9709/51 Oct/Nov 2024 — Mathematics Solved Past Paper

Q1MediumDiscrete Random Variables

Nicola throws an ordinary fair six-sided dice. The random variable $X$ is the number of throws that she takes to obtain a 6.

(a)

Find $P(X < 8)$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $X$ is the number of throws until the first 6, $X$ has a geometric distribution with success probability $p = \frac{1}{6}$ . The event $X < 8$ means the first 6 appears on one of the first 7 throws. It is easier to use the complement: $X \geq 8$ means no 6 appears in the first 7 throws.

Working

\mathrm{P}(X < 8) = 1 - \mathrm{P}(X \geq 8)

No 6 in any of the first 7 throws has probability:

\mathrm{P}(X \geq 8) = \left(\frac{5}{6}\right)^7

Therefore:

\mathrm{P}(X < 8) = 1 - \left(\frac{5}{6}\right)^7

= 1 - \frac{78125}{279936} = \frac{201811}{279936} = 0.720918\ldots

Answer

\mathrm{P}(X < 8) = 0.721

Final answer

0.721

Detailed explanation

Walkthrough

Nicola keeps throwing until she gets a 6, so each throw is independent with probability $\frac{1}{6}$ of success and $\frac{5}{6}$ of failure. The random variable $X$ is geometric. For $\mathrm{P}(X < 8)$ , we need the first 6 to happen on throw 1, 2, 3, 4, 5, 6, or 7. Instead of adding seven geometric probabilities, use the complement: $X \geq 8$ means she fails to get a 6 on all of the first 7 throws. That probability is $\left(\frac{5}{6}\right)^7$ . Subtracting from 1 gives the required probability. The exact value is $\frac{201811}{279936}$ , which rounds to 0.721.

Key Takeaways

A waiting time for the first success is modelled by a geometric distribution.
The complement is often quicker than summing many terms.
$\mathrm{P}(X < 8)$ includes throws 1 through 7, not 1 through 8.

Common Mistakes

Using $1 - \left(\frac{5}{6}\right)^8$ instead of $1 - \left(\frac{5}{6}\right)^7$ , which would calculate $\mathrm{P}(X < 9)$ .
Adding only six terms instead of seven.
Forgetting to subtract from 1 and giving $\left(\frac{5}{6}\right)^7$ .
Writing only the decimal without showing the complement method may lose method marks; the mark scheme gives only special-case credit for an unsupported correct value.

Things to Be Careful About

The inequality is strict: $X < 8$ excludes $X = 8$ .
The complement event is no 6 in the first 7 throws.
Give the answer to 3 significant figures as 0.721, or the exact fraction $\frac{201811}{279936}$ .

Techniques used

model the waiting time with a geometric distributionapply the complement ruleevaluate a power and subtract from 1

(b)

Find the probability that Nicola obtains a 6 for the second time on her 8th throw.

2M Similar

DifficultyMedium

Worked solution

Approach

For the second 6 to occur on the 8th throw, the 8th throw must be a 6, and exactly one of the first 7 throws must also be a 6. Choose which of the first 7 throws is the first 6, then multiply the probabilities of the 8 independent throws.

Working

Choose the position of the first 6 among the first 7 throws:

\binom{7}{1} = 7

There is one 6 and six non-6s in the first 7 throws, and the 8th throw is a 6:

\mathrm{P} = \binom{7}{1}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^6\left(\frac{1}{6}\right)

= 7\left(\frac{5}{6}\right)^6\left(\frac{1}{6}\right)^2

= \frac{109375}{1679616} = 0.065119\ldots

Answer

\mathrm{P} = 0.0651

Final answer

0.0651

Detailed explanation

Walkthrough

We want the second 6 to occur exactly on the 8th throw. This means the 8th throw must be a 6, and among the first 7 throws there must be exactly one 6. The first 6 could be in any of the 7 positions, so there are $\binom{7}{1} = 7$ choices. For a fixed choice, the probability of one 6 and six non-6s in the first 7 throws is $\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^6$ , and the 8th throw being a 6 contributes another $\frac{1}{6}$ . Multiplying gives $7\left(\frac{5}{6}\right)^6\left(\frac{1}{6}\right)^2$ , which is approximately 0.0651.

Key Takeaways

For the second success to occur on a given throw, fix the last throw as a success and place the first success among the earlier throws.
Use a binomial coefficient to count the positions of the earlier success.
This is a negative-binomial/geometric-type counting problem.

Common Mistakes

Using $\binom{8}{2}$ instead of $\binom{7}{1}$ , which would allow the second 6 to occur before the 8th throw.
Forgetting to multiply by 7, giving only the probability of one particular arrangement.
Writing $\left(\frac{5}{6}\right)^7\left(\frac{1}{6}\right)$ for the first success on the 8th throw instead of the second success.
Adding probabilities instead of multiplying independent throw probabilities; the mark scheme allows no inappropriate addition.

Things to Be Careful About

The 8th throw must definitely be a 6, so it is not part of the choice.
Exactly one of the first 7 throws is a 6; the other six are not 6s.
Round the final probability to 0.0651, or give the exact fraction $\frac{109375}{1679616}$ .

Techniques used

fix the final throw as the required successcount the possible positions of the earlier successmultiply independent probabilities

Q2MediumDiscrete Random Variables

The random variable $X$ takes the values $-2, -1, 0, 2, 3$ . It is given that $P(X = x) = k(x^2 + 2)$ , where $k$ is a positive constant.

(a)

Draw up the probability distribution table for $X$ , giving the probabilities as numerical fractions.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

For a probability distribution, the sum of all the probabilities is $1$ . Substitute each value of $x$ into $k(x^2+2)$ , sum the resulting expressions, and solve for $k$ . Then present the probabilities as fractions.

Working

For the values $x = -2, -1, 0, 2, 3$ :

P(X=-2)=6k,\quad P(X=-1)=3k,\quad P(X=0)=2k,\quad P(X=2)=6k,\quad P(X=3)=11k

Since the probabilities sum to $1$ :

6k+3k+2k+6k+11k=1

28k=1

k=\frac{1}{28}

The probability distribution table is:

$x$	$-2$	$-1$	$0$	$2$	$3$
$P(X=x)$	$\frac{6}{28}$	$\frac{3}{28}$	$\frac{2}{28}$	$\frac{6}{28}$	$\frac{11}{28}$

Equivalently, the probabilities are $\frac{3}{14}, \frac{3}{28}, \frac{1}{14}, \frac{3}{14}, \frac{11}{28}$ .

Answer

$k=\frac{1}{28}$

$x$	$-2$	$-1$	$0$	$2$	$3$
$P(X=x)$	$\frac{6}{28}$	$\frac{3}{28}$	$\frac{2}{28}$	$\frac{6}{28}$	$\frac{11}{28}$

Final answer

k = 1/28; probabilities are 6/28, 3/28, 2/28, 6/28, 11/28

Detailed explanation

Walkthrough

The stem says that $X$ takes the values $-2, -1, 0, 2, 3$ and that $P(X=x)=k(x^2+2)$ . To find a valid probability distribution, substitute each value of $x$ into the formula. This gives the probabilities $6k, 3k, 2k, 6k, 11k$ . Since the total probability must be $1$ , add these five probabilities and set the sum equal to $1$ . This gives $28k=1$ , so $k=\frac{1}{28}$ . Substituting this value back into each probability gives the required numerical fractions. The mark scheme awards the first mark for identifying $k$ , a method mark for a table with the correct outcomes and at least two correct probabilities, and the final mark for a fully correct table.

Key Takeaways

A probability distribution table must list all possible outcomes and their probabilities, and the probabilities must sum to $1$ . This question uses the total-sum condition to find an unknown constant in a probability distribution. Once $k$ is found, each probability can be written as a fraction.

Common Mistakes

Forgetting to include all five possible values of $x$ in the sum, giving an incorrect total such as $22k$ .
Leaving the probabilities in terms of $k$ instead of substituting $k=\frac{1}{28}$ .
Including extra values of $x$ that are not in the given set. The mark scheme does not allow extra values unless their probability is $0$ .
Rounding probabilities to decimals too early; the question asks for numerical fractions.

Things to Be Careful About

The total probability must equal exactly $1$ , so $k$ is found from $28k=1$ .
The mark scheme accepts a table with probabilities written as $6k$ , $3k$ , $2k$ , $6k$ , $11k$ or $\frac{6k}{28}$ etc. as part of the method, but the final table should contain numerical fractions.
Decimal answers are accepted by the mark scheme only if given to at least 3 significant figures, but the question specifically asks for fractions.

Techniques used

substitute each outcome into the probability formulasum probabilities to solve for the constantconstruct the probability distribution table

(b)

Find the value of $\text{Var}(X)$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the probability distribution from part (a). First calculate $E(X) = \sum x P(X=x)$ , then calculate $E(X^2) = \sum x^2 P(X=x)$ , and use $\text{Var}(X) = E(X^2) - [E(X)]^2$ .

Working

E(X) = (-2)\frac{6}{28} + (-1)\frac{3}{28} + (0)\frac{2}{28} + (2)\frac{6}{28} + (3)\frac{11}{28}

E(X) = \frac{-12 - 3 + 0 + 12 + 33}{28} = \frac{30}{28} = \frac{15}{14}

E(X^2) = (-2)^2\frac{6}{28} + (-1)^2\frac{3}{28} + (0)^2\frac{2}{28} + (2)^2\frac{6}{28} + (3)^2\frac{11}{28}

E(X^2) = \frac{24 + 3 + 0 + 24 + 99}{28} = \frac{150}{28} = \frac{75}{14}

Then

\text{Var}(X) = \frac{75}{14} - \left(\frac{15}{14}\right)^2

\text{Var}(X) = \frac{1050}{196} - \frac{225}{196} = \frac{825}{196}

Thus $\text{Var}(X) = \frac{825}{196} = 4\frac{41}{196} \approx 4.21$ .

Answer

\text{Var}(X) = \frac{825}{196} = 4\frac{41}{196} \approx 4.21

Final answer

825/196 = 4 41/196 ≈ 4.21

Detailed explanation

Walkthrough

Using the distribution from part (a), first find the expected value $E(X)$ by summing each value of $x$ multiplied by its probability. This gives $\frac{15}{14}$ . Next find $E(X^2)$ by summing each value of $x^2$ multiplied by its probability. This gives $\frac{75}{14}$ . The variance is then $E(X^2) - [E(X)]^2$ . The mark scheme awards a method mark for the calculation of $E(X)$ , a second method mark for using the correct variance formula, and the final mark for the simplified answer.

Key Takeaways

The expectation of a discrete random variable is $E(X) = \sum x P(X=x)$ .
The variance is $\text{Var}(X) = E(X^2) - [E(X)]^2$ .
When probabilities are fractions, keeping all terms over a common denominator avoids rounding errors.
The variance may be written as an improper fraction, a mixed number, or a decimal.

Common Mistakes

Using $E(X^2) - E(X)$ instead of $E(X^2) - [E(X)]^2$ .
Forgetting to square the values of $x$ when computing $E(X^2)$ .
Forgetting to multiply each $x^2$ by its probability.
Making arithmetic errors when subtracting the fractions.

Things to Be Careful About

The method marks allow follow-through from the table in part (a), provided the probabilities are between $0$ and $1$ and sum to $1$ .
The mark scheme accepts the variance as $\frac{825}{196}$ , $4\frac{41}{196}$ , or $4.21$ .
If the final answer is correct with no working, the mark scheme gives a special case mark of B1, but full method marks require showing both $E(X)$ and the variance formula.

Techniques used

compute the expectation from the distribution tablecompute the second momentapply the variance formula

Q3MediumRepresentation of Data

The time taken, in minutes, to walk to school was recorded for 200 pupils at a certain school. These times are summarised in the following table.

Time taken ( $t$ minutes)	$t \leq 15$	$t \leq 25$	$t \leq 30$	$t \leq 40$	$t \leq 50$	$t \leq 70$
Cumulative frequency	18	46	88	140	176	200

(a)

Draw a cumulative frequency graph to illustrate the data.

Fig. 3.1

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

To draw the cumulative frequency graph, we plot the cumulative frequency against the upper bound of each time interval. We start at $(0, 0)$ because at time $t=0$ , the cumulative frequency is 0. The points to plot are $(15, 18)$ , $(25, 46)$ , $(30, 88)$ , $(40, 140)$ , $(50, 176)$ , and $(70, 200)$ . A smooth S-shaped curve (ogive) is drawn through these points, starting from $(0,0)$ .

Working

The coordinates to plot are:

\begin{aligned} &(0, 0) \\ &(15, 18) \\ &(25, 46) \\ &(30, 88) \\ &(40, 140) \\ &(50, 176) \\ &(70, 200) \end{aligned}

Plot these points on the grid with time $t$ on the horizontal axis (0 to 70) and cumulative frequency on the vertical axis (0 to 200). Join them with a smooth curve.

Answer

Cumulative frequency graph drawn with points $(0,0), (15,18), (25,46), (30,88), (40,140), (50,176), (70,200)$ and a smooth curve.

Final answer

Graph plotted with points (0,0), (15,18), (25,46), (30,88), (40,140), (50,176), (70,200) and smooth curve

Detailed explanation

Walkthrough

A cumulative frequency graph (or ogive) is used to represent cumulative data. The horizontal axis represents the variable (time $t$ in minutes), and the vertical axis represents the cumulative frequency (number of pupils).

Identify points: The table gives cumulative frequencies at upper bounds. We add $(0,0)$ as the starting point. The points are $(15, 18)$ , $(25, 46)$ , $(30, 88)$ , $(40, 140)$ , $(50, 176)$ , $(70, 200)$ .
Plot and draw: Plot these points on the grid. Join them with a smooth curve that starts at $(0,0)$ and ends at $(70, 200)$ . The curve should be S-shaped, generally increasing.

Key Takeaways

Cumulative frequency graphs plot cumulative frequency against upper class boundaries.
Always include the origin $(0,0)$ if the data starts from 0.
The curve must be smooth, not a series of straight line segments (unless specified as a polygon).

Common Mistakes

Forgetting to plot the point $(0,0)$ .
Joining points with straight lines instead of a smooth curve.
Incorrect axis scales (e.g., not using enough of the grid).

Things to Be Careful About

Ensure the axes are labelled correctly: 'cumulative frequency' and 'time taken / minutes'.
The scale must be linear and large enough to use at least half the grid.

Techniques used

plot cumulative frequency pointsdraw smooth ogive curve

(b)

Use your graph to estimate the median and the interquartile range of the data.

3M Similar

DifficultyMedium

Worked solution

Approach

The total number of pupils is $n = 200$ .

The median is the value at cumulative frequency $\frac{n}{2} = 100$ .
The lower quartile (LQ) is the value at cumulative frequency $\frac{n}{4} = 50$ .
The upper quartile (UQ) is the value at cumulative frequency $\frac{3n}{4} = 150$ .
The interquartile range (IQR) is $\text{UQ} - \text{LQ}$ .

Working

From the cumulative frequency graph (or using the estimated values from the mark scheme):

Median: Find $t$ when $\text{cf} = 100$ . Looking at the graph, $t \approx 33$ .
$\text{Median} \approx 33 \text{ minutes}$
Quartiles: Find $t$ when $\text{cf} = 50$ and $\text{cf} = 150$ .
- For $\text{cf} = 50$ , $t \approx 26$ . So $\text{LQ} \approx 26$ .
- For $\text{cf} = 150$ , $t \approx 42$ . So $\text{UQ} \approx 42$ .
IQR:
$\text{IQR} = \text{UQ} - \text{LQ} = 42 - 26 = 16$

Answer

Median $= 33$ minutes, IQR $= 16$ minutes.

Final answer

Median = 33, IQR = 16

Detailed explanation

Walkthrough

To estimate the median and interquartile range from a cumulative frequency graph, we use the total frequency $n=200$ .

Median: The median corresponds to the $\frac{n}{2}$ -th value, which is the 100th value. Draw a horizontal line from $\text{cf}=100$ to the curve, then drop a vertical line to the time axis. The value is approximately 33 minutes.
Lower Quartile (LQ): Corresponds to $\frac{n}{4} = 50$ . Draw a line from $\text{cf}=50$ to the curve and down to the axis. Value is approximately 26 minutes.
Upper Quartile (UQ): Corresponds to $\frac{3n}{4} = 150$ . Draw a line from $\text{cf}=150$ to the curve and down to the axis. Value is approximately 42 minutes.
IQR: The difference between the upper and lower quartiles: $42 - 26 = 16$ .

Note: Values read from graphs have some tolerance (typically $\pm 0.5$ square on the axis). The mark scheme accepts median around 33, LQ between 25-27, UQ between 41-43.

Key Takeaways

Median is at $n/2$ cumulative frequency.
Quartiles are at $n/4$ and $3n/4$ cumulative frequencies.
IQR is UQ - LQ.
Reading from graphs requires care with scales and interpolation.

Common Mistakes

Using $n$ instead of $n/2$ for the median (i.e., looking for cf=200).
Calculating IQR as UQ + LQ or UQ * LQ.
Reading the wrong axis (time vs frequency).

Things to Be Careful About

Ensure you are reading the time value from the horizontal axis, not the cumulative frequency.
Graph reading estimates will vary slightly between students; mark schemes allow tolerance.

Techniques used

read median from cumulative frequency graphread quartiles from cumulative frequency graphcalculate interquartile range

(c)

Calculate an estimate for the mean value of the times taken by the 200 pupils to walk to school.

3M Similar

DifficultyMedium

Worked solution

Approach

To estimate the mean from grouped data, we need the frequency ( $f$ ) and midpoint ( $x$ ) for each class interval. The frequencies are found by subtracting consecutive cumulative frequencies. The midpoints are the average of the lower and upper bounds of each interval.

$\text{Mean} = \frac{\sum fx}{\sum f}$

Working

First, determine the class intervals, frequencies, and midpoints:

Time interval	Cumulative Frequency	Frequency ( $f$ )	Midpoint ( $x$ )	$fx$
$0 < t \leq 15$	18	18	$\frac{0+15}{2} = 7.5$	$18 \times 7.5 = 135$
$15 < t \leq 25$	46	$46 - 18 = 28$	$\frac{15+25}{2} = 20$	$28 \times 20 = 560$
$25 < t \leq 30$	88	$88 - 46 = 42$	$\frac{25+30}{2} = 27.5$	$42 \times 27.5 = 1155$
$30 < t \leq 40$	140	$140 - 88 = 52$	$\frac{30+40}{2} = 35$	$52 \times 35 = 1820$
$40 < t \leq 50$	176	$176 - 140 = 36$	$\frac{40+50}{2} = 45$	$36 \times 45 = 1620$
$50 < t \leq 70$	200	$200 - 176 = 24$	$\frac{50+70}{2} = 60$	$24 \times 60 = 1440$

Total frequency $\sum f = 200$ .

Total $\sum fx = 135 + 560 + 1155 + 1820 + 1620 + 1440 = 6730$ .

$\text{Mean} = \frac{6730}{200} = 33.65$

Answer

$\text{Mean} = 33.65 \text{ minutes}$

Final answer

33.65

Detailed explanation

Walkthrough

The data is given in cumulative frequency form. To calculate the mean, we must convert this to a frequency distribution table.

Frequencies: Subtract each cumulative frequency from the next to get the frequency for that interval. For the first interval, the frequency is just the first cumulative frequency (assuming it starts from 0).
- $0 < t \leq 15$ : $f = 18$
- $15 < t \leq 25$ : $f = 46 - 18 = 28$
- $25 < t \leq 30$ : $f = 88 - 46 = 42$
- $30 < t \leq 40$ : $f = 140 - 88 = 52$
- $40 < t \leq 50$ : $f = 176 - 140 = 36$
- $50 < t \leq 70$ : $f = 200 - 176 = 24$
Midpoints: Calculate the midpoint $x$ for each interval as $\frac{\text{lower} + \text{upper}}{2}$ .
- $7.5, 20, 27.5, 35, 45, 60$ .
Mean Calculation: Multiply each frequency by its midpoint ( $fx$ ), sum these products, and divide by the total frequency ( $\sum f = 200$ ).
- $\sum fx = 18(7.5) + 28(20) + 42(27.5) + 52(35) + 36(45) + 24(60) = 6730$ .
- Mean $= 6730 / 200 = 33.65$ .

Key Takeaways

Cumulative frequency can be converted to grouped frequency data by subtraction.
The mean of grouped data is estimated using class midpoints.
Formula: $\bar{x} = \frac{\sum fx}{\sum f}$ .

Common Mistakes

Using the upper bound of the interval as the midpoint instead of the average of bounds.
Using cumulative frequencies in the mean formula instead of actual frequencies.
Arithmetic errors in calculating $\sum fx$ .

Things to Be Careful About

Ensure the first interval starts at 0 (implied by $t \leq 15$ having cf 18, usually meaning $0 < t \leq 15$ ).
The midpoint of $50 < t \leq 70$ is 60, not 50 or 70.
Acceptable answers include 33.7 or $33 \frac{13}{20}$ .

Techniques used

calculate frequencies from cumulative frequenciesdetermine class midpointsapply grouped mean formula

Q4MediumProbability

Rahul has two bags, $X$ and $Y$ . Bag $X$ contains 4 red marbles and 2 blue marbles. Bag $Y$ contains 3 red marbles and 4 blue marbles. Rahul also has a coin which is biased so that the probability of obtaining a head when it is thrown is $\frac{1}{4}$ .

Rahul throws the coin.

If he obtains a head, he chooses at random a marble from bag $X$ . He notes the colour and replaces the marble in bag $X$ . He then chooses at random a second marble from bag $X$ .
If he obtains a tail, he chooses at random a marble from bag $Y$ . He notes the colour and discards the marble. He then chooses at random a second marble from bag $Y$ .

(a)

Find the probability that the two marbles that Rahul chooses are the same colour.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $H$ and $T$ denote the outcomes of the biased coin. If $H$ occurs, two draws are made from bag $X$ with replacement; if $T$ occurs, two draws are made from bag $Y$ without replacement. The two marbles are the same colour in four mutually exclusive ways: $HRR$ , $TRR$ , $HBB$ , $TBB$ . Multiply along each branch and add the results.

Working

The coin probabilities are

P(H)=\frac{1}{4}, \quad P(T)=\frac{3}{4}

For bag $X$ (with replacement):

P(R\mid H)=\frac{4}{6}=\frac{2}{3}, \quad P(B\mid H)=\frac{2}{6}=\frac{1}{3}

For bag $Y$ (without replacement):

P(R\mid T)=\frac{3}{7}, \quad P(B\mid T)=\frac{4}{7}

After a red marble is taken from bag $Y$ , $2$ red and $4$ blue remain, so

P(R\mid \text{first red from }Y)=\frac{2}{6}=\frac{1}{3}

After a blue marble is taken from bag $Y$ , $3$ red and $3$ blue remain, so

P(B\mid \text{first blue from }Y)=\frac{3}{6}=\frac{1}{2}

Now calculate the four same-colour probabilities:

P(HRR)=\frac{1}{4}\times\frac{4}{6}\times\frac{4}{6}=\frac{16}{144}=\frac{4}{36}

P(TRR)=\frac{3}{4}\times\frac{3}{7}\times\frac{2}{6}=\frac{18}{168}=\frac{3}{28}

P(HBB)=\frac{1}{4}\times\frac{2}{6}\times\frac{2}{6}=\frac{4}{144}=\frac{1}{36}

P(TBB)=\frac{3}{4}\times\frac{4}{7}\times\frac{3}{6}=\frac{36}{168}=\frac{6}{28}=\frac{3}{14}

These four events are mutually exclusive, so add them:

P(\text{same colour}) = \frac{4}{36}+\frac{3}{28}+\frac{1}{36}+\frac{6}{28}

= \frac{5}{36}+\frac{9}{28}

Using a common denominator of $252$ :

= \frac{35}{252}+\frac{81}{252}=\frac{116}{252}=\frac{29}{63}

Answer

\frac{29}{63} \quad (\approx 0.460)

Final answer

29/63 or 0.460

Detailed explanation

Walkthrough

Start with the coin. It is biased, so the probability of heads is $\frac{1}{4}$ and the probability of tails is $\frac{3}{4}$ . If heads occurs, Rahul uses bag $X$ and replaces the marble after the first draw, so the second draw has exactly the same probabilities as the first: $\frac{4}{6}$ red and $\frac{2}{6}$ blue. If tails occurs, Rahul uses bag $Y$ and discards the first marble, so the second draw depends on the colour of the first marble.

The four ways to get two marbles of the same colour are: head then red-red, tail then red-red, head then blue-blue, and tail then blue-blue. For each way, multiply the probabilities along the route. Since these four routes are mutually exclusive, add them together.

Key Takeaways

This question tests the multiplication law for probabilities, the addition law for mutually exclusive events, and the important difference between sampling with replacement and sampling without replacement.

Common Mistakes

Forgetting to include the coin probability in each branch.
Treating bag $Y$ as if the marble were replaced; after the first draw from $Y$ , the second draw has denominator $6$ , not $7$ .
Adding probabilities that are not mutually exclusive, or missing one of the four same-colour routes.

Things to Be Careful About

In bag $X$ , with replacement, the probabilities are $\frac{4}{6}$ red and $\frac{2}{6}$ blue for both draws.
In bag $Y$ , without replacement, after a red is drawn there are $2$ red and $4$ blue left; after a blue is drawn there are $3$ red and $3$ blue left.
Show the unsimplified products, for example $\frac{1}{4}\times\frac{4}{6}\times\frac{4}{6}$ , because the mark scheme awards a method mark for two clearly identified unsimplified probabilities.
The final answer can be given as $\frac{29}{63}$ or $0.460$ to 3 significant figures.

Techniques used

multiply probabilities along branches of a probability treeadd probabilities of mutually exclusive outcomesdistinguish sampling with and without replacement

(b)

Find the probability that the two marbles that Rahul chooses are both from bag $Y$ given that both marbles are blue.

3M Similar

DifficultyMedium

Worked solution

Approach

Let $BB$ denote the event that both marbles are blue. We need $P(T \mid BB)$ , the probability that the marbles came from bag $Y$ given that they are both blue. Using

P(T \mid BB)=\frac{P(T \cap BB)}{P(BB)}

The numerator is $P(TBB)$ from part (a), and the denominator is $P(HBB)+P(TBB)$ .

Working

From part (a),

P(T \cap BB)=P(TBB)=\frac{3}{4}\times\frac{4}{7}\times\frac{3}{6}=\frac{36}{168}=\frac{3}{14}

Also from part (a),

P(HBB)=\frac{1}{4}\times\frac{2}{6}\times\frac{2}{6}=\frac{4}{144}=\frac{1}{36}

So the total probability that both marbles are blue is

P(BB)=P(HBB)+P(TBB)=\frac{1}{36}+\frac{3}{14}

= \frac{7}{252}+\frac{54}{252}=\frac{61}{252}

Therefore

P(T \mid BB)=\frac{\frac{3}{14}}{\frac{61}{252}}

= \frac{3}{14}\times\frac{252}{61}=\frac{54}{61}

Answer

\frac{54}{61} \quad (\approx 0.885)

Final answer

54/61 or 0.885

Detailed explanation

Walkthrough

This part asks for a conditional probability: the probability that the marbles came from bag $Y$ , given that both are blue. Write the conditioning event as $BB$ . The formula is

P(T \mid BB)=\frac{P(T \cap BB)}{P(BB)}

The numerator is the probability that the coin is a tail and both marbles are blue. This is exactly $P(TBB)$ from part (a), which is $\frac{3}{14}$ . The denominator is the total probability that both marbles are blue, and it includes both the head route and the tail route: $P(HBB)+P(TBB)=\frac{1}{36}+\frac{3}{14}=\frac{61}{252}$ . Dividing gives the required conditional probability.

Key Takeaways

Conditional probability is a renormalisation: restrict attention to the cases where the conditioning event occurs, then compare the probability of the intersection with that total. Here the conditioning event is 'both marbles are blue', so we divide by the total probability of both blue.

Common Mistakes

Using $P(T)=\frac{3}{4}$ as the numerator instead of $P(T \cap BB)$ .
Forgetting to include $P(HBB)$ in the denominator.
Confusing the numerator and denominator when writing the fraction.

Things to Be Careful About

The numerator and denominator must both refer to the event 'both marbles are blue'.
The denominator is $P(BB)=P(HBB)+P(TBB)$ , not just $P(TBB)$ .
The final answer can be given as $\frac{54}{61}$ or $0.885$ to 3 significant figures.
If a previous part was answered incorrectly, follow-through marks allow using the candidate's $P(TBB)$ from part (a).

Techniques used

apply the conditional probability formulacompute the intersection probabilitydivide by the total probability of the conditioning event

Q5MediumThe Normal DistributionDiscrete Random Variables

The weights of the green apples sold by a shop are normally distributed with mean 90 grams and standard deviation 8 grams.

(a)

Find the probability that a randomly chosen green apple weighs between 83 grams and 95 grams.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $X$ grams be the weight of a green apple. Since $X$ is normally distributed, standardise using $Z = \frac{X - \mu}{\sigma}$ and use the standard normal distribution table.

Working

Given $X \sim N(90, 8^2)$ , the required probability is

P(83 < X < 95) = P\left(\frac{83-90}{8} < Z < \frac{95-90}{8}\right)

= P(-0.875 < Z < 0.625)

Using symmetry:

P(-0.875 < Z < 0.625) = \Phi(0.625) - \Phi(-0.875)

= \Phi(0.625) + \Phi(0.875) - 1

From the standard normal table:

\Phi(0.625) = 0.7340

\Phi(0.875) = 0.8092

P = 0.7340 + 0.8092 - 1 = 0.5432

Answer

0.543 \quad (3\text{ s.f.})

Final answer

0.543

Detailed explanation

Walkthrough

The question gives a normal distribution for weights, so the variable is continuous. The probability is the area under the normal curve between 83 and 95 grams. We standardise each boundary with $Z = \frac{X - \mu}{\sigma}$ .

With $\mu = 90$ and $\sigma = 8$ , the lower boundary becomes $(83 - 90)/8 = -0.875$ and the upper boundary becomes $(95 - 90)/8 = 0.625$ .

Because one z-value is negative, tables for positive values are used with symmetry: the area to the left of $-0.875$ is $1 - \Phi(0.875)$ . Therefore the required interval area is

\Phi(0.625) - (1 - \Phi(0.875)) = \Phi(0.625) + \Phi(0.875) - 1.

Substituting the tabulated values gives $0.7340 + 0.8092 - 1 = 0.5432$ , rounded to $0.543$ .

Key Takeaways

This question tests standardisation of a normal random variable, converting an interval into a standard normal interval, and using symmetry to handle negative z-values. It also checks that you know which value is the mean and which is the standard deviation.

Common Mistakes

Dividing by the variance $64$ instead of the standard deviation $8$ in the standardisation formula.
Treating $\Phi(-0.875)$ incorrectly; it must be converted using symmetry before subtracting.
Adding or subtracting the two table probabilities in the wrong order.
Applying a continuity correction to the normal distribution itself; it is only needed when approximating a discrete distribution.
Giving unsupported or insufficiently rounded final probabilities.

Things to Be Careful About

Use $\mu = 90$ and $\sigma = 8$ , not $\sigma^2$ .

The two z-values are $-0.875$ and $0.625$ . The final probability should be greater than $0.5$ , which is a useful check because the interval is close to the mean and covers a large central region.

The mark scheme expects the standardisation formula to be shown with 90 and 8; if the method is not shown, only special-credit marks may be available.

Techniques used

standardise a normal variable to Zexpress an interval probability using cumulative probabilitiesuse symmetry of the normal curve for negative Z-valueslook up values in the standard normal table

(b)

The shop also sells red apples. 60% of the red apples sold by the shop weigh more than 80 grams. 160 red apples are chosen at random from the shop.

Use a suitable approximation to find the probability that fewer than 105 of the chosen red apples weigh more than 80 grams.

5M Similar

DifficultyMedium

Worked solution

Approach

Let $Y$ be the number of red apples, out of 160, that weigh more than 80 grams. Each apple independently does this with probability $p = 0.6$ , so $Y \sim B(160, 0.6)$ . Since $np = 96$ and $n(1-p) = 64$ are both greater than 5, the normal approximation is appropriate. Use a continuity correction because $Y$ is discrete.

Working

For the binomial distribution:

\mu = np = 160 \times 0.6 = 96

\sigma^2 = np(1-p) = 160 \times 0.6 \times 0.4 = 38.4

Approximately, $Y \sim N(96, 38.4)$ .

"Fewer than 105" means $Y \leq 104$ , so with continuity correction:

P(Y < 105) \approx P\left(Z < \frac{104.5 - 96}{\sqrt{38.4}}\right)

z = \frac{8.5}{\sqrt{38.4}} = \frac{8.5}{6.19677\ldots} = 1.372

Using the standard normal table:

P(Z < 1.372) = \Phi(1.372) = 0.915

Answer

0.915 \quad (3\text{ s.f.})

Final answer

0.915

Detailed explanation

Walkthrough

There are 160 independent red apples, each with the same probability $p = 0.6$ of weighing more than 80 grams, so the count $Y$ has a binomial distribution $B(160, 0.6)$ . A normal approximation is suitable because $np = 96$ and $n(1-p) = 64$ are both large enough.

For the approximation we need the binomial mean and variance:

\mu = np = 96, \qquad \sigma^2 = np(1-p) = 38.4.

Since $Y$ is discrete, the event $Y < 105$ is the same as $Y \leq 104$ . The continuity correction halfway between 104 and 105 is 104.5, so we standardise 104.5 using mean 96 and standard deviation $\sqrt{38.4}$ .

z = \frac{104.5 - 96}{\sqrt{38.4}} \approx 1.372.

The required probability is therefore $\Phi(1.372) = 0.915$ . It is greater than 0.5 because 104.5 is above the mean 96 but not far into the upper tail.

Key Takeaways

This question combines the binomial distribution with the normal approximation. You must know the binomial mean and variance, recognise when a normal approximation is valid, and apply a continuity correction correctly when converting a discrete event to a continuous one.

Common Mistakes

Omitting the continuity correction entirely.
Using 105.5 instead of 104.5 for a "fewer than 105" probability; 104.5 is correct because the event is equivalent to $Y \leq 104$ .
Writing the variance as $6.197$ and then using it as if it were still the variance; the standard deviation is $\sqrt{38.4} \approx 6.197$ .
Confusing the success category: here success is "weighs more than 80 grams", so $p = 0.6$ .
Not checking that $np$ and $n(1-p)$ are large enough for the normal approximation.

Things to Be Careful About

Use $\mu = 96$ and $\sigma^2 = 38.4$ . In the standardisation formula the denominator is $\sqrt{38.4}$ , not $38.4$ .

The boundary for "fewer than 105" is 104.5. Keep enough decimal places when computing $z$ so the final table reading is accurate. The final probability should be about 0.915 and should be greater than 0.5, which is a good sanity check.

The mark scheme allows unsimplified forms such as $160 \times 0.6$ and $160 \times 0.6 \times 0.4$ , but if the variance is clearly identified as the standard deviation the mean/variance mark may be lost.

Techniques used

modell the count with a binomial distributioncompute the mean and variance of a binomial distributionapply the binomial-to-normal approximationapply a continuity correctionstandardise and look up a normal probability

Q6MediumThe Normal DistributionDiscrete Random Variables

The heights of the female students at Breven college are normally distributed:

90% of the female students have heights less than 182.7 cm.
40% of the female students have heights less than 162.5 cm.

(a)

Find the mean and the standard deviation of the heights of the female students at Breven college.

5M Similar

DifficultyMedium

Worked solution

Approach

Let the heights be $X \sim N(\mu, \sigma^2)$ . Convert each given percentage into a standard normal $z$ -value, then write the two standardisation equations and solve them for $\mu$ and $\sigma$ .

Working

90% of the students have height less than 182.7 cm, so

z = 1.282

and

\frac{182.7 - \mu}{\sigma} = 1.282.

40% of the students have height less than 162.5 cm, so

z = -0.253

and

\frac{162.5 - \mu}{\sigma} = -0.253.

Rewrite both equations:

182.7 - \mu = 1.282\sigma

162.5 - \mu = -0.253\sigma

Subtract the second equation from the first:

20.2 = (1.282 + 0.253)\sigma = 1.535\sigma

\sigma = \frac{20.2}{1.535} = 13.16

Substitute back to find $\mu$ :

\mu = 182.7 - 1.282(13.16) = 165.83

Answer

The mean is $\mu = 165.8$ cm and the standard deviation is $\sigma = 13.2$ cm.

Final answer

mean = 165.8 cm, standard deviation = 13.2 cm

Detailed explanation

Walkthrough

We are given two cumulative probabilities from a normal distribution. The key idea is to convert each height into a standard normal score using the known inverse $z$ -values. For 90% below, the $z$ -value is $1.282$ ; for 40% below, the $z$ -value is $-0.253$ . Writing each height as $\mu + z\sigma$ gives two linear equations in $\mu$ and $\sigma$ . Subtracting one equation from the other eliminates $\mu$ , allowing us to solve for $\sigma$ . Substituting back gives $\mu$ .

Key Takeaways

This question tests the relationship between a normal distribution and the standard normal distribution. It shows how to recover the mean and standard deviation from two cumulative probability statements. The key skill is converting a percentage into a $z$ -value and then solving simultaneous equations.

Common Mistakes

Using the probabilities 0.9 and 0.4 directly instead of the corresponding $z$ -values.
Using $\sigma^2$ or $\sqrt{\sigma}$ in the standardisation formula.
Making a sign error: a percentage below the mean gives a negative $z$ -value.
Rounding too early, which can lead to final answers that are not accurate enough.

Things to Be Careful About

The mark scheme requires the critical values $1.282$ and $-0.253$ to be seen. Ensure the two equations are written with $\mu$ and $\sigma$ on the same side before solving. Give the final answers to at least one decimal place because the context is real measurements.

Techniques used

convert given normal values to standard normal valuesset up two standardisation equationssolve simultaneous equations for the mean and standard deviation

(b)

Ten female students are chosen at random from those at Breven college.

Find the probability that fewer than 8 of these 10 students have heights more than 162.5 cm.

3M Similar

DifficultyMedium

Worked solution

Approach

From part (a), $P(X < 162.5) = 0.4$ , so the probability that one student has height more than 162.5 cm is $0.6$ . Let $Y$ be the number of the 10 students with height more than 162.5 cm. Then $Y \sim B(10, 0.6)$ . We need $P(Y < 8)$ , which is easier to find using the complement $P(Y = 8) + P(Y = 9) + P(Y = 10)$ .

Working

P(Y < 8) = 1 - P(Y \geq 8)

P(Y \geq 8) = P(Y=8) + P(Y=9) + P(Y=10)

Using $Y \sim B(10, 0.6)$ :

P(Y \geq 8) = {}^{10}C_8(0.6)^8(0.4)^2 + {}^{10}C_9(0.6)^9(0.4)^1 + (0.6)^{10}

P(Y=8) = 0.12093, \quad P(Y=9) = 0.04031, \quad P(Y=10) = 0.00605

P(Y \geq 8) = 0.12093 + 0.04031 + 0.00605 = 0.16729

Therefore

P(Y < 8) = 1 - 0.16729 = 0.83271 \approx 0.833

Answer

P(\text{fewer than 8}) = 0.833

Final answer

0.833

Detailed explanation

Walkthrough

From part (a), 40% of the students have height less than 162.5 cm, so 60% have height more than 162.5 cm. Therefore the probability of success for each randomly chosen student is $p = 0.6$ . With 10 students, the number of successes follows a binomial distribution. Let $Y$ be the number of students with height more than 162.5 cm. “Fewer than 8” means $Y = 0, 1, 2, 3, 4, 5, 6$ or $7$ . It is easier to calculate the complement: the probability that 8, 9 or 10 students have height more than 162.5 cm. Each binomial term is computed using ${}^{10}C_x p^x (1-p)^{10-x}$ , and the three terms are summed and subtracted from 1.

Key Takeaways

This part combines a normal distribution result with a binomial distribution. The important step is identifying $p = 0.6$ from the 40% probability of being below 162.5 cm. The complement rule is useful when “fewer than” would require summing many terms.

Common Mistakes

Using $p = 0.4$ instead of $p = 0.6$ .
Misreading “fewer than 8” as “8 or fewer”; it means $0$ to $7$ .
Forgetting the combinatorial coefficients ${}^{10}C_8$ , ${}^{10}C_9$ , and ${}^{10}C_{10}=1$ .
Omitting one of the three complement terms, or subtracting the complement incorrectly.

Things to Be Careful About

Make sure the binomial probability uses the correct exponents for $p$ and $1-p$ . Since $p = 0.6$ , the term for exactly 10 successes is simply $(0.6)^{10}$ . Give the final answer to three decimal places, matching the mark scheme value $0.833$ .

Techniques used

model the count with a binomial distributionuse the complementary probability for fewer than 8evaluate binomial probabilities and sum them

Q7MediumPermutations and CombinationsProbability

(a)

How many different arrangements are there of the 9 letters in the word INTELLECT in which the two Ts are together?

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Treat the two T's as a single block, since they must be together. This leaves 8 items to arrange, but the two E's and the two L's are identical, so divide by $2!$ for each repeated pair.

Working

The letters of INTELLECT are

\text{I, N, T, T, E, E, L, L, C}

With the two T's joined as one block, arrange:

(\text{TT}),\ \text{I, N, E, E, L, L, C}

Number of arrangements:

\frac{8!}{2!\,2!} = \frac{40320}{4} = 10080

Answer

10080

Final answer

10080

Detailed explanation

Walkthrough

The phrase "the two T's are together" means the two T's must appear as the single block TT somewhere in the word. Once you join them, the problem becomes arranging eight items: TT, I, N, E, E, L, L, C.

If all eight of those items were different, the answer would be $8!$ . They are not all different: there are two E's and two L's. For any arrangement, swapping the two E's gives the same word, and swapping the two L's gives the same word. These two swaps are independent, so each distinct word has been counted $2! \times 2!$ times. Dividing $8!$ by $2!2!$ gives 10080.

Why don't we also divide by $2!$ for the T's? Because the T's are treated as one block; they are not being arranged separately inside the count. The only order inside the block is TT, so there is no extra factor.

Key Takeaways

A constraint that two letters are adjacent can be handled by forming a block and arranging one fewer object.
Repeated letters reduce the number of distinct arrangements: divide by the factorial of each repeated letter frequency.

Common Mistakes

Counting $9!$ directly without considering repeated letters.
Writing an extra division by $2!$ for the T's after blocking them; the T's should contribute no separate factorial.
Forgetting to divide by $2!$ for both E and L.

Things to Be Careful About

The standard calculation is $\frac{8!}{2!2!}=10080$ . The mark scheme also accepts equivalent forms such as $\frac{k!}{2!m!}$ with $k=7$ or $8$ and $m=1$ or $2$ , but the simplest is to treat TT as one block and divide by the repeated E and L factorials.

Techniques used

treat forced adjacent letters as one blockarrange the reduced set of objectsdivide by factorials for repeated letters

(b)

How many different arrangements are there of the 9 letters in the word INTELLECT in which there is a T at each end and the two Es are not next to each other?

3M Similar

DifficultyMedium

Worked solution

Approach

Fix the two T's at the ends, one at each end, so only the 7 middle letters need to be arranged. Start with all middle arrangements and subtract those in which the two E's are adjacent.

Working

Middle letters:

\text{I, N, E, E, L, L, C}

All middle arrangements:

\frac{7!}{2!\,2!} = \frac{5040}{4} = 1260

If the two E's are together, treat them as one block. The middle items become:

(\text{EE}),\ \text{I, N, L, L, C}

So the number with the E's together is

\frac{6!}{2!} = \frac{720}{2} = 360

Required number:

1260 - 360 = 900

Answer

900

Final answer

900

Detailed explanation

Walkthrough

With a T fixed at each end, the two end positions are already occupied. Since the two T's are identical, there is no extra factor for choosing which T goes to which end.

The middle seven letters are I, N, E, E, L, L, C. Count their arrangements without any restriction on the E's:

\frac{7!}{2!2!} = 1260

Now subtract the forbidden arrangements in which the two E's are next to each other. Treat EE as one block, so the middle items become EE, I, N, L, L, C. There are six items, with L repeated twice, giving

\frac{6!}{2!} = 360

So the number satisfying the condition is $1260 - 360 = 900$ .

An alternative gap method is also valid: arrange the non-E middle letters, choose two gaps for the E's, and multiply; it gives the same result.

Key Takeaways

When identical letters must be placed at the two ends, fix them first and do not multiply by an extra factor.
Counting arrangements that avoid adjacency is often easiest by subtracting the arrangements where the two letters ARE adjacent.

Common Mistakes

Forgetting that the two T's are identical, so there is no factor of 2 for swapping the end T's.
Subtracting the E-together case incorrectly, e.g. using $6!$ without dividing by $2!$ for the repeated L.
Counting the middle as 8 positions instead of 7.

Things to Be Careful About

The mark scheme accepts either the subtraction method or the gap method. In the subtraction method, the E-together block is just one item among the middle six, and the repeated L still requires a division by $2!$ .

Techniques used

fix identical letters at the endsarrange the middle positionssubtract arrangements with a repeat pair adjacenttreat adjacent identical letters as a block

(c)

Four letters are selected at random from the 9 letters in the word INTELLECT.

Find the percentage of the possible selections which contain at least one E and exactly one T.

4M Similar

DifficultyMedium

Worked solution

Approach

Count selections of 4 letters that contain exactly one T and at least one E. The word contains 2 T's, 2 E's, and 5 other letters. Split into selections with exactly one E and selections with two E's. Divide the favourable count by the total number of 4-letter selections, then convert to a percentage.

Working

Case 1: exactly one T and exactly one E

{}^2\mathrm{C}_1 \times {}^2\mathrm{C}_1 \times {}^5\mathrm{C}_2 = 2 \times 2 \times 10 = 40

Case 2: exactly one T and two E's

{}^2\mathrm{C}_1 \times {}^2\mathrm{C}_2 \times {}^5\mathrm{C}_1 = 2 \times 1 \times 5 = 10

Therefore the number of favourable selections is

40 + 10 = 50

The total number of selections of 4 letters is

{}^9\mathrm{C}_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126

Required percentage:

\frac{50}{126} \times 100 = 39.6825\ldots \approx 39.7\%

Answer

39.7\%

Final answer

39.7%

Detailed explanation

Walkthrough

We are choosing 4 letters, not arranging them, so combinations are appropriate. The denominator is ${}^9\mathrm{C}_4$ because there are 9 physical letter positions and any 4 can be chosen.

For the favourable count, exactly one T means choose 1 of the two T's. The condition "at least one E" can happen in two mutually exclusive ways.

Exactly one E: choose 1 T, 1 E, and then 2 letters from the 5 letters that are neither T nor E. These 5 letters are L, L, I, N, C. This gives ${}^2\mathrm{C}_1 \times {}^2\mathrm{C}_1 \times {}^5\mathrm{C}_2 = 40$ .

Exactly two E's: choose 1 T, both E's, and then 1 of the remaining 5 letters. This gives ${}^2\mathrm{C}_1 \times {}^2\mathrm{C}_2 \times {}^5\mathrm{C}_1 = 10$ .

Adding the two disjoint cases gives 50 favourable selections. The total possible selections are ${}^9\mathrm{C}_4 = 126$ . Therefore the percentage is

\frac{50}{126} \times 100 = 39.7\%

Key Takeaways

"Selections" means combinations, not arrangements.
Repeated letters are still counted as separate physical letter positions when using ${}^n\mathrm{C}_r$ .
"At least one E" is handled by considering the disjoint cases: exactly one E and exactly two E's.

Common Mistakes

Using arrangements rather than combinations.
Counting only one E copy or only one T copy, forgetting that the word has two of each.
Double-counting the case with two E's by also including it in the one-E case.
Forgetting to convert the probability fraction into a percentage.

Things to Be Careful About

The mark scheme requires the denominator to be written as ${}^9\mathrm{C}_4$ . The five non-T, non-E letters are L, L, I, N and C, so there are 5 choices for the remaining spots. The final percentage should be in the range $39.68 \leqslant p \leqslant 39.7$ , so $39.7\%$ is correct.

Techniques used

count favourable selections by caseschoose from repeated identical letter positionsdivide by the total number of selectionsconvert a probability to a percentage

Mathematics 9709/51 — October/November 2024

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