9709/52 May/June 2025 — Mathematics Solved Past Paper

Q13MMedium-EasyDiscrete Random Variables

Rachel has three coins. The first coin is biased so that the probability of obtaining a head when it is thrown is $\frac{1}{3}$ . The second coin is biased so that the probability of obtaining a head when it is thrown is $\frac{1}{4}$ . The third coin is fair.

Rachel throws the three coins at the same time. The random variable $X$ is the number of tails that she obtains.

Draw up the probability distribution table for $X$ .

DifficultyMedium-Easy

Worked solution

Approach

For each coin, write down the probability of tails (and heads). Since the coins are thrown independently, multiply the probabilities for each combination of outcomes. Group the combinations by the total number of tails to find $P(X=0)$ , $P(X=1)$ , $P(X=2)$ and $P(X=3)$ , then present them in a table.

Working

Let $T_1, T_2, T_3$ denote tails on coins 1, 2 and 3. Then

P(T_1)=\frac{2}{3},\quad P(H_1)=\frac{1}{3}

P(T_2)=\frac{3}{4},\quad P(H_2)=\frac{1}{4}

P(T_3)=\frac{1}{2},\quad P(H_3)=\frac{1}{2}

$X=0$ (all heads)

P(X=0)=P(H_1H_2H_3)=\frac{1}{3}\cdot\frac{1}{4}\cdot\frac{1}{2}=\frac{1}{24}

$X=1$ (exactly one tail)

\begin{aligned} P(X=1)&=P(T_1H_2H_3)+P(H_1T_2H_3)+P(H_1H_2T_3)\\ &=\frac{2}{3}\cdot\frac{1}{4}\cdot\frac{1}{2}+\frac{1}{3}\cdot\frac{3}{4}\cdot\frac{1}{2}+\frac{1}{3}\cdot\frac{1}{4}\cdot\frac{1}{2}\\ &=\frac{2}{24}+\frac{3}{24}+\frac{1}{24}=\frac{6}{24}=\frac{1}{4} \end{aligned}

$X=2$ (exactly two tails)

\begin{aligned} P(X=2)&=P(T_1T_2H_3)+P(T_1H_2T_3)+P(H_1T_2T_3)\\ &=\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{1}{2}+\frac{2}{3}\cdot\frac{1}{4}\cdot\frac{1}{2}+\frac{1}{3}\cdot\frac{3}{4}\cdot\frac{1}{2}\\ &=\frac{6}{24}+\frac{2}{24}+\frac{3}{24}=\frac{11}{24} \end{aligned}

$X=3$ (all tails)

P(X=3)=P(T_1T_2T_3)=\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{1}{2}=\frac{6}{24}=\frac{1}{4}

Check that the probabilities sum to 1:

\frac{1}{24}+\frac{6}{24}+\frac{11}{24}+\frac{6}{24}=\frac{24}{24}=1

Answer

$x$	0	1	2	3
$P(X=x)$	$\frac{1}{24}$	$\frac{6}{24}$	$\frac{11}{24}$	$\frac{6}{24}$
decimal	0.0417	0.25	0.4583	0.25

Final answer

P(X=0)=1/24, P(X=1)=1/4, P(X=2)=11/24, P(X=3)=1/4

Detailed explanation

Walkthrough

We need the probability distribution of $X$ , the number of tails. First identify the tail probability for each coin. Coin 1 has $P(H_1)=\frac{1}{3}$ , so $P(T_1)=\frac{2}{3}$ . Coin 2 has $P(H_2)=\frac{1}{4}$ , so $P(T_2)=\frac{3}{4}$ . Coin 3 is fair, so $P(H_3)=P(T_3)=\frac{1}{2}$ . Because the coins are thrown independently, the probability of any particular combination of outcomes is the product of the individual probabilities.

For $X=0$ , all coins must show heads. Multiply the three head probabilities:

P(X=0)=\frac{1}{3}\cdot\frac{1}{4}\cdot\frac{1}{2}=\frac{1}{24}

For $X=1$ , there are three mutually exclusive ways: coin 1 tail and coins 2 and 3 heads; coin 2 tail and coins 1 and 3 heads; coin 3 tail and coins 1 and 2 heads. Compute each product and add them:

P(X=1)=\frac{2}{24}+\frac{3}{24}+\frac{1}{24}=\frac{6}{24}=\frac{1}{4}

For $X=2$ , list the three ways with exactly two tails, compute each product and add:

P(X=2)=\frac{6}{24}+\frac{2}{24}+\frac{3}{24}=\frac{11}{24}

For $X=3$ , all coins must show tails:

P(X=3)=\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{1}{2}=\frac{6}{24}=\frac{1}{4}

Finally, check that the probabilities sum to 1:

\frac{1}{24}+\frac{6}{24}+\frac{11}{24}+\frac{6}{24}=1

This confirms that every possible value of $X$ has been accounted for.

Key Takeaways

A probability distribution table lists every possible value of a random variable together with its probability.
For independent events, multiply probabilities to find the probability of a combined outcome.
When several distinct outcomes give the same value of $X$ , add their probabilities because the outcomes are mutually exclusive.
The probabilities in any probability distribution must sum to 1.

Common Mistakes

Using the given head probabilities instead of the tail probabilities. Since $X$ counts tails, use $P(T_1)=\frac{2}{3}$ , $P(T_2)=\frac{3}{4}$ and $P(T_3)=\frac{1}{2}$ .
Missing one of the three arrangements for $X=1$ or $X=2$ . There are three distinct orders for exactly one tail and exactly two tails.
Adding probabilities without first multiplying the three coin probabilities for each outcome.
Not checking that the probabilities sum to 1, which can hide an arithmetic error.

Things to Be Careful About

The fair coin has $P(T_3)=\frac{1}{2}$ and $P(H_3)=\frac{1}{2}$ .
When presenting the table, make sure each probability is linked to the correct value of $x$ . The mark scheme accepts non-exact decimals correct to at least 3 significant figures.
The mark scheme allows probabilities not in the table if they are clearly identified, but a table is the clearest presentation.
Always verify the total probability is 1.

Techniques used

identify tail probabilities for each coinlist all possible tail-count outcomesmultiply independent probabilities for each outcomesum probabilities for each value of Xverify probabilities sum to 1

Approach

Let $X$ be the number of residents who own a bicycle. Since $X \sim \mathrm{B}(160, 0.7)$ and $n$ is large, we use the normal approximation $X \sim N(np, np(1-p))$ . A continuity correction is needed because we are approximating a discrete distribution with a continuous one.

Working

Calculate the mean:

\mu = np = 160 \times 0.7 = 112

Calculate the variance:

\sigma^2 = np(1-p) = 160 \times 0.7 \times 0.3 = 33.6

So the standard deviation is:

\sigma = \sqrt{33.6} = 5.79655\ldots

Apply the continuity correction. The discrete event $X > 120$ becomes $X > 120.5$ in the continuous approximation.

Standardise:

z = \frac{120.5 - 112}{\sqrt{33.6}} = \frac{8.5}{5.79655\ldots} = 1.466

Find the probability using the standard normal distribution table:

P(Z > 1.466) = 1 - \Phi(1.466) = 1 - 0.9287 = 0.0713

Answer

0.0713

Final answer

0.0713

Detailed explanation

Walkthrough

We start by identifying the distribution of the number of residents who own a bicycle. Since each resident either owns a bicycle or not, and the sample is random, $X$ follows a binomial distribution with $n = 160$ and $p = 0.7$ . Because $n$ is large and both $np = 112$ and $n(1-p) = 48$ are greater than 5, the binomial distribution can be approximated by a normal distribution with mean $np$ and variance $np(1-p)$ .

We calculate the mean and variance of this normal distribution. The mean is $112$ and the variance is $33.6$ , so the standard deviation is $\sqrt{33.6} \approx 5.7966$ .

Next, we apply the continuity correction. The question asks for the probability that more than 120 residents own a bicycle. In the discrete binomial distribution, this is $P(X > 120)$ . When we move to the continuous normal distribution, we must adjust the boundary by 0.5 to account for the gap between integer values. Since we want values strictly greater than 120, the continuous boundary becomes $120.5$ .

We then standardise this boundary value to a Z-score using the formula $z = \frac{x - \mu}{\sigma}$ . Substituting $x = 120.5$ , $\mu = 112$ , and $\sigma = \sqrt{33.6}$ gives $z = 1.466$ .

We look up the standard normal distribution table to find the probability that $Z$ is greater than 1.466. The table gives $\Phi(1.466) = 0.9287$ , which is the probability that $Z$ is less than 1.466. Since the total area under the normal curve is 1, the probability that $Z$ is greater than 1.466 is $1 - 0.9287 = 0.0713$ .

Key Takeaways

When $n$ is large and both $np$ and $n(1-p)$ are greater than 5, the binomial distribution $B(n, p)$ can be approximated by the normal distribution $N(np, np(1-p))$ .
The continuity correction is essential when approximating a discrete distribution with a continuous one. For $P(X > x)$ , use $x + 0.5$ ; for $P(X < x)$ , use $x - 0.5$ .
The standardisation formula $z = \frac{x - \mu}{\sigma}$ converts any normal distribution to the standard normal distribution, allowing the use of standard normal tables.

Common Mistakes

Forgetting to apply the continuity correction. Using $120$ instead of $120.5$ will give an incorrect probability.
Using the variance instead of the standard deviation in the standardisation formula. The standard deviation is $\sqrt{33.6}$ , not $33.6$ .
Not subtracting the tabulated probability from 1 when finding $P(Z > z)$ . The table gives $P(Z \leq z)$ , so $P(Z > z) = 1 - \Phi(z)$ .
Confusing the mean and variance: the mean is $112$ , and the variance is $33.6$ .

Things to Be Careful About

Always check that both $np > 5$ and $n(1-p) > 5$ before using the normal approximation.
Ensure the continuity correction is applied in the correct direction. For $P(X > 120)$ , use $120.5$ . For $P(X \geq 120)$ , use $119.5$ .
When reading the standard normal table, be careful about whether it gives $P(Z \leq z)$ or $P(Z \geq z)$ . Most tables give $P(Z \leq z)$ , so you must subtract from 1 for upper-tail probabilities.
The final answer must be a probability between 0 and 1. Since 120 is above the mean of 112, the probability should be less than 0.5; a value greater than 0.5 indicates a mistake.

Techniques used

calculate the mean and variance of a binomial distributionapply the continuity correctionstandardise to a Z-scoreuse normal distribution tables to find the probability

Approach

Use a two-stage probability tree. The outcomes 'both blue' and 'both red' are mutually exclusive, so multiply probabilities along each branch and then add the two branch probabilities.

Working

Let $B_1$ and $R_1$ denote the colours on the first draw, and let $B_2$ and $R_2$ denote the colours on the second draw.

Both marbles are the same colour when both are blue or both are red:

P(\text{both same}) = P(B_1 \cap B_2) + P(R_1 \cap R_2)

If the first marble is blue, it is replaced, so the second draw is again from 4 blue and 12 red marbles:

P(B_1 \cap B_2) = \frac{4}{16} \times \frac{4}{16}

If the first marble is red, it is not replaced, so the second draw is from 15 marbles, of which 11 are red:

P(R_1 \cap R_2) = \frac{12}{16} \times \frac{11}{15}

Therefore:

P(\text{both same}) = \frac{4}{16} \times \frac{4}{16} + \frac{12}{16} \times \frac{11}{15}

Simplify:

= \frac{1}{16} + \frac{11}{20} = \frac{5}{80} + \frac{44}{80} = \frac{49}{80}

Answer

P(\text{both same}) = \frac{49}{80} = 0.6125

Final answer

49/80 (0.6125)

Detailed explanation

Walkthrough

Model the two draws with a probability tree. From the first draw, $P(\text{blue}) = \frac{4}{16}$ and $P(\text{red}) = \frac{12}{16}$ . If the first marble is blue, it is put back, so the second draw still has 4 blue and 12 red marbles. If the first marble is red, it is not put back, so the second draw has 15 marbles remaining, with 11 red and 4 blue. The event 'both same colour' consists of the two disjoint outcomes 'blue then blue' and 'red then red'. Multiply along each branch to get the probability of each outcome, then add the two results because mutually exclusive events cannot happen together. This gives $\frac{4}{16}\times\frac{4}{16}$ for blue-blue and $\frac{12}{16}\times\frac{11}{15}$ for red-red. Adding and simplifying gives $\frac{49}{80}$ .

Key Takeaways

The probability tree keeps track of how the probabilities change after the first draw. A replacement causes the second draw to use the original composition, while a non-replacement reduces the total and removes one marble of the drawn colour. Mutually exclusive outcomes are combined by addition.

Common Mistakes

Using $\frac{12}{16}$ for the second red after a red has already been removed, instead of $\frac{11}{15}$ , is the most common error. Another common mistake is adding the two products without using a common denominator, or treating blue-blue and red-red as not mutually exclusive.

Things to Be Careful About

The replacement rule depends on the first marble: blue marbles are replaced, red marbles are not. Therefore the second-draw probabilities are different on the two branches. The final answer may be given as an exact fraction $\frac{49}{80}$ or as the exact decimal $0.6125$ .

Techniques used

multiply probabilities along dependent branchesadd probabilities of mutually exclusive outcomes

(b)

Find the probability that the first marble is blue given that the second marble is red.

3M Similar

DifficultyMedium

Worked solution

Approach

Use the conditional probability formula $P(F \mid S) = \frac{P(F \cap S)}{P(S)}$ . The numerator is the probability that the first marble is blue and the second is red; the denominator is the total probability that the second marble is red, found by adding the two paths ending in red.

Working

Let $F$ be the event that the first marble is blue, and let $S$ be the event that the second marble is red.

P(F \mid S) = \frac{P(F \cap S)}{P(S)}

The first marble is blue and replaced, so the second draw still has 12 red marbles out of 16:

P(F \cap S) = \frac{4}{16} \times \frac{12}{16}

The second marble is red in two ways: first blue then red, or first red then red:

P(S) = \frac{4}{16} \times \frac{12}{16} + \frac{12}{16} \times \frac{11}{15}

Hence:

P(F \mid S) = \frac{\frac{4}{16} \times \frac{12}{16}}{\frac{4}{16} \times \frac{12}{16} + \frac{12}{16} \times \frac{11}{15}}

Simplify the numerator:

\frac{4}{16} \times \frac{12}{16} = \frac{48}{256} = \frac{3}{16}

Simplify the denominator:

\frac{3}{16} + \frac{11}{20} = \frac{15}{80} + \frac{44}{80} = \frac{59}{80}

Therefore:

P(F \mid S) = \frac{\frac{3}{16}}{\frac{59}{80}} = \frac{3}{16} \times \frac{80}{59} = \frac{15}{59}

Answer

P(F \mid S) = \frac{15}{59} \approx 0.254

Final answer

15/59 (0.254)

Detailed explanation

Walkthrough

The question asks for $P(\text{first blue} \mid \text{second red})$ , a conditional probability. Start with the formula $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ . The numerator is the probability both events happen: first draw is blue and second draw is red. Since a blue is replaced, the second draw still has 12 red marbles out of 16, so the numerator is $\frac{4}{16} \times \frac{12}{16}$ . The denominator is the probability that the second marble is red. This can happen in two disjoint ways: first blue and second red, or first red and second red. The first of these is the numerator just found; the second is $\frac{12}{16} \times \frac{11}{15}$ because a red is not replaced. Add these to get $P(S)$ . Then divide the numerator by this denominator, simplifying to $\frac{15}{59}$ .

Key Takeaways

Conditional probability reverses the conditioning: knowing that the second marble is red changes the probability that the first was blue. The denominator is not 1 but the total probability of the condition occurring. This is a standard application of the conditional probability formula combined with the addition law for disjoint paths.

Common Mistakes

A common error is to use only the numerator as the answer, forgetting to divide by the probability that the second marble is red. Another is to omit the branch 'first red then red' from the denominator. Using $\frac{12}{16}$ for the second red after a red has been removed is also wrong; it should be $\frac{11}{15}$ .

Things to Be Careful About

The denominator in the conditional formula is $\frac{59}{80}$ , not $1$ or the probability from part (a) alone. Keep the exact fractions throughout; do not round until the final step. The exact answer is $\frac{15}{59}$ , which is approximately $0.254$ , and answers should be given to at least 3 significant figures if a decimal is used.

Techniques used

apply the conditional probability formulacalculate the numerator as a product of branch probabilitiescalculate the denominator as a sum of mutually exclusive pathssimplify a compound fraction

Q4Medium-EasyProbabilityDiscrete Random Variables

Vehicles approaching a certain road junction from Bromley must go either left, right or straight on. Over time, it is known that 30% turn left, 25% turn right and 45% go straight on. The driver of each vehicle chooses a direction independently of all other drivers.

(a)

Find the probability that the next three vehicles approaching this junction from Bromley all go in different directions.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

We need the probability that all three vehicles go in different directions. Since each vehicle chooses independently, the probability of a particular ordering, such as left, right, straight, is the product of the three individual probabilities. There are $3! = 6$ possible orderings of the three different directions, so we multiply by 6.

Working

For one particular ordering, e.g. left, right, straight:

P = 0.3 \times 0.25 \times 0.45

There are $3! = 6$ possible orderings, so:

P = 6 \times 0.3 \times 0.25 \times 0.45

Compute step by step:

0.3 \times 0.25 = 0.075

0.075 \times 0.45 = 0.03375

6 \times 0.03375 = 0.2025

Answer

P = 0.2025 = \frac{81}{400}

Final answer

0.2025

Detailed explanation

Walkthrough

The problem asks for the probability that all three vehicles go to different directions. Because the drivers choose independently, we can multiply the individual probabilities. Consider one fixed ordering, such as left, right, straight: its probability is $0.3 \times 0.25 \times 0.45$ . There are $3! = 6$ orderings of the three distinct directions, and each has the same probability, so we multiply by 6.

Key Takeaways

Independent events allow probabilities to be multiplied.
When order matters, multiply by the number of permutations.
The result can be written exactly as a fraction or a decimal.

Common Mistakes

Forgetting to multiply by $6$ , which gives only the probability of one specific order.
Using $3^3$ or $3^2$ instead of $3!$ for the number of orderings.
Rounding intermediate products too early.

Things to Be Careful About

The three directions have different probabilities, so the product is not $3! \times p^3$ but $3! \times 0.3 \times 0.25 \times 0.45$ . The multiplication by $3!$ is necessary because the three vehicles are distinct and any assignment of the three directions to them is allowed.

Techniques used

multiply independent probabilitiesmultiply by the number of permutations of the three directionscombine probabilities for a single ordering

(b)

Find the probability that, from the vehicles approaching this junction from Bromley today, the 1st vehicle to go left is before the 9th vehicle.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The first left turn occurs before the 9th vehicle if at least one of the first 8 vehicles turns left. The complement is that none of the first 8 vehicles turns left. Since each vehicle independently turns left with probability $0.3$ , the probability that a vehicle does not turn left is $1 - 0.3 = 0.7$ . Hence the probability that none of the first 8 turns left is $0.7^8$ , and the required probability is $1 - 0.7^8$ .

Working

Probability a vehicle does not turn left:

0.7

Probability none of the first 8 turns left:

0.7^8

Therefore:

P = 1 - 0.7^8

Compute:

0.7^8 = 0.05764801

P = 1 - 0.05764801 = 0.94235199

Answer

P \approx 0.942

Final answer

0.942

Detailed explanation

Walkthrough

The first left turn must happen before the 9th vehicle, meaning at least one of the first 8 vehicles must turn left. It is easier to consider the complement: none of the first 8 vehicles turns left. Each vehicle turns left with probability 0.3, so it does not turn left with probability 0.7. Because the choices are independent, the probability that all 8 do not turn left is $0.7^8$ . Subtracting this from 1 gives the probability that at least one of the first 8 turns left.

Key Takeaways

The first success before a given trial is a geometric distribution problem.
Using the complement is often simpler than summing a geometric series.
The probability of "at least one" is $1 - P(\text{none})$ .

Common Mistakes

Using $0.3^8$ instead of $0.7^8$ for the complement.
Forgetting to subtract from 1, giving $0.7^8$ as the answer.
Confusing "before the 9th" with "on the 9th" or "by the 9th".

Things to Be Careful About

"Before the 9th" means the first left must occur among the first 8 vehicles.
The complement is "no left among the first 8", not "no left among the first 9".
Round the final answer to 3 significant figures, giving 0.942.

Techniques used

recognise a geometric distribution for the first successuse the complement of no success in the first 8 trialscompute the complement probability

(c)

Find the probability that, from the vehicles approaching this junction from Bromley today, the 2nd vehicle to go left is the 7th vehicle.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

We need the 2nd left turn to be the 7th vehicle. This means exactly one of the first 6 vehicles goes left, and the 7th vehicle also goes left. The number of ways to choose which of the first 6 vehicles is left is $\binom{6}{1} = 6$ . For each such arrangement, the probability is $0.3^2 \times 0.7^5$ : two left turns (one in the first six and the 7th) and five non-left turns (the other five of the first six).

Working

Number of ways to choose the left-turning vehicle among the first six:

\binom{6}{1} = 6

Probability of exactly one left in the first six and a left on the 7th:

6 \times (0.3)^2 (0.7)^5

Compute:

(0.3)^2 = 0.09

(0.7)^5 = 0.16807

0.09 \times 0.16807 = 0.0151263

6 \times 0.0151263 = 0.0907578

Answer

P \approx 0.0908

Final answer

0.0908

Detailed explanation

Walkthrough

We want the 2nd left turn to happen exactly on the 7th vehicle. Therefore, among the first 6 vehicles there must be exactly one left turn, and the 7th vehicle must also be left. The other five of the first six must not turn left. First, choose which of the first 6 vehicles is the left turn: there are $\binom{6}{1} = 6$ choices. For each choice, the probability is $0.3^2 \times 0.7^5$ : $0.3$ for the left turn in the first six, $0.7^5$ for the five non-left turns, and another $0.3$ for the 7th vehicle. Multiplying by the 6 choices gives the required probability.

Key Takeaways

This is a negative binomial type problem: the 2nd success occurs on the 7th trial.
The binomial coefficient counts the possible positions of the successes before the final trial.
The final trial must be a success, so its probability is included separately.

Common Mistakes

Forgetting to multiply by 6, which counts the possible positions of the first left turn.
Using $0.7^6$ instead of $0.7^5$ for the non-left turns.
Treating the 7th vehicle as not necessarily left, which would give $6 \times 0.3 \times 0.7^6$ .

Things to Be Careful About

The 7th vehicle must be left, so it contributes a factor of $0.3$ .
Among the first 6 vehicles, exactly one is left, so the other five contribute $0.7^5$ .
The binomial coefficient is $\binom{6}{1} = 6$ , not $\binom{6}{2}$ .

Techniques used

choose the position of the first left among the first six vehiclesmultiply the probabilities for two lefts and five non-leftsapply the binomial coefficient

Q5MediumRepresentation of Data

The times taken, $t$ minutes, by 300 students to travel to Hollowton College are recorded. The results are summarised in the table below.

Time ( $t$ minutes)	$t \leq 10$	$t \leq 20$	$t \leq 30$	$t \leq 40$	$t \leq 60$	$t \leq 90$
Cumulative frequency	34	86	142	208	265	300

(a)

On the grid, draw a cumulative frequency graph to illustrate this information.

Cumulative Frequency Grid

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Identify the coordinates to plot from the cumulative frequency table. The points are $(t, \text{cf})$ : $(10, 34)$ , $(20, 86)$ , $(30, 142)$ , $(40, 208)$ , $(60, 265)$ , $(90, 300)$ . Also include the origin $(0, 0)$ . Plot these points on the grid and join them with a smooth curve (ogive).

Working

Coordinates to plot:

\begin{aligned} &(0, 0) \\ &(10, 34) \\ &(20, 86) \\ &(30, 142) \\ &(40, 208) \\ &(60, 265) \\ &(90, 300) \end{aligned}

Graph construction:

Draw axes: horizontal axis for time $t$ (minutes) from 0 to 90, vertical axis for cumulative frequency from 0 to 300.
Plot the 6 points listed above.
Draw a smooth increasing curve passing through all points and starting at $(0,0)$ . Do not use straight line segments.

Answer

Cumulative frequency graph drawn with points $(0,0)$ , $(10,34)$ , $(20,86)$ , $(30,142)$ , $(40,208)$ , $(60,265)$ , $(90,300)$ joined by a smooth curve.

Final answer

Cumulative frequency graph drawn with points (0,0), (10,34), (20,86), (30,142), (40,208), (60,265), (90,300) joined by a smooth curve.

Detailed explanation

Walkthrough

First, we extract the data points from the cumulative frequency table. The table gives the number of students who take at most $t$ minutes. This directly gives us the cumulative frequency at the upper boundary of each class interval. We also know that at $t=0$ , the cumulative frequency is 0. So we plot the points $(0, 0)$ , $(10, 34)$ , $(20, 86)$ , $(30, 142)$ , $(40, 208)$ , $(60, 265)$ , and $(90, 300)$ . After plotting, we draw a smooth curve (an ogive) through these points. The curve must be increasing and smooth, not made of straight line segments.

Key Takeaways

A cumulative frequency graph (ogive) plots cumulative frequency against the upper class boundary.
The curve must be smooth and increasing.
Always include the starting point $(0, 0)$ if the data starts from 0.

Common Mistakes

Drawing straight lines between points instead of a smooth curve.
Forgetting to plot the origin $(0,0)$ .
Incorrectly scaling the axes.

Things to Be Careful About

Ensure axes are labelled: time $t$ (minutes) and cumulative frequency.
The curve must be smooth; bar charts or histograms are not accepted for cumulative frequency graphs.

Techniques used

plot cumulative frequency pointsdraw smooth cumulative frequency curve

(b)

120 students take more than $k$ minutes to travel to college. Use your graph to estimate the value of $k$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The problem states that 120 students take more than $k$ minutes. Since the total number of students is 300, the number of students who take at most $k$ minutes (the cumulative frequency at $k$ ) is $300 - 120 = 180$ . We use the cumulative frequency graph to find the time $t$ corresponding to a cumulative frequency of 180.

Working

Step 1: Find the target cumulative frequency.

\text{cf} = 300 - 120 = 180

Step 2: Read from the graph.

Draw a horizontal line from $\text{cf} = 180$ on the vertical axis to meet the curve.
From the intersection point, draw a vertical line down to the horizontal axis to read the value of $t$ .
Looking at the graph, $\text{cf} = 180$ lies between $\text{cf} = 142$ (at $t=30$ ) and $\text{cf} = 208$ (at $t=40$ ).
Reading from the curve, the corresponding time is approximately $t = 35$ .

Answer

k = 35

Final answer

k = 35

Detailed explanation

Walkthrough

We are given that 120 students take more than $k$ minutes. The cumulative frequency graph shows the number of students who take at most $t$ minutes. Therefore, we need to find the cumulative frequency value that corresponds to the remaining students: $300 - 120 = 180$ . We locate 180 on the cumulative frequency (vertical) axis, move horizontally to the curve, and then move vertically down to read the time $k$ on the horizontal axis. From the graph, this value is approximately 35 minutes.

Key Takeaways

'More than' means we subtract from the total to get the 'at most' (cumulative) frequency.
Always read the graph correctly: horizontal from cf axis, vertical from curve to time axis.

Common Mistakes

Reading 120 directly from the cf axis instead of calculating $300 - 120 = 180$ .
Reading the axes in the wrong direction.

Things to Be Careful About

The value read from the graph is an estimate. Values around 35 are acceptable (e.g., 34-36 depending on graph accuracy).

Techniques used

calculate complementary cumulative frequencyread value from cumulative frequency graph

(c)

Calculate estimates of the mean and standard deviation of the times taken to travel to college by the 300 students.

6M Similar

DifficultyMedium

Worked solution

Approach

To estimate the mean and standard deviation for grouped data, we first need the frequency for each class interval and the midpoint of each class. Then we apply the formulas:

\text{Mean} = \frac{\sum f \cdot m}{\sum f}

\text{Variance} = \frac{\sum f \cdot m^2}{\sum f} - (\text{Mean})^2

\text{Standard deviation} = \sqrt{\text{Variance}}

Working

Step 1: Find frequencies and midpoints.

Time $t$ (min)	Cumulative Frequency	Frequency $f$	Midpoint $m$	$f \cdot m$	$f \cdot m^2$
$t \leq 10$	34	34	5	170	850
$t \leq 20$	86	$86 - 34 = 52$	15	780	11700
$t \leq 30$	142	$142 - 86 = 56$	25	1400	35000
$t \leq 40$	208	$208 - 142 = 66$	35	2310	80850
$t \leq 60$	265	$265 - 208 = 57$	50	2850	142500
$t \leq 90$	300	$300 - 265 = 35$	75	2625	196875
Total	300	300		10135	467775

Step 2: Calculate the mean.

\text{Mean} = \frac{\sum f \cdot m}{\sum f} = \frac{10135}{300} = 33.7833\ldots \approx 33.8 \text{ minutes}

Step 3: Calculate the variance.

\text{Variance} = \frac{\sum f \cdot m^2}{\sum f} - (\text{Mean})^2

\text{Variance} = \frac{467775}{300} - \left(\frac{10135}{300}\right)^2

\text{Variance} = 1559.25 - 1141.3102\ldots = 417.9398\ldots

Step 4: Calculate the standard deviation.

\text{Standard deviation} = \sqrt{417.9398\ldots} = 20.4435\ldots \approx 20.4 \text{ minutes}

Answer

Mean $= 33.8$ minutes, Standard deviation $= 20.4$ minutes.

Final answer

Mean = 33.8 minutes, Standard deviation = 20.4 minutes

Detailed explanation

Walkthrough

First, we determine the frequency for each class by subtracting consecutive cumulative frequencies. For example, the frequency for $10 < t \leq 20$ is $86 - 34 = 52$ . Next, we find the midpoint $m$ of each class interval: $(0+10)/2 = 5$ , $(10+20)/2 = 15$ , etc. Note that the last class is $60 < t \leq 90$ , so its midpoint is $(60+90)/2 = 75$ . We then calculate $f \cdot m$ and $f \cdot m^2$ for each row, sum them up, and apply the formulas. The mean is $\sum(f \cdot m) / \sum f = 10135 / 300 \approx 33.8$ . The variance is $\sum(f \cdot m^2) / \sum f - \text{mean}^2 = 1559.25 - 33.7833^2 \approx 417.94$ . The standard deviation is the square root of the variance, $\sqrt{417.94} \approx 20.4$ .

Key Takeaways

Frequencies for grouped data from a cumulative frequency table are found by subtraction.
Midpoints are used as representative values for each class.
The variance formula $\frac{\sum f m^2}{\sum f} - \bar{x}^2$ is efficient for calculation.

Common Mistakes

Using upper or lower class boundaries instead of midpoints.
Incorrectly calculating frequencies (e.g., using cumulative frequencies directly as frequencies).
Rounding the mean too early before calculating the variance, leading to rounding errors.

Things to Be Careful About

The last class interval is $60 < t \leq 90$ , which has a width of 30, not 10. Its midpoint is 75, not 65.
Keep full precision in intermediate calculations (use fractions or at least 4-5 decimal places) to avoid rounding errors in the final standard deviation.

Techniques used

calculate class frequencies from cumulative frequenciesfind class midpointsapply mean formula for grouped dataapply variance formula for grouped data

Q6MediumPermutations and Combinations

(a)

Find the number of different ways in which the 10 letters in the word AMALGAMATE can be arranged so that there is an M at the beginning, an M at the end and no As are together.

3M Similar

DifficultyMedium

Worked solution

Approach

Fix the two M's at the ends. The remaining 8 positions contain 4 identical A's and 4 distinct letters (L, G, T, E). To ensure no two A's are adjacent, arrange the 4 distinct letters first, then place the 4 A's into the 5 gaps around them.

Working

With the M's fixed, the arrangement has the form M _ _ _ _ _ _ _ _ M.

Arrange the 4 distinct letters L, G, T, E:

4! = 24

These 4 letters create 5 gaps: before the first letter, between each pair, and after the last letter. Choose 4 of these 5 gaps for the identical A's:

{}^5C_4 = 5

Total number of arrangements:

4! \times {}^5C_4 = 24 \times 5 = 120

Final answer

120

Detailed explanation

Walkthrough

The word AMALGAMATE has 4 A's, 2 M's and four single letters L, G, T, E. The condition fixes an M at each end, so the two M's are no longer free to move. We are left with 8 middle positions to fill with 4 identical A's and 4 distinct letters.

The key idea is the gap method. If we arrange the 4 distinct letters first, there are 5 gaps around them: one before the first letter, one between each pair, and one after the last letter. Placing at most one A in each gap guarantees no two A's are adjacent. Since we need to place all 4 A's, we choose 4 of the 5 gaps. The A's are identical, so their order does not matter; this is a combination. Then multiply by 4! for arranging the distinct letters.

Key Takeaways

Identical letters reduce the number of arrangements: divide by the factorial of the repetition count.
The gap method is the standard tool for ensuring no two identical objects are together.
Fixing letters at specified positions removes them from the rearrangement.

Common Mistakes

Forgetting that the A's are identical and multiplying by 4! for their order.
Choosing gaps as ${}^5C_4$ but then also arranging the A's in $4!$ ways, which overcounts.
Forgetting that the two M's are fixed, and counting arrangements of the M's too.

Things to Be Careful About

The 4 non-M letters are all distinct, so they contribute $4!$ , not a fraction.
${}^5C_4 = {}^5C_1 = 5$ ; either form is acceptable in the mark scheme.
The final answer is an integer count, not a probability.

Techniques used

fix identical M letters at the endsarrange distinct remaining lettersuse the gap method to place identical A's non-adjacently

(b)

Find the number of different ways in which the 10 letters in the word AMALGAMATE can be arranged with exactly 3 letters between the two Ms.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The two M's must have exactly 3 letters between them, so their positions differ by 4. Count the possible placements of the M pair, then arrange the remaining 8 letters (4 identical A's and 4 distinct letters) in the 8 remaining positions.

Working

If the first M is in position $i$ , the second M must be in position $i+4$ . For the pair to fit in a 10-letter line:

i = 1, 2, 3, 4, 5, 6

so there are 6 possible placements.

For each placement, the remaining 8 positions are filled with 4 identical A's and L, G, T, E:

\frac{8!}{4!} = 1680

Total number of arrangements:

6 \times \frac{8!}{4!} = 6 \times 1680 = 10080

Final answer

10080

Detailed explanation

Walkthrough

We need exactly 3 letters between the two M's. If the first M is in position $i$ , the second M must be in position $i+4$ . Since there are 10 positions, $i$ can be 1 through 6, giving 6 possible placements of the M's.

For each placement, the remaining 8 positions must be filled with the 4 A's and the letters L, G, T, E. The A's are identical, so the number of arrangements is $\frac{8!}{4!}$ . Multiplying by the 6 possible M placements gives the total.

Key Takeaways

For identical repeated letters, divide by the factorial of the repetition count.
"Exactly k letters between two identical items" translates to a fixed position difference.
Counting positions first, then arranging the remaining letters, avoids overcounting.

Common Mistakes

Multiplying by 2! for the two M's: they are identical, so no extra factor is needed.
Counting 7 placements instead of 6: positions $i$ and $i+4$ allow $i = 1, 2, 3, 4, 5, 6$ .
Forgetting to divide by 4! for the repeated A's.

Things to Be Careful About

The 8 remaining letters include 4 identical A's and 4 distinct letters, so the arrangement count is $\frac{8!}{4!}$ .
The 6 placements are independent of the arrangement of the remaining letters.

Techniques used

count possible placements of the separated M pairarrange remaining letters with repeated identical A's

(c)

Five letters are selected from the 10 letters in the word AMALGAMATE.

Find the number of different selections in which the five letters include at least one M and at least two As.

3M Similar

DifficultyMedium

Worked solution

Approach

Count unordered selections of 5 letters. Since at least one M and at least two A's are needed, split into cases by the number of M's (1 or 2) and the number of A's (2, 3 or 4), then choose the remaining letters from the 4 distinct letters L, G, T, E.

Working

Case 1: two M's and three A's:

\text{MMAAA}: 1

Case 2: two M's, two A's and one other letter:

\text{MMAA} + {}^4C_1 = 4

Case 3: one M, two A's and two other letters:

\text{MAA} + {}^4C_2 = 6

Case 4: one M, three A's and one other letter:

\text{MAAA} + {}^4C_1 = 4

Case 5: one M and four A's:

\text{MAAAA}: 1

Total number of selections:

1 + 4 + 6 + 4 + 1 = 16

Final answer

16

Detailed explanation

Walkthrough

We are selecting 5 letters, not arranging them, so order does not matter. The multiset contains 4 A's, 2 M's, and 4 distinct letters L, G, T, E. We need at least one M and at least two A's.

Since only 5 letters are selected, the number of M's can be 1 or 2, and the number of A's can be 2, 3 or 4. Enumerate every combination that satisfies both conditions:

Two M's and three A's: 1 way.
Two M's, two A's and one other letter: choose 1 of the 4 distinct letters.
One M, two A's and two other letters: choose 2 of the 4 distinct letters.
One M, three A's and one other letter: choose 1 of the 4 distinct letters.
One M and four A's: 1 way.

Adding these cases gives the total number of selections.

Key Takeaways

"Selections" means combinations, not permutations.
When a word has repeated letters, count distinct choices by the number of each repeated letter.
The four non-repeated letters are distinct, so choosing $r$ of them gives ${}^4C_r$ .

Common Mistakes

Treating selections as arrangements and multiplying by factorials.
Forgetting cases such as MAAAA or MMAAA.
Counting the identical A's and M's as distinguishable.
Double-counting by choosing "at least one M" and "at least two A's" separately instead of using cases.

Things to Be Careful About

The 4 distinct letters are L, G, T, E; there are no other distinct letters.
The maximum number of A's is 4 and the maximum number of M's is 2, so the cases are limited.
The mark scheme also accepts the alternative method using MAA, MAAA and MAAAA with choices from the 4 distinct letters.

Techniques used

enumerate cases by numbers of M's and A'schoose remaining distinct letters using combinationssum the case counts

Q7MediumThe Normal DistributionDiscrete Random Variables

Kestrels are birds whose adult wingspans are normally distributed with mean 74.8 cm and standard deviation 3.2 cm. A random sample of 120 adult kestrels is selected.

(a)

How many of these 120 adult kestrels would you expect to have wingspan between 72.4 cm and 76.3 cm?

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Standardize both wingspan limits to $z$ -scores using the mean $\mu = 74.8$ and standard deviation $\sigma = 3.2$ , then use the standard normal table to find the probability that $Z$ lies between the two $z$ -values. Multiply this probability by the sample size 120 to obtain the expected number of kestrels.

Working

Standardize the lower limit $x = 72.4$ :

z_1 = \frac{72.4 - 74.8}{3.2} = \frac{-2.4}{3.2} = -0.75

Standardize the upper limit $x = 76.3$ :

z_2 = \frac{76.3 - 74.8}{3.2} = \frac{1.5}{3.2} = 0.46875

The required probability is:

P(72.4 < X < 76.3) = P(-0.75 < Z < 0.46875)

Since the standard normal table gives cumulative probabilities for positive $z$ , write:

P(-0.75 < Z < 0.46875) = \Phi(0.46875) + \Phi(0.75) - 1

From the standard normal table:

\Phi(0.75) = 0.7734, \quad \Phi(0.46875) \approx 0.6804

Therefore:

P = 0.7734 + 0.6804 - 1 = 0.4538

Expected number:

0.4538 \times 120 = 54.456

So the expected number is 54 (or 55).

Answer

0.4538 \times 120 = 54.456

So the expected number is 54 (or 55).

Final answer

54 (or 55)

Detailed explanation

Walkthrough

We are given that the wingspans are normally distributed with mean 74.8 cm and standard deviation 3.2 cm. To find the probability that a wingspan is between 72.4 cm and 76.3 cm, we standardize each limit.

The lower limit 72.4 is 2.4 cm below the mean, which is $\frac{2.4}{3.2} = 0.75$ standard deviations below the mean, so $z_1 = -0.75$ . The upper limit 76.3 is 1.5 cm above the mean, which is $\frac{1.5}{3.2} = 0.46875$ standard deviations above the mean, so $z_2 = 0.46875$ .

We want the area between these two $z$ -values. Since the standard normal table gives cumulative probabilities to the left, we compute $\Phi(0.46875)$ and $\Phi(0.75)$ , add them, and subtract 1 to avoid double-counting the area on the left of $-0.75$ .

The table gives $\Phi(0.75) = 0.7734$ and $\Phi(0.46875) \approx 0.6804$ . Adding and subtracting 1 gives $0.4538$ .

Finally, the expected number in a sample of 120 is $0.4538 \times 120 = 54.456$ , which rounds to 54 (or 55).

Key Takeaways

The standardisation formula $z = \frac{x - \mu}{\sigma}$ converts a normal value into a standard normal $z$ -score.

The probability of a value between two limits is the difference of the cumulative probabilities.

The expected count in a sample is the probability multiplied by the sample size.

Common Mistakes

Forgetting to subtract 1 when combining $\Phi$ values on opposite sides of the mean.

Using the $z$ -values as probabilities directly.

Giving a non-integer answer for the expected number.

Things to Be Careful About

The probability must be between 0 and 1; the mark scheme expects a value less than 0.5.

The expected number must be a single integer (54 or 55).

The mark scheme awards marks for the method and the probability area calculation, so show every step.

Techniques used

standardize values to z-scoresuse the standard normal tableconvert probability to expected count

(b)

The masses of adult kestrels are normally distributed with mean $\mu$ kg and standard deviation $\sigma$ kg. It is known that 20% of adult kestrels have mass greater than 0.202 kg and 28% have mass less than 0.185 kg.

Find the value of $\mu$ and the value of $\sigma$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Convert the two known percentage conditions into $z$ -scores using the standard normal table, then write the standardisation formula for each condition and solve the resulting simultaneous equations for $\mu$ and $\sigma$ .

Working

Since 20% of adult kestrels have mass greater than 0.202 kg:

P(X > 0.202) = 0.20 \quad \Rightarrow \quad P(Z < z_1) = 0.80

So $z_1 = 0.842$ .

Using the standardisation formula:

\frac{0.202 - \mu}{\sigma} = 0.842

Since 28% have mass less than 0.185 kg:

P(X < 0.185) = 0.28 \quad \Rightarrow \quad z_2 = -0.583

Thus:

\frac{0.185 - \mu}{\sigma} = -0.583

Rearrange both equations:

0.202 - \mu = 0.842\sigma

0.185 - \mu = -0.583\sigma

Subtract the second equation from the first:

0.017 = 1.425\sigma

\sigma = \frac{0.017}{1.425} = 0.0119

Substitute back to find $\mu$ :

\mu = 0.202 - 0.842 \times 0.0119 = 0.192

Answer

\mu = 0.192 \text{ kg}, \quad \sigma = 0.0119 \text{ kg}

Final answer

mu = 0.192 kg, sigma = 0.0119 kg

Detailed explanation

Walkthrough

We are told that 20% of adult kestrels have mass greater than 0.202 kg. This means the $z$ -score for 0.202 kg has a cumulative probability of 0.80 to its left. The standard normal table gives $z = 0.842$ for a cumulative probability of 0.80.

We are also told that 28% have mass less than 0.185 kg, so the $z$ -score for 0.185 kg has a cumulative probability of 0.28, which is $z = -0.583$ .

Using the standardisation formula $z = \frac{x - \mu}{\sigma}$ , we write the two equations:

\frac{0.202 - \mu}{\sigma} = 0.842, \quad \frac{0.185 - \mu}{\sigma} = -0.583

Rearranging gives two linear equations in $\mu$ and $\sigma$ . Subtracting the second from the first eliminates $\mu$ and gives $\sigma = 0.0119$ . Substituting back gives $\mu = 0.192$ .

Key Takeaways

Percentages of a normal distribution correspond to $z$ -scores through the standard normal table.

The standardisation formula links a value, the mean, and the standard deviation.

Two conditions give two equations, which can be solved simultaneously for $\mu$ and $\sigma$ .

Common Mistakes

Using the wrong $z$ -value (e.g., 0.20 instead of 0.842).

Sign errors in the equations (e.g., using a positive $z$ -value for the 28% condition).

Solving the equations incorrectly or losing a value.

Things to Be Careful About

The $z$ -value for a cumulative probability less than 0.5 is negative.

The mark scheme requires consistency of signs in the solution.

The final values should be rounded to 3 significant figures: $\mu = 0.192$ , $\sigma = 0.0119$ .

Techniques used

convert percentages to z-valuesset up simultaneous equationssolve for mean and standard deviation

(c)

10 adult kestrels are selected at random.

Find the probability that fewer than 3 have masses greater than 0.202 kg.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

From part (b), the probability that a randomly selected adult kestrel has mass greater than 0.202 kg is $p = 0.20$ . With $n = 10$ kestrels, the number with mass greater than 0.202 kg follows $X \sim \text{B}(10, 0.2)$ . We need $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$ .

Working

Using the binomial formula $P(X = x) = {}^{10}C_x \, (0.2)^x (0.8)^{10-x}$ :

P(X = 0) = 0.8^{10} = 0.107374

P(X = 1) = {}^{10}C_1 \, (0.2)^1 (0.8)^9 = 10 \times 0.2 \times 0.134218 = 0.268435

P(X = 2) = {}^{10}C_2 \, (0.2)^2 (0.8)^8 = 45 \times 0.04 \times 0.167772 = 0.301989

Summing:

P(X < 3) = 0.107374 + 0.268435 + 0.301989 = 0.677798 \approx 0.678

Answer

0.678

Final answer

0.678

Detailed explanation

Walkthrough

From part (b), the probability that an adult kestrel has mass greater than 0.202 kg is 0.20. When 10 kestrels are selected at random, the number with mass greater than 0.202 kg follows a binomial distribution with $n = 10$ and $p = 0.2$ .

"Fewer than 3" means 0, 1, or 2. We compute each probability using the binomial formula:

P(X = k) = {}^{10}C_k \, (0.2)^k (0.8)^{10-k}

Adding the three probabilities gives $0.107374 + 0.268435 + 0.301989 = 0.677798$ , which rounds to 0.678.

Key Takeaways

The number of successes in a fixed number of independent trials follows a binomial distribution.

"Fewer than 3" means $X = 0, 1, 2$ , not including $X = 3$ .

Binomial probabilities are computed with the formula $P(X = x) = {}^{n}C_x \, p^x (1-p)^{n-x}$ .

Common Mistakes

Using $p = 0.8$ instead of $p = 0.2$ (confusing the success and failure probabilities).

Forgetting to include $X = 0$ in the sum.

Using $P(X \le 3)$ instead of $P(X < 3)$ .

Things to Be Careful About

The answer should be rounded to 3 significant figures: 0.678.

The mark scheme accepts any value in the range 0.677 to 0.678.

The binomial terms must show the combination, the power of $p$ , and the power of $(1-p)$ clearly.

Techniques used

identify binomial distributioncompute individual binomial probabilitiessum probabilities for fewer than 3

Mathematics 9709/52 — May/June 2025

Approach

Working

X=0X=0X=0 (all heads)

X=1X=1X=1 (exactly one tail)

X=2X=2X=2 (exactly two tails)

X=3X=3X=3 (all tails)

Answer

Walkthrough

Key Takeaways

Common Mistakes

Things to Be Careful About

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Working

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Working

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Working

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Working

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Working

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Working

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Working

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Working

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Working

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Approach

Working

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Key Takeaways

Common Mistakes

$X=0$ (all heads)

$X=1$ (exactly one tail)

$X=2$ (exactly two tails)

$X=3$ (all tails)