeaways

• For $n = 19$, the median is the 10th value, the lower quartile is the 5th value, and the upper quartile is the 15th value.
• A back-to-back stem-and-leaf diagram already lists the data in ascending order within each stem, which makes locating summary statistics straightforward.
• The IQR is a robust measure of spread, less affected by extreme values than the standard deviation.

Common Mistakes

• Forgetting the $100 part of the key, so the answer comes out 100 times too small (e.g. writing the median as 320 instead of 32000).
• Using the wrong position for the quartiles, for example taking the 4th and 16th values rather than the 5th and 15th.
• Including the median inside either half when locating the quartiles, which would shift the positions.

Things to Be Careful About

• Always use the Petral half of the key ($2\mid31\mid5 \Rightarrow$ $31,200), not the Ravon half.
• Make sure the count of leaves on the Petral side adds to 19 before assuming the median is the 10th value.

Walkthrough

A box-and-whisker plot is built from a five-number summary. For each company, list the minimum, lower quartile, median, upper quartile and maximum. For Petral, the minimum (30000) and maximum (36800) are the smallest and largest leaves in the stem-and-leaf diagram, and the quartiles and median are the values found in part (a). For Ravon, the question gives the lower quartile (32000), median (33800) and upper quartile (34400); the minimum (30200) is the smallest leaf on the Ravon side and the maximum (36900) is the largest. The two plots are then drawn as horizontal boxes on the same scale, with the whiskers extending to the extremes and a clear vertical line inside each box marking the median.

Key Takeaways

• The five-number summary fully determines a box-and-whisker plot.
• Putting two box plots on a common axis makes it easy to compare centre, spread and symmetry of the two distributions visually.
• A scale of at least $1$ cm = $1000 is required so the plots can be read accurately.

Common Mistakes

• Forgetting the $100 part of the key when reading the extremes from the stem-and-leaf.
• Drawing the whiskers through the box or at the corners of the box instead of stopping at the box edges.
• Using a non-linear or compressed scale that prevents the quartile values being plotted accurately.

Things to Be Careful About

• The label "salary ($)" or equivalent must appear on the axis so the scale is unambiguous.
• Each box must be clearly labelled (R and P) so the two plots cannot be confused.
• The horizontal width of each box has no statistical meaning — only the positions of the box edges, the median line and the whisker ends do.

Walkthrough

A measure of central tendency should reflect a typical observation. The Petral salaries are mostly clustered between $30,000 and $34,100, with a single salary of $36,800 sitting noticeably above the rest. Such an extreme value drags the arithmetic mean upwards, away from where the bulk of the data lie. The median, being the middle value, ignores the size of that extreme observation and remains within the main cluster. So when the data are skewed or contain an outlier, the median is the more robust summary.

Key Takeaways

• The mean is sensitive to extreme values; the median is resistant to them.
• When a data set contains an outlier, the median gives a fairer representation of a typical value than the mean does.
• Justifying the choice of measure requires naming the feature of the data (outlier, skew) that makes one measure preferable to the other.

Common Mistakes

• Saying "the median because there are extreme values" — the mark scheme requires identifying the extreme value itself (e.g. $36,800) or the direction of the skew, not a vague reference to "values".
• Choosing the mean because the data are numerical, with no consideration of the outlier.

Things to Be Careful About

• The Petral data are positively skewed (long right tail), which on its own is enough to justify the median; the single $36,800 salary is the most concrete reason.

Walkthrough

Jasmine has one $5 coin, two $2 coins, and two $1 coins. She chooses two coins at random. We want the probability that their total value is $7. The only way to get $7 is to pick the $5 coin and one of the two $2 coins. There are $1 \times 2 = 2$ favourable pairs. The total number of pairs of coins she could choose is $\binom{5}{2} = 10$. So the probability is $\frac{2}{10} = 0.2$. This is a "show that" question, so we must state both the favourable count and the total count.

Key Takeaways

• When selecting objects without replacement, use combinations to count selections.
• Probability = favourable outcomes / total outcomes.
• For "show that" questions, include enough working to justify the given value.

Common Mistakes

• Forgetting to count both $2 coins, so only getting $\frac{1}{10}$ instead of $\frac{2}{10}$.
• Using ordered selections and forgetting to multiply by 2 for the two orders (5 then 2, or 2 then 5).
• Not showing the total number of selections.

Things to Be Careful About

• The two $2 coins are distinct coins even though they have the same value.
• If using ordered probability, the probability is $\frac{1}{5}\times\frac{2}{4} + \frac{2}{5}\times\frac{1}{4} = 0.2$.
• The mark scheme requires the value 7, the coins 5 and 2, and the probabilities linked to the appropriate value.

Walkthrough

List all possible pairs of coins and their sums. There are 5 coins: $5, $2, $2, $1, $1. The possible totals are 2, 3, 4, 6, 7. Count the number of pairs for each total:

• $1 + $1 = 2: only one pair (the two $1 coins).
• $2 + $1 = 3: each of the two $2 coins can pair with each of the two $1 coins, so 4 pairs.
• $2 + $2 = 4: one pair (the two $2 coins).
• $5 + $1 = 6: the $5 coin with each of the two $1 coins, so 2 pairs.
• $5 + $2 = 7: the $5 coin with each of the two $2 coins, so 2 pairs.
  Total pairs = $1 + 4 + 1 + 2 + 2 = 10$. Divide each count by 10 to get probabilities: 0.1, 0.4, 0.1, 0.2, 0.2.

Key Takeaways

• A probability distribution table must list every possible value and the probabilities must sum to 1.
• Identical coin values still represent distinct coins when counting outcomes.
• Systematic listing avoids missing outcomes.

Common Mistakes

• Missing the $X = 3$ outcome (the most common one) or miscounting it as 2 instead of 4.
• Including a $X = 5$ row, which is impossible.
• Forgetting that probabilities must sum to 1.

Things to Be Careful About

• The two $2 coins and two $1 coins are distinct objects, so there are 4 ways to get $3.
• The mark scheme allows extra $x$ values only if their probabilities are stated as 0.
• If writing outcomes, label them clearly so the probabilities can be linked to the correct totals.

Walkthrough

Use the formula $\operatorname{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2$. First compute $\mathrm{E}(X)$ by multiplying each value by its probability and summing: $0.1 \times 2 + 0.4 \times 3 + 0.1 \times 4 + 0.2 \times 6 + 0.2 \times 7 = 4.4$. Then compute $\mathrm{E}(X^2)$ by multiplying each squared value by its probability: $0.1 \times 4 + 0.4 \times 9 + 0.1 \times 16 + 0.2 \times 36 + 0.2 \times 49 = 22.6$. Finally subtract $[\mathrm{E}(X)]^2 = 4.4^2 = 19.36$, giving $22.6 - 19.36 = 3.24$.

Key Takeaways

• Variance measures spread and is calculated as $\mathrm{E}(X^2) - [\mathrm{E}(X)]^2$.
• Always compute $\mathrm{E}(X)$ first because it is needed for both the mean and the variance.
• Use the probability distribution table from part (b) consistently.

Common Mistakes

• Forgetting to square the values when computing $\mathrm{E}(X^2)$.
• Subtracting $\mathrm{E}(X)$ instead of $[\mathrm{E}(X)]^2$.
• Using the wrong probabilities from the table.
• Rounding intermediate values too early.

Things to Be Careful About

• The mark scheme allows follow-through from an incorrect table, but the final answer 3.24 is exact (CAO).
• If your table is wrong, your variance may still earn method marks as long as you use the correct variance formula.
• $4.4^2 = 19.36$; be careful with the subtraction $22.6 - 19.36$.

Walkthrough

We start by recognising that the number of residents who classify the service as good in a sample of 110 follows a binomial distribution: $X \sim B(110, 0.25)$. Because $n$ is large, we can approximate this with a normal distribution. The mean of the approximation is $np = 27.5$ and the variance is $np(1-p) = 20.625$.

Since the binomial variable is discrete and we want $P(X < 22)$, we apply a continuity correction: $X < 22$ is equivalent to $X \le 21$, so we use 21.5 as the boundary. We then standardise using $Z = \frac{X - \mu}{\sigma}$, giving $z = \frac{21.5 - 27.5}{\sqrt{20.625}} = -1.3212$. The required probability is the area to the left of this z-score. Because the z-score is negative, we use symmetry: $P(Z < -1.3212) = 1 - \Phi(1.3212)$. From the table, $\Phi(1.3212) = 0.9068$, so the probability is $1 - 0.9068 = 0.0932$.

Key Takeaways

This question tests the normal approximation to the binomial distribution. The key skills are: checking that the approximation is appropriate, calculating the mean and variance correctly, applying the continuity correction, and standardising to use the normal table.

Common Mistakes

• Forgetting the continuity correction and using 22 instead of 21.5.
• Using the standard deviation instead of the variance in the standardisation formula, or vice versa.
• Reading the wrong tail of the normal distribution and giving a probability greater than 0.5.
• Not showing the mean and variance, which are required for the first mark.

Things to Be Careful About

The continuity correction is essential: $P(X < 22)$ becomes $P(X \le 21)$, so the boundary is 21.5. The standard deviation is $\sqrt{20.625}$, not 20.625. When the z-score is negative, remember to take $1 - \Phi(|z|)$. The final answer should be less than 0.5.

Walkthrough

First combine the probabilities for 'good' and 'satisfactory': $0.25 + 0.60 = 0.85$. The number of such residents in a sample of 10 is binomial, $Y \sim B(10, 0.85)$. We need $P(Y < 8)$, i.e. $P(Y \le 7)$. Computing all eight terms from 0 to 7 would be tedious, so use the complement: $P(Y \le 7) = 1 - P(Y \ge 8)$. The complement consists of exactly 8, 9 or 10 successes. Each is a binomial term: ${}^{10}C_k (0.85)^k (0.15)^{10-k}$. Sum these three terms and subtract from 1 to obtain 0.180.

Key Takeaways

This question tests the binomial distribution and the use of the complement rule. Recognising when to use the complement can greatly simplify the calculation. It also reinforces the formula $P(X = x) = {}^{n}C_x p^x (1-p)^{n-x}$.

Common Mistakes

• Forgetting to subtract from 1 and giving the probability of 8, 9 or 10 instead.
• Using $p = 0.25$ or $p = 0.60$ instead of $p = 0.85$.
• Misreading 'fewer than 8' as $\le 8$ instead of $\le 7$.
• Rounding intermediate terms too early, which can change the final answer.

Things to Be Careful About

'Fewer than 8' means $Y \le 7$, not $Y \le 8$. The complement is $P(Y \ge 8)$, which includes exactly 8, 9 and 10. Use the full binomial formula, not just $p^k$, because the ${}^{n}C_k$ factor and the $(1-p)^{n-k}$ factor are both needed. The final answer should be rounded to 0.180.

Walkthrough

We need the probability that among three randomly selected residents, one falls into each category. For a specific order, say good first, satisfactory second, poor third, the probability is $0.25 \times 0.60 \times 0.15$. However, the one good resident could be any of the three positions, so there are $3! = 6$ possible orders. Multiply by 6 to account for all of them: $0.25 \times 0.60 \times 0.15 \times 6 = 0.135$.

Key Takeaways

This question tests the multiplication law for independent events and the idea of counting arrangements. When selecting one item from each of several categories, multiply the individual probabilities and then multiply by the number of possible orderings.

Common Mistakes

• Forgetting to multiply by 6 and giving $0.0225$.
• Adding the probabilities instead of multiplying them.
• Using only one order and not considering the different arrangements.

Things to Be Careful About

The three selections are independent, so probabilities multiply. The number of arrangements is $3! = 6$, not 3. The final answer is $0.135$, which can also be written as $\frac{27}{200}$.

Walkthrough

We need to count distinct arrangements of the 10 letters in REGENERATE. If all 10 letters were different, there would be $10!$ arrangements. But the word has repeated letters: R appears twice and E appears four times. Swapping the two R's does not produce a new arrangement, and swapping any of the four E's does not produce a new arrangement. Therefore we divide by $2!$ for the repeated R's and by $4!$ for the repeated E's.

Key Takeaways

This question tests the standard formula for arrangements of a word with repeated letters. The key idea is that identical items should not be counted as different orders, so we divide by the factorial of each repeated count.

Common Mistakes

• Using $10!$ without dividing by the repeated-letter factorials.
• Dividing by $2!$ but forgetting to divide by $4!$, or vice versa.
• Treating the repeated letters as distinct, which overcounts.

Things to Be Careful About

Count the letters carefully: REGENERATE has 10 letters, with R repeated twice and E repeated four times. The total $75600$ will be used as the denominator in part (c).

Walkthrough

First make the four E's behave as one block, say $X$. This guarantees that the E's are together. Now we arrange seven units: $X$, R, R, G, N, A, T.

The condition says the two R's must have exactly 3 letters between them. The block $X$ represents 4 actual letters, so it cannot be one of the three letters between the R's; otherwise there would be too many letters between them. Therefore the three middle letters must be chosen from G, N, A and T. We choose and order three of these four letters: $P(4,3)=4!$.

Next, where can the two R's be placed among the seven units so that exactly three units lie between them? The possible position pairs are (1,5), (2,6) and (3,7), giving 3 choices. For each choice, the remaining one of G, N, A, T and the block X occupy the two outside positions, which can be done in $2!$ ways.

So the total is $4! \times 3 \times 2 = 144$. Since $3 \times 2 = 3!$, this matches the compact expression $4! \times 3!$.

Key Takeaways

This problem combines two ideas: treating identical letters as a single block, and imposing a positional restriction between two identical letters. It shows how to count arrangements by fixing positions rather than listing all cases.

Common Mistakes

• Forgetting that the E-block counts as 4 letters, so it cannot be placed between the R's.
• Forgetting that there are 3 possible position pairs for the R's.
• Treating the two R's as distinct and multiplying by an extra $2!$.
• Multiplying by $4!$ for the internal order of the E's; the E's are identical, so the block has only one internal arrangement.

Things to Be Careful About

The three letters between the R's must be actual letters, not units. Since $X$ represents four E's, it would make the gap too large. Also, the two R's are identical, so placing R at position 1 and R at position 5 is the same as the reverse; the position-pair method already accounts for this.

Walkthrough

For consonants and vowels to alternate, the 10 positions must be filled in one of two patterns: consonant-vowel-consonant-vowel... or vowel-consonant-vowel-consonant... . Since there are exactly 5 consonants and 5 vowels, both patterns use all letters.

For each pattern, arrange the 5 consonants in the consonant slots. The consonants are G, N, R, R, T, so the number of arrangements is $5!/2!$ because R is repeated. Similarly, the vowels are A, E, E, E, E, so the number of vowel arrangements is $5!/4!$ because E is repeated.

There are two patterns, so multiply by 2. This gives $600$ favourable arrangements. The total number of arrangements of all 10 letters is $75600$ from part (a). Therefore the probability is $600/75600 = 1/126 \approx 0.00794$.

Key Takeaways

This question links counting arrangements with probability. The numerator counts arrangements satisfying the alternating condition, and the denominator is the total number of equally likely arrangements. Repeated letters must be accounted for in both counts.

Common Mistakes

• Using $5!$ instead of $5!/2!$ for the consonants because R is repeated.
• Using $5!$ instead of $5!/4!$ for the vowels because E is repeated.
• Forgetting to multiply by 2 for the two alternating patterns.
• Using $10!$ as the denominator without accounting for repeated letters; this requires a different numerator ($5! \times 5! \times 2$) but gives the same probability.

Things to Be Careful About

There are exactly two alternating patterns because there are equal numbers of consonants and vowels. If the denominator is taken from part (a), the numerator must be $600$. If instead the denominator is $10!$, the numerator must be $28800$; either way the probability simplifies to $1/126$. The mark scheme accepts $600/75600$ as an equivalent final answer.