A bag contains 10 marbles, of which 4 are red and 6 are blue. Four marbles are selected from the bag at random, without replacement. The random variable $X$ denotes the number of blue marbles selected. (a) Show that $P(X = 2) = \frac{3}{7}$. (b) Draw up the probability distribution table for $X$. (c) Find the probability that at least 2 of the marbles chosen are blue, given that at least 1 red marble and at least 1 blue marble are chosen.

(a) P(X=2) = 3/7 · (b) x = 0: 1/210, x = 1: 4/35, x = 2: 3/7, x = 3: 8/21, x = 4: 1/14 · (c) 85/97 (approximately 0.876)

9709/51 May/June 2025 — Mathematics Solved Past Paper

Q13MMedium-EasyThe Normal Distribution

The masses of the bags of rice made by a company are normally distributed with mean $\mu\text{ kg}$ and standard deviation $0.14\text{ kg}$ . The probability that the mass of a randomly chosen bag of this rice is less than $1.48\text{ kg}$ is $0.22$ .

Find the value of $\mu$ .

DifficultyMedium-Easy

Worked solution

Approach

Let $X$ be the mass of a randomly chosen bag of rice. We are told that $X \sim N(\mu, 0.14^2)$ and that $\mathrm{P}(X < 1.48) = 0.22$ . Standardise $1.48$ using the $z$ -score formula, find the corresponding $z$ -value from the normal distribution table, then solve for $\mu$ .

Working

Standardising:

\mathrm{P}\left(Z < \frac{1.48 - \mu}{0.14}\right) = 0.22

From the normal distribution table, the value of $z$ such that $\mathrm{P}(Z < z) = 0.22$ is:

z = -0.772

Therefore:

\frac{1.48 - \mu}{0.14} = -0.772

Multiply both sides by $0.14$ :

1.48 - \mu = -0.772 \times 0.14 = -0.10808

Solve for $\mu$ :

\mu = 1.48 + 0.10808 = 1.58808

Answer

\mu \approx 1.59\text{ kg}

Final answer

1.59 kg

Detailed explanation

Walkthrough

We have a normal distribution with a known standard deviation of $0.14$ kg and an unknown mean $\mu$ . The statement $\mathrm{P}(X < 1.48) = 0.22$ means that $1.48$ kg lies below the mean, because the probability of being less than it is only $0.22$ , which is less than $0.5$ . Therefore the corresponding $z$ -score should be negative.

To use the normal distribution table, we standardise $X$ :

z = \frac{1.48 - \mu}{0.14}

We know that $\mathrm{P}(Z < z) = 0.22$ . Looking up the inverse normal table for a left-tail probability of $0.22$ gives $z \approx -0.772$ . So we set:

\frac{1.48 - \mu}{0.14} = -0.772

Multiplying both sides by $0.14$ gives $1.48 - \mu = -0.10808$ , and rearranging gives $\mu = 1.48 + 0.10808 = 1.58808$ . Rounding to two decimal places gives $\mu \approx 1.59$ kg.

Key Takeaways

A normal distribution probability can be converted to a standard normal probability by using $z = \frac{x - \mu}{\sigma}$ .
When the probability is less than $0.5$ , the required $z$ -score is negative.
The normal table gives $\mathrm{P}(Z < z)$ for a given $z$ , and can also be used backwards to find $z$ from a given probability.
After finding the $z$ -score, solving for the unknown mean is a simple linear equation.

Common Mistakes

Using a positive $z$ -value instead of a negative one. Since $\mathrm{P}(X < 1.48) = 0.22 < 0.5$ , the $z$ -score must be negative.
Using $0.22$ or $0.78$ directly as the $z$ -value instead of looking up the inverse normal value.
Forgetting to divide by $0.14$ when standardising, or forgetting to multiply by $0.14$ when solving for $\mu$ .
Rounding the $z$ -value too early, which can change the final mean slightly.

Things to Be Careful About

The standard deviation is given as $0.14$ kg; do not use $0.14^2$ as the standard deviation. The mark scheme condones using $\sigma^2$ or $\sqrt{\sigma}$ but the standard method uses $\sigma = 0.14$ .
Do not apply a continuity correction; this is not a binomial approximation, and the mark scheme allows it only as a condoned error, not a required step.
The accepted range for the answer is $1.585 < \mu \leqslant 1.59$ , so $\mu = 1.59$ kg is the expected final answer.

Techniques used

standardise a normal distributionuse inverse normal table to find a z-valuesolve a linear equation for the unknown mean

Approach

The word KANGAROO has 8 letters: K, A, N, G, A, R, O, O. We need arrangements where the two As are together and the two Os are NOT together.

Use the subtraction method: first count the arrangements with the As together, then subtract those where both the As and the Os are together.

Working

Treat "AA" as a single block. The resulting units are AA, K, N, G, R, O, O. The two O's are identical, so the number of arrangements is:

\frac{7!}{2!} = \frac{5040}{2} = 2520

Now treat both "AA" and "OO" as blocks. The units are K, N, G, R, AA, OO, all distinct, so the number is:

6! = 720

The required count is the difference:

2520 - 720 = 1800

Answer

1800

Final answer

1800

Detailed explanation

Walkthrough

We need to count arrangements of the 8 letters of KANGAROO where the two As are together and the two Os are separated.

The block method is the key. Treat the two As as one block 'AA', so that we have 7 items: AA, K, N, G, R, O, O. Since the two O's are identical, we divide by $2!$ to get $\frac{7!}{2!} = 2520$ . Then we subtract the arrangements where the Os are also together: treating both 'AA' and 'OO' as blocks gives 6 distinct items, so $6! = 720$ . The difference $2520 - 720 = 1800$ is exactly the count we need.

Key Takeaways

This question teaches the block method for 'together' constraints and the subtraction principle for 'not together' constraints. Recognising that identical items require division by factorials is essential.

Common Mistakes

Forgetting to divide by $2!$ for the two identical O's.
Counting the arrangements where both pairs are together incorrectly.
Treating the subtraction as an addition.

Things to Be Careful About

The two As are identical, so the block 'AA' has no internal ordering.
The two O's are identical, so the block 'OO' has no internal ordering.
The subtraction removes exactly the arrangements where both conditions hold.

Techniques used

treat the two As as a single blockdivide by factorial for identical itemssubtract arrangements where both pairs are together

(b)

A fair 8-sided dice has faces labelled K, A, N, G, A, R, O, O. The dice is rolled repeatedly.

Find the probability that fewer than 6 rolls of this dice are required to obtain an A.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The dice has 8 faces, two of which show A, so $P(A) = \frac{2}{8} = \frac{1}{4}$ and $P(\text{not A}) = \frac{6}{8} = \frac{3}{4}$ . 'Fewer than 6 rolls' means an A appears on the 1st, 2nd, 3rd, 4th, or 5th roll. This is the complement of no A in the first 5 rolls.

Working

The probability of rolling an A is:

P(A) = \frac{2}{8} = \frac{1}{4}

The probability of not rolling an A is:

P(\text{not A}) = \frac{6}{8} = \frac{3}{4}

The probability that no A appears in the first 5 rolls is:

\left(\frac{3}{4}\right)^5 = \frac{243}{1024}

Therefore:

P(X < 6) = 1 - \left(\frac{3}{4}\right)^5 = 1 - \frac{243}{1024} = \frac{781}{1024} \approx 0.763

Answer

\frac{781}{1024} \approx 0.763

Final answer

781/1024 ≈ 0.763

Detailed explanation

Walkthrough

We need the probability that fewer than 6 rolls are needed to get an A. Each roll is independent, and the probability of an A on any roll is $\frac{1}{4}$ . The complement of 'no A in the first 5 rolls' is exactly 'an A appears within the first 5 rolls', which is 'fewer than 6 rolls'. So we compute $\left(\frac{3}{4}\right)^5$ and subtract from 1.

Key Takeaways

This is a classic complement-rule application for repeated independent trials. Recognising that 'fewer than $k$ ' is the complement of 'no success in the first $k-1$ trials' simplifies the calculation.

Common Mistakes

Using $\left(\frac{3}{4}\right)^5$ as the final answer instead of $1 - \left(\frac{3}{4}\right)^5$ .
Using $P(A) = \frac{1}{8}$ instead of $\frac{1}{4}$ .
Confusing 'fewer than 6' with 'fewer than or equal to 6'.

Things to Be Careful

The dice has 8 faces, two of which are A, so $P(A) = \frac{2}{8} = \frac{1}{4}$ .
'Fewer than 6' means rolls 1 through 5, so the complement is 5 consecutive non-A rolls.
The rolls are independent, so the multiplication law applies.

Techniques used

compute the probability of a single faceapply the complement rule for repeated trialsuse the multiplication law for independent rolls

(c)

Find the probability that the second A is obtained on the 6th roll of the dice.

2M Similar

DifficultyMedium

Worked solution

Approach

The second A is obtained on the 6th roll, so exactly one A must appear in the first 5 rolls, and the 6th roll must be an A.

Working

The probability of exactly one A in the first 5 rolls is:

\binom{5}{1} \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^4 = 5 \times \frac{1}{4} \times \frac{81}{256} = \frac{405}{1024}

The probability that the 6th roll is A is:

\frac{1}{4}

Therefore the required probability is:

\frac{405}{1024} \times \frac{1}{4} = \frac{405}{4096} \approx 0.0989

Answer

\frac{405}{4096} \approx 0.0989

Final answer

405/4096 ≈ 0.0989

Detailed explanation

Walkthrough

The second A on the 6th roll requires exactly one A in the first 5 rolls and an A on the 6th. The binomial coefficient $\binom{5}{1}$ counts the number of ways to place the single A among the first 5 rolls. The probability of exactly one A in 5 rolls is $\binom{5}{1} \times \frac{1}{4} \times \left(\frac{3}{4}\right)^4$ . We then multiply by $\frac{1}{4}$ for the 6th roll to get the final probability.

Key Takeaways

This is a negative-binomial distribution scenario: the $r$ -th success occurs on a specific trial. The key is to recognise the pattern and use the binomial coefficient to count the arrangements of the earlier successes.

Common Mistakes

Forgetting the factor $\binom{5}{1} = 5$ .
Using $\left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^4$ without the binomial coefficient.
Confusing this with the probability of the first A on the 6th roll.

Things to Be Careful

The factor 5 accounts for the five possible positions of the single A among the first 5 rolls.
The 6th roll must be A, so we multiply by $\frac{1}{4}$ .
The events are independent, so the multiplication law applies.

Techniques used

identify the pattern of exactly one success in the first five trialsapply the binomial coefficient to count arrangementsmultiply by the probability of success on the sixth trial

Q3MediumRepresentation of Data

Last Sunday, teams of runners took part in a charity event. The time taken, in seconds, to run $50\text{ m}$ was recorded, correct to 1 decimal place, for each runner. The times recorded for 11 runners from each of the Gulls and the Herons are shown in the table.

Gulls	7.9	8.2	8.3	8.6	8.6	8.8	9.2	9.7	9.8	10.0	10.4
Herons	9.5	9.9	8.5	8.1	9.2	10.8	8.3	9.7	9.3	9.9	8.7

(a)

Draw a back-to-back stem-and-leaf diagram to represent this information, with Gulls on the left-hand side.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Sort the times for each team, then draw a back-to-back stem-and-leaf diagram with the stem as the common integer part and the leaves as the tenths digit, with Gulls on the left.

Working

Gulls times sorted ascending: $7.9,\ 8.2,\ 8.3,\ 8.6,\ 8.6,\ 8.8,\ 9.2,\ 9.7,\ 9.8,\ 10.0,\ 10.4$

Herons times sorted ascending: $8.1,\ 8.3,\ 8.5,\ 8.7,\ 9.2,\ 9.3,\ 9.5,\ 9.7,\ 9.9,\ 9.9,\ 10.8$

Using the integer part as the stem and the tenths digit as the leaf:

Answer

The back-to-back stem-and-leaf diagram is shown above, with Gulls on the left and Herons on the right.

Final answer

Back-to-back stem-and-leaf diagram with Gulls on the left and Herons on the right.

Detailed explanation

Walkthrough

A stem-and-leaf diagram needs ordered data. First sort the Gulls and Herons times separately. The stem is the integer part of the time, so the stems are 7, 8, 9 and 10; the leaf is the tenths digit. For Gulls, place each leaf on the left of the corresponding stem, writing the leaves in increasing order as they move away from the stem. For Herons, place the leaves on the right of the stem, again in increasing order. Finally, add a key that explains how to read the diagram, for example 7|9|5 means 9.7 seconds for Gulls and 9.5 seconds for Herons.

Key Takeaways

A back-to-back stem-and-leaf diagram is a compact way to compare two data sets. The stem is shared, one data set appears on the left and the other on the right, and the leaves must be ordered. A key is essential because it explains how the stem and leaves combine to give the original times.

Common Mistakes

Placing the Gulls leaves on the wrong side of the stem. Writing leaves in the wrong order or with commas or other punctuation. Forgetting to include a key or not stating the units. Using a split stem without adjusting the layout correctly.

Things to Be Careful About

The mark scheme requires Gulls on the left, leaves ordered and vertically aligned, and no commas or punctuation between leaves. If a split stem is used, the diagram is still valid, but the stem and leaf marks are awarded differently. The key must identify both teams and state 'sec' or 's'.

Techniques used

sort data into ascending orderconstruct a back-to-back stem-and-leaf diagramprovide a key for the diagram

(b)

Find the median and the interquartile range of the times of the runners from the Gulls.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

With 11 Gulls times, the median is the 6th value. Split the ordered list into the lower and upper halves of 5 values to find the lower and upper quartiles, then subtract them.

Working

Ordered Gulls times:

7.9,\ 8.2,\ 8.3,\ 8.6,\ 8.6,\ 8.8,\ 9.2,\ 9.7,\ 9.8,\ 10.0,\ 10.4

Median position:

\frac{11+1}{2}=6

So the median is the 6th value:

8.8

Lower 5 values: $7.9,\ 8.2,\ 8.3,\ 8.6,\ 8.6$ , so the lower quartile is the 3rd value:

8.3

Upper 5 values: $9.2,\ 9.7,\ 9.8,\ 10.0,\ 10.4$ , so the upper quartile is the 3rd value:

9.8

Interquartile range:

\text{IQR}=9.8-8.3=1.5

Answer

Median $= 8.8$ seconds; IQR $= 1.5$ seconds.

Final answer

Median = 8.8 seconds; IQR = 1.5 seconds

Detailed explanation

Walkthrough

There are 11 Gulls times, so after ordering them the median is the 6th value. The lower half contains the first 5 values, and its 3rd value is the lower quartile. The upper half contains the last 5 values, and its 3rd value is the upper quartile. The interquartile range is the difference between the upper and lower quartiles, which measures the spread of the middle 50% of the data.

Key Takeaways

For an odd-sized ordered data set, the median is the middle value. The lower and upper quartiles are found from the lower and upper halves, and the IQR is the difference between them.

Common Mistakes

Forgetting to sort the data before finding the median and quartiles. Taking the wrong positions for the quartiles. Reporting the median but not clearly identifying the quartiles before calculating the IQR.

Things to Be Careful About

The mark scheme accepts a small range for the quartile values, but the final IQR should be 1.5. Clearly label the median, lower quartile and upper quartile so the marks for each are awarded.

Techniques used

identify the median positionfind the lower and upper quartilescalculate the interquartile range

(c)

Two other teams of runners, the Eagles and the Swifts, also took part in the event. The recorded times in seconds for 20 runners from the Eagles and 30 runners from the Swifts are denoted by $x$ and $y$ respectively.

It is given that $\sum x = 175.0$ and that the mean of $y$ is $8.4$ .

Find the mean of the times taken by all 50 runners.

2M Similar

(d)

It is given that $\sum x^2 = 1823.0$ .

It is also known that the standard deviation of the times taken by all 50 runners is $1.38$ seconds.

Find the value of $\sum y^2$ , correct to 1 decimal place.

3M Similar

DifficultyMedium

Worked solution

Approach

Use the variance formula for all 50 runners: the variance is the mean of the squares minus the square of the mean. The standard deviation is 1.38, so the variance is $1.38^2$ . The total of all times is $427.0$ and the total sum of squares is $1823.0 + \sum y^2$ . Substitute these values and solve for $\sum y^2$ .

Working

Total of all times:

\sum_{\text{all}} t = 175.0 + 30 \times 8.4 = 427.0

Variance formula:

1.38^2 = \frac{1823.0 + \sum y^2}{50} - \left(\frac{427.0}{50}\right)^2

Rearrange:

\frac{1823.0 + \sum y^2}{50} = 1.38^2 + \left(\frac{427.0}{50}\right)^2

Evaluate:

1.38^2 = 1.9044

\left(\frac{427.0}{50}\right)^2 = 8.54^2 = 72.9316

\frac{1823.0 + \sum y^2}{50} = 1.9044 + 72.9316 = 74.836

So:

1823.0 + \sum y^2 = 50 \times 74.836 = 3741.8

\sum y^2 = 3741.8 - 1823.0 = 1918.8

Answer

$\sum y^2 = 1918.8$ , correct to 1 decimal place.

Final answer

1918.8

Detailed explanation

Walkthrough

The variance of a data set is the mean of the squares minus the square of the mean. For all 50 runners, the standard deviation is 1.38, so the variance is $1.38^2$ . The total of all times is $427.0$ from part (c). The total sum of squares is the Eagles' sum of squares, $1823.0$ , plus the Swifts' unknown sum of squares, $\sum y^2$ . Substitute these into the variance formula and solve the resulting equation for $\sum y^2$ .

Key Takeaways

Variance is the mean of the squares minus the square of the mean. When the standard deviation and all but one sum of squares are known, the unknown sum of squares can be found by rearranging the variance formula.

Common Mistakes

Forgetting to square the standard deviation when using the variance formula. Using 175.0 as the total of all times instead of 427.0. Forgetting to subtract 1823.0 after multiplying by 50.

Things to Be Careful About

Use the population variance formula with denominator 50, not a sample denominator. Keep the intermediate values accurate; the final answer is 1918.8 to 1 decimal place.

Techniques used

substitute values into the variance formulasolve for the unknown sum of squaresuse the standard deviation squared

Q4Medium-HardThe Normal DistributionDiscrete Random Variables

Every Saturday, a particular community holds a ‘Puzzle’ event to raise money for a new Leisure Centre. Competitors attempt to solve a puzzle as quickly as possible.

Last Saturday, 600 competitors took part. The times taken to complete the puzzle were normally distributed with mean $32.4$ minutes and standard deviation $2.5$ minutes.

(a)

How many competitors would you expect to have times within $1.2$ minutes of the mean time?

4M Similar

DifficultyMedium

Worked solution

Approach

Standardise the two boundary times using $z = \frac{x - \mu}{\sigma}$ , use the symmetry of the standard normal distribution to find the probability of being within 1.2 minutes of the mean, then multiply by the number of competitors.

Working

The times are normally distributed with $\mu = 32.4$ minutes and $\sigma = 2.5$ minutes.

Times within 1.2 minutes of the mean satisfy $31.2 < X < 33.6$ .

Standardising:

z = \frac{31.2 - 32.4}{2.5} = -0.48, \qquad z = \frac{33.6 - 32.4}{2.5} = 0.48

So

\text{P}(31.2 < X < 33.6) = \text{P}(-0.48 < Z < 0.48)

By symmetry:

\text{P}(-0.48 < Z < 0.48) = 2\Phi(0.48) - 1 = 2(0.6844) - 1 = 0.3688

Expected number:

0.3688 \times 600 = 221.28

Answer

221 or 222 competitors.

Final answer

221 or 222 competitors

Detailed explanation

Walkthrough

We are told the times are normally distributed with mean 32.4 and standard deviation 2.5. "Within 1.2 minutes of the mean" means between 32.4 - 1.2 = 31.2 and 32.4 + 1.2 = 33.6 minutes. To use the standard normal table, we convert these boundaries to z-scores using $z = \frac{x - \mu}{\sigma}$ . This gives -0.48 and 0.48. Because the normal curve is symmetric, the probability of being between -0.48 and 0.48 is twice the probability of being below 0.48 minus 1 (or equivalently, $2\Phi(0.48) - 1$ ). From the table, $\Phi(0.48) = 0.6844$ , so the probability is $2(0.6844) - 1 = 0.3688$ . Finally, to find the expected number out of 600, multiply: $0.3688 \times 600 = 221.28$ , which rounds to 221 or 222.

Key Takeaways

This question tests the ability to standardise a normal variable and use the symmetry of the standard normal distribution. It also connects a probability to an expected frequency by multiplying by the sample size.

Common Mistakes

Using the variance instead of the standard deviation in the standardisation formula (the mark scheme explicitly rejects $\sigma^2$ ).
Forgetting to use both limits (31.2 and 33.6) symmetrically.
Using a continuity correction — not needed here because the underlying times are continuous.
Not rounding the expected count to a whole number of competitors.

Things to Be Careful About

The standardisation must use $\sigma = 2.5$ , not $\sigma^2 = 6.25$ .
The final expected number must be a single integer (221 or 222); the mark scheme allows either.
The probability must be quoted to 4 significant figures (0.3688) before multiplying.

Techniques used

standardise the normal variable using z = (x - μ)/σuse symmetry of the standard normal curve to find the probabilitymultiply the probability by the sample size

(b)

In this Saturday’s event, $60\%$ of the competitors had times less than $36.0$ minutes.

9 competitors who took part in this Saturday’s event are selected at random.

Find the probability that at least 2 and fewer than 8 of these competitors had times less than $36.0$ minutes.

3M Similar

DifficultyMedium

Worked solution

Approach

Let $X$ be the number of competitors (out of 9) with times less than 36.0 minutes. Then $X \sim \text{B}(9, 0.6)$ . The required probability is $\text{P}(2 \leq X < 8)$ , which is easier to compute as $1 - \text{P}(X = 0, 1, 8, 9)$ .

Working

$p = 0.6$ , so $1 - p = 0.4$ .

\text{P}(2 \leq X < 8) = 1 - \text{P}(X = 0) - \text{P}(X = 1) - \text{P}(X = 8) - \text{P}(X = 9)

= 1 - \left(0.4^9 + {}^9\text{C}_1(0.4)^8(0.6) + {}^9\text{C}_8(0.4)(0.6)^8 + 0.6^9\right)

= 1 - (0.000262144 + 0.00353894 + 0.060466176 + 0.010077696)

= 1 - 0.074345\ldots = 0.92565\ldots

Answer

0.926

Final answer

0.926

Detailed explanation

Walkthrough

We know 60% of competitors had times less than 36.0 minutes, so for each randomly selected competitor, the probability of success (time less than 36.0) is $p = 0.6$ . With 9 competitors, the number $X$ with times less than 36.0 follows a binomial distribution $\text{B}(9, 0.6)$ . We need $\text{P}(2 \leq X < 8)$ , i.e. $X = 2, 3, 4, 5, 6,$ or $7$ . Rather than summing six terms, it is easier to use the complement: $\text{P}(2 \leq X < 8) = 1 - \text{P}(X = 0, 1, 8, 9)$ . Each term uses the binomial formula $\text{C}(n,x)\,p^x(1-p)^{n-x}$ . Substituting $p = 0.6$ and $1-p = 0.4$ gives the four terms, which sum to about 0.074345. Subtracting from 1 gives 0.92565..., which rounds to 0.926.

Key Takeaways

This question tests the binomial distribution and the use of the complement rule to simplify probability calculations. It also requires careful evaluation of binomial probabilities.

Common Mistakes

Using $p = 0.4$ instead of 0.6 (confusing success and failure).
Computing $\text{P}(2 < X < 8)$ instead of $\text{P}(2 \leq X < 8)$ , which would exclude 2 and 7.
Forgetting to include both $X = 8$ and $X = 9$ in the complement.
Arithmetic errors when evaluating the powers and combinations.

Things to Be Careful About

The complement must include $X = 0, 1, 8,$ and $9$ — all four values, not just the two extremes.
The final answer must be in the range $0.925 < p \leq 0.926$ per the mark scheme.
Use the correct binomial coefficient notation and evaluate carefully.

Techniques used

identify the binomial distribution B(9, 0.6)use the complement rule to avoid summing six termscompute binomial probabilities

(c)

80 competitors who took part in this Saturday’s event are selected at random.

Use a suitable approximation to find the probability that more than 50 of these competitors had times less than $36.0$ minutes.

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Let $X$ be the number of competitors (out of 80) with times less than 36.0 minutes. Then $X \sim \text{B}(80, 0.6)$ . Since $n$ is large, approximate $X$ by a normal distribution with mean $np$ and variance $np(1-p)$ . Apply a continuity correction because $X$ is discrete, then standardise and use the normal tables.

Working

$n = 80$ , $p = 0.6$ .

\text{Mean} = np = 80 \times 0.6 = 48

\text{Variance} = np(1-p) = 80 \times 0.6 \times 0.4 = 19.2

So $X \approx \text{N}(48, 19.2)$ .

We need $\text{P}(X > 50)$ . With the continuity correction, $X > 50$ becomes $X \geq 51$ , so we use 50.5:

\text{P}(X > 50) \approx \text{P}\left(Z > \frac{50.5 - 48}{\sqrt{19.2}}\right)

= \text{P}(Z > 0.571)

= 1 - \Phi(0.571) = 1 - 0.7160 = 0.284

Answer

0.284

Final answer

0.284

Detailed explanation

Walkthrough

Again, $X$ is the number of competitors with times less than 36.0 minutes, now out of 80, so $X \sim \text{B}(80, 0.6)$ . Because $n = 80$ is large and $np = 48$ and $n(1-p) = 32$ are both greater than 5, the normal approximation is suitable. The mean is $np = 48$ and the variance is $np(1-p) = 19.2$ . Since $X$ is discrete, we apply a continuity correction: $\text{P}(X > 50)$ becomes $\text{P}(X \geq 51)$ , which we approximate using the boundary 50.5. Standardising gives $z = \frac{50.5 - 48}{\sqrt{19.2}} = 0.571$ . From the normal table, $\Phi(0.571) = 0.7160$ , so $\text{P}(Z > 0.571) = 1 - 0.7160 = 0.284$ .

Key Takeaways

This question tests the normal approximation to the binomial distribution, including the continuity correction. It combines the binomial mean/variance formulas with standardisation.

Common Mistakes

Forgetting the continuity correction (the mark scheme requires 49.5 or 50.5).
Using the wrong boundary: for $\text{P}(X > 50)$ the correct boundary is 50.5, not 49.5 (49.5 would be for $\text{P}(X \geq 50)$ ).
Using variance instead of standard deviation when standardising (dividing by 19.2 instead of $\sqrt{19.2}$ ).
Reading the wrong tail of the normal table.

Things to Be Careful About

The continuity correction must use 50.5 because "more than 50" means $X \geq 51$ , so we use the boundary halfway between 50 and 51.
The mark scheme awards B1 for mean = 48 and variance = 19.2 (CAO), and requires the standardisation to use their values.
The final answer should be less than 0.5 (since 50 is slightly above the mean of 48, the probability of exceeding 50 is less than half).

Techniques used

compute mean and variance of the binomial distributionapply the continuity correctionstandardise using the normal approximationuse the standard normal table

Q5MediumPermutations and Combinations

In a group of 20 musicians, there are 9 guitarists, 6 pianists and 5 drummers.

6 musicians are selected from these 20 to perform at a concert.

(a)

Find the number of different ways in which the 6 musicians can be selected if there must be at least 3 guitarists, at most 2 pianists and exactly 1 drummer.

4M Similar

DifficultyMedium

Worked solution

Approach

We need exactly 1 drummer, so choose 1 drummer from the 5 drummers. The remaining 5 musicians must be chosen from the 9 guitarists and 6 pianists. The conditions "at least 3 guitarists" and "at most 2 pianists" give exactly three possible cases:

3 guitarists, 2 pianists, 1 drummer
4 guitarists, 1 pianist, 1 drummer
5 guitarists, 0 pianists, 1 drummer

Count each case using combinations, then add the results.

Working

Exactly 1 drummer from 5:

\binom{5}{1}

Case 1: 3 guitarists, 2 pianists, 1 drummer.

\binom{9}{3}\binom{6}{2}\binom{5}{1} = 84 \cdot 15 \cdot 5 = 6300

Case 2: 4 guitarists, 1 pianist, 1 drummer.

\binom{9}{4}\binom{6}{1}\binom{5}{1} = 126 \cdot 6 \cdot 5 = 3780

Case 3: 5 guitarists, 0 pianists, 1 drummer.

\binom{9}{5}\binom{6}{0}\binom{5}{1} = 126 \cdot 1 \cdot 5 = 630

Total number of ways:

6300 + 3780 + 630 = 10710

Answer

10710

Final answer

10710

Detailed explanation

Walkthrough

We have 9 guitarists, 6 pianists and 5 drummers, and we must choose 6 musicians. The condition says exactly 1 drummer, so we first choose 1 drummer from the 5 drummers. After that, the other 5 musicians must be chosen from the guitarists and pianists only.

The condition "at least 3 guitarists" and "at most 2 pianists" means the number of guitarists chosen must be 3, 4 or 5. If we chose 2 guitarists, we would need 3 pianists, which violates "at most 2 pianists". If we chose 6 guitarists, there would be no room for a drummer, so that is impossible. Therefore the only possible splits are:

3 guitarists and 2 pianists
4 guitarists and 1 pianist
5 guitarists and 0 pianists

For each split, we multiply the number of ways to choose the guitarists, the pianists and the drummer. Because these cases are mutually exclusive, we add their totals.

Key Takeaways

When order does not matter, use combinations.
"At least" and "at most" conditions can be handled by listing all valid cases.
When cases are mutually exclusive, add their counts.

Common Mistakes

Forgetting to choose the drummer at all.
Missing the case with 5 guitarists and 0 pianists.
Including invalid cases such as 2 guitarists and 3 pianists.
Multiplying the case totals instead of adding them.

Things to Be Careful About

Exactly 1 drummer means one of the 5 drummers must be selected.
"At most 2 pianists" includes the possibility of 0 pianists.
The total number of musicians selected must always be 6.

Techniques used

enumerate the valid cases satisfying the constraintscount each case using combinationssum the case totals

(b)

Three bands will be selected from the original group of 20 musicians. Each band will consist of 3 guitarists, 1 pianist and 1 drummer. No musician can be in more than one band. The first band selected will play at a concert in France, the second band selected will play in Italy and the third band selected will play in Spain.

Find the number of different ways in which these three bands can be selected.

3M Similar

DifficultyMedium

Worked solution

Approach

The three bands are ordered: France, Italy, Spain. Choose the first band from the original 20 musicians, then choose the second band from the remaining musicians, then choose the third band from the musicians left after that. Each band needs 3 guitarists, 1 pianist and 1 drummer, and no musician can be used more than once.

Working

First band (France): choose 3 guitarists from 9, 1 pianist from 6 and 1 drummer from 5.

\binom{9}{3}\binom{6}{1}\binom{5}{1} = 84 \cdot 6 \cdot 5 = 2520

After the first band, the remaining musicians are 6 guitarists, 5 pianists and 4 drummers.

Second band (Italy):

\binom{6}{3}\binom{5}{1}\binom{4}{1} = 20 \cdot 5 \cdot 4 = 400

After the second band, the remaining musicians are 3 guitarists, 4 pianists and 3 drummers.

Third band (Spain):

\binom{3}{3}\binom{4}{1}\binom{3}{1} = 1 \cdot 4 \cdot 3 = 12

Total number of ways:

2520 \times 400 \times 12 = 12096000

Answer

$12096000$

Final answer

12096000

Detailed explanation

Walkthrough

We have three bands, and each band must contain 3 guitarists, 1 pianist and 1 drummer. The bands are labelled by their concert location: France, Italy and Spain, so the order matters.

Start with the first band. Choose 3 guitarists from the 9 available, 1 pianist from the 6 available, and 1 drummer from the 5 available. This gives 2520 ways.

Since no musician can be in more than one band, those musicians are now unavailable. So for the second band there are 6 guitarists, 5 pianists and 4 drummers left. Choose 3 guitarists, 1 pianist and 1 drummer from these, giving 400 ways.

After the second band, there are 3 guitarists, 4 pianists and 3 drummers left. For the third band, we must choose all 3 remaining guitarists, 1 of the 4 pianists and 1 of the 3 drummers, giving 12 ways.

Finally, multiply the three numbers because all three selections must happen together. The result is 12,096,000.

Key Takeaways

When selections are made without replacement and the groups are ordered, multiply the sequential combination counts.

The order of the bands is already accounted for by choosing first, second and third, so no extra factorial is needed.

Use combinations to choose within each category because the musicians within a band are not arranged.

Common Mistakes

Forgetting that musicians used in the first band cannot be used again.
Using the original counts for every band instead of reducing them after each selection.
Multiplying by 3! after already counting the bands in order, which would overcount.
Forgetting that the third band must take all 3 remaining guitarists, giving exactly 1 way for the guitarists.

Things to Be Careful About

After the first band, the remaining counts are 6 guitarists, 5 pianists and 4 drummers.

After the second band, the remaining counts are 3 guitarists, 4 pianists and 3 drummers.

The final product is 2520 × 400 × 12 = 12,096,000.

Techniques used

choose the first band from the original groupupdate the remaining counts after each bandmultiply the sequential selection counts

Q6MediumProbabilityDiscrete Random Variables

A bag contains 10 marbles, of which 4 are red and 6 are blue. Four marbles are selected from the bag at random, without replacement. The random variable $X$ denotes the number of blue marbles selected.

(a)

Show that $P(X = 2) = \frac{3}{7}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Total selections of 4 marbles from 10 are $\binom{10}{4}$ . For $X=2$ , choose 2 blue from 6 and 2 red from 4.

Working

\binom{10}{4} = 210

The number of favourable selections is

\binom{6}{2}\binom{4}{2} = 15 \times 6 = 90

Therefore

P(X=2) = \frac{\binom{6}{2}\binom{4}{2}}{\binom{10}{4}} = \frac{90}{210} = \frac{3}{7}

Answer

P(X=2) = \frac{3}{7}

Final answer

P(X=2) = 3/7

Detailed explanation

Walkthrough

We are choosing marbles without replacement, so the order does not matter and each selection is counted by a combination. The total number of equally likely selections of 4 marbles from 10 is $\binom{10}{4}=210$ . To get exactly two blue marbles, we choose 2 of the 6 blue marbles and 2 of the 4 red marbles, giving $\binom{6}{2}\binom{4}{2}=15\times6=90$ favourable selections. Since all selections are equally likely, the probability is $\frac{90}{210}=\frac{3}{7}$ .

Key Takeaways

When a probability question asks for selections without replacement, combinations are usually the simplest tool. If the selection must contain certain numbers from different categories, multiply the combinations for each category.

Common Mistakes

Counting only the blue choices and forgetting to choose the red marbles.
Mixing permutations and combinations: the numerator and denominator must use the same counting convention.
Stopping at $\frac{90}{210}$ without simplifying, even though the question requires showing it equals $\frac{3}{7}$ .

Things to Be Careful About

This is an answer-given (AG) question: the final value is stated, so full working must be shown to earn the marks. Any valid method (combinations, or ordered selections with a consistent arrangement factor) is accepted, but the working must be consistent and complete.

Techniques used

count total selections using combinationscount favourable selections by categorysimplify probability fraction

(b)

Draw up the probability distribution table for $X$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

The possible values of $X$ are $0,1,2,3,4$ . For each $x$ , select $x$ blue marbles and $4-x$ red marbles. Use combinations to calculate each probability.

Working

The total number of selections is

\binom{10}{4} = 210

For $x=0$ (all red):

P(X=0)=\frac{\binom{4}{4}\binom{6}{0}}{210}=\frac{1}{210}

For $x=1$ :

P(X=1)=\frac{\binom{6}{1}\binom{4}{3}}{210}=\frac{24}{210}=\frac{4}{35}

For $x=2$ :

P(X=2)=\frac{\binom{6}{2}\binom{4}{2}}{210}=\frac{90}{210}=\frac{3}{7}

For $x=3$ :

P(X=3)=\frac{\binom{6}{3}\binom{4}{1}}{210}=\frac{80}{210}=\frac{8}{21}

For $x=4$ (all blue):

P(X=4)=\frac{\binom{6}{4}\binom{4}{0}}{210}=\frac{15}{210}=\frac{1}{14}

The probability distribution table is:

$x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{1}{210}$	$\frac{4}{35}$	$\frac{3}{7}$	$\frac{8}{21}$	$\frac{1}{14}$

Check the probabilities sum to 1:

\frac{1}{210}+\frac{4}{35}+\frac{3}{7}+\frac{8}{21}+\frac{1}{14}=1

Answer

$x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{1}{210}$	$\frac{4}{35}$	$\frac{3}{7}$	$\frac{8}{21}$	$\frac{1}{14}$

Final answer

x = 0: 1/210, x = 1: 4/35, x = 2: 3/7, x = 3: 8/21, x = 4: 1/14

Detailed explanation

Walkthrough

There are 4 marbles chosen from 10, so the total number of outcomes is $\binom{10}{4}=210$ . For each possible value $x$ of blue marbles, we choose $x$ of the 6 blue marbles and $4-x$ of the 4 red marbles. For $x=0$ , choose 0 blue and 4 red: $\binom{4}{4}=1$ , so $P(X=0)=1/210$ . For $x=1$ , choose 1 blue and 3 red: $6\times4=24$ , giving $4/35$ . For $x=2$ , choose 2 blue and 2 red: $15\times6=90$ , giving $3/7$ . For $x=3$ , choose 3 blue and 1 red: $20\times4=80$ , giving $8/21$ . For $x=4$ , choose 4 blue and 0 red: $15$ , giving $1/14$ . Place these in a table and note they sum to 1.

Key Takeaways

The probability distribution of a discrete random variable lists every possible value and its probability. The probabilities must be non-negative and sum to exactly 1, which provides a check on the work.

Common Mistakes

Missing the extreme values $x=0$ or $x=4$ .
Forgetting that when choosing $x$ blue marbles you must also choose $4-x$ red marbles.
Giving probabilities that do not sum to 1, which indicates an arithmetic or counting error.

Things to Be Careful About

The values in the table can be given as fractions or decimals, but should match the accuracy expected by the mark scheme. Use the same denominator when checking the sum. The mark scheme allows probabilities not placed in a table if clearly identified, but a neat table is expected.

Techniques used

list all possible values of Xcalculate each probability using combinationstabulate probabilities and check the sum is 1

(c)

Find the probability that at least 2 of the marbles chosen are blue, given that at least 1 red marble and at least 1 blue marble are chosen.

3M Similar

DifficultyMedium

Worked solution

Approach

The event "at least 1 red and at least 1 blue" means $X$ cannot be 0 or 4, so the conditioning event is $X \in \{1,2,3\}$ . We want $P(X \geq 2)$ conditional on this.

Working

First find the probability of the conditioning event:

P(X=1)+P(X=2)+P(X=3)

= \frac{4}{35}+\frac{3}{7}+\frac{8}{21}

= \frac{12}{105}+\frac{45}{105}+\frac{40}{105} = \frac{97}{105}

The intersection event "at least 2 blue, and at least 1 red and at least 1 blue" is $X \in \{2,3\}$ :

P(X=2)+P(X=3)=\frac{3}{7}+\frac{8}{21}=\frac{17}{21}

Therefore

P(X \geq 2 \mid \text{at least 1 red and at least 1 blue}) = \frac{\frac{17}{21}}{\frac{97}{105}}

= \frac{17}{21}\times\frac{105}{97} = \frac{85}{97} \approx 0.876

Answer

\frac{85}{97} \approx 0.876

Final answer

85/97 (approximately 0.876)

Detailed explanation

Walkthrough

We are asked for a conditional probability. The condition is that at least one red and at least one blue marble are chosen. This means the selection cannot be all red ( $X=0$ ) or all blue ( $X=4$ ), so the possible values are $X=1$ , $X=2$ or $X=3$ . The denominator is therefore $P(X=1)+P(X=2)+P(X=3)=\frac{4}{35}+\frac{3}{7}+\frac{8}{21}=\frac{97}{105}$ . The numerator is the part of this condition where at least two marbles are blue, namely $X=2$ or $X=3$ , so $P(X=2)+P(X=3)=\frac{17}{21}$ . Dividing gives $\frac{17/21}{97/105}=\frac{85}{97}\approx0.876$ .

Key Takeaways

Conditional probability is found by restricting the sample space to the given event. Here the given event is best expressed as a set of values of $X$ . Mutually exclusive outcomes such as $X=2$ and $X=3$ have probabilities that can be added.

Common Mistakes

Including $X=0$ or $X=4$ in the denominator; both fail the condition that there is at least one marble of each colour.
Using $P(X\geq2 \mid X\geq1)$ instead of also imposing the red-marble condition.
Forgetting to use values from the distribution table, or carrying forward an incorrect probability from part (b) without following through.

Things to Be Careful About

Write the conditional probability as the ratio of the required intersection probability to the probability of the given condition. The mark scheme gives follow-through marks for using their probabilities from part (b), but the final simplified fraction must be correct for full marks. $\frac{85}{97}$ is already in lowest terms; it can also be given as $0.876$ to 3 significant figures.

Techniques used

identify the conditional event in terms of Xadd probabilities of mutually exclusive outcomesapply the conditional probability formula

Mathematics 9709/51 — May/June 2025

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