Mathematics 9709/52 — February/March 2024

Walkthrough

There are 12 marbles initially, 9 blue and 3 red.

For both marbles to be red, the first choice must be red: probability $\frac{3}{12}$. Since the first marble is red, it is not returned, so only 11 marbles remain and 2 of them are red. Therefore the probability that the second marble is also red is $\frac{2}{11}$.

Multiply these along the same branch: $\frac{3}{12} \times \frac{2}{11} = \frac{6}{132} = \frac{1}{22}$.

Key Takeaways

This question tests the multiplication law for dependent events. The key is to notice that the denominator changes from 12 to 11 only when the first marble is red and not replaced.

Common Mistakes

• Using $\frac{3}{12}$ for the second red marble instead of $\frac{2}{11}$, forgetting that the red marble was not replaced.
• Adding probabilities instead of multiplying them.
• Not simplifying $\frac{6}{132}$; although $\frac{1}{22}$ is not required for the mark, it is the cleanest form.

Things to Be Careful About

The mark scheme accepts $\frac{6}{132}$, $\frac{1}{22}$, or $0.04545...$ to at least three significant figures. Show the product clearly to earn the method mark.

Walkthrough

We are asked for the probability that the first marble was blue, given that the second marble is red. This is a conditional probability, so we use

The numerator is the probability that the first marble is blue and the second is red. If the first is blue, it is replaced, so the bag still has 12 marbles and 3 red marbles for the second draw. Thus

The denominator is the total probability that the second marble is red. This can happen in two ways: first blue then red, or first red then red. The first branch is the numerator above. The second branch is $\frac{3}{12} \times \frac{2}{11}$, because a red first marble is not replaced. So

Dividing the numerator by the denominator gives $\frac{33}{41}$.

Key Takeaways

Conditional probability is a ratio: the probability of the intersection divided by the probability of the conditioning event. The law of total probability is needed to find the denominator when the second event can occur through more than one first-step outcome.

Common Mistakes

• Writing the numerator as $\frac{9}{12} \times \frac{2}{11}$; since a blue marble is replaced, the second draw still has 3 red marbles out of 12.
• Forgetting the branch first red then red in the denominator.
• Confusing $P(B_1 | R_2)$ with $P(R_2 | B_1)$.
• Not showing the sum of the two branches, which loses the method marks.

Things to Be Careful About

The mark scheme awards M1 for seeing $\frac{9}{12} \times \frac{3}{12}$ or $\frac{3}{16}$ as numerator or denominator, M1 for adding the second branch, and A1 for the final answer $\frac{33}{41}$. If the final answer is correct without working, only SC B1 may be awarded, so show the full conditional-probability setup.

Walkthrough

We are told that Sam scores with probability $0.7$ on each attempt, independently. This is a classic binomial situation: a fixed number of independent trials, each with the same probability of success. Here the number of trials is $10$ and the success probability is $p = 0.7$.

We want the probability that she scores on fewer than 8 attempts, i.e. $X = 0, 1, \dots, 7$. Adding all eight of those probabilities would be tedious, so we use the complement. The only outcomes not included in $X < 8$ are $X = 8, 9, 10$. Therefore

Each probability is found from the binomial formula $\binom{10}{x}(0.7)^x(0.3)^{10-x}$. Substituting $x = 8, 9, 10$ gives the three terms shown in the working. Their sum is about $0.3828$, so the required probability is $1 - 0.3828 = 0.6172$, which rounds to $0.617$.

Key Takeaways

This question tests the binomial distribution model $X \sim B(n, p)$ and the use of the complement rule when a range of values is needed. It also tests the ability to evaluate binomial probabilities using the formula $\binom{n}{x}p^x(1-p)^{n-x}$.

Common Mistakes

• Trying to add $P(0)$ through $P(7)$ directly; this is longer and more prone to error.
• Forgetting to include $X = 10$ in the complement. The complement of $X < 8$ is $X \geq 8$, which includes $8, 9$ and $10$.
• Using $0.7$ and $0.7$ instead of $0.7$ and $0.3$ in the binomial formula.

Things to Be Careful About

• The phrase 'fewer than 8' means $X \leq 7$, not $X \leq 8$.
• The mark scheme accepts $0.617 \leq p < 0.6175$; any answer in that range is fine.
• Show the unsimplified binomial expression before evaluating, because the method mark is awarded for a correct term such as $\binom{10}{8}(0.7)^8(0.3)^2$.

Walkthrough

We are looking for the probability that Sam's first successful attempt happens before her 5th attempt. That means the first success must occur on attempt 1, 2, 3 or 4.

One way is to add the probabilities of success on attempt 1, or first failure then success, or two failures then success, or three failures then success:

A quicker way is to use the complement. The only way the first success is not before the 5th attempt is if Sam fails on all of the first 4 attempts. Since each failure has probability $0.3$, this happens with probability $0.3^4 = 0.0081$. Therefore

Key Takeaways

This is a geometric distribution problem: we count the number of trials until the first success. The complement trick $1 - (1-p)^k$ is very useful for 'first success within the first $k$ attempts'.

Common Mistakes

• Thinking 'before her 5th attempt' includes the 5th attempt. It does not; it means attempts 1, 2, 3 and 4 only.
• Using $0.7^4$ instead of $0.3^4$ for the complement. The complement is all four attempts fail, so the failure probability $0.3$ must be used.
• Forgetting that the attempts are independent, so probabilities multiply.

Things to Be Careful About

• The mark scheme allows $0.9919$ or $\frac{9919}{10000}$, and condones $0.992$.
• If using the direct addition method, make sure the last term is $(0.3)^3(0.7)$, not $(0.3)^4(0.7)$.
• The answer is close to 1, so a small arithmetic slip can change the final digits; keep at least 4 decimal places.

Walkthrough

Wei keeps attempting until he scores 3 goals. We want the probability that the 3rd goal happens exactly on the 7th attempt.

For this to happen, two conditions must be met:

1. In the first 6 attempts, Wei must score exactly 2 goals (and therefore fail 4 times).
2. On the 7th attempt, Wei must score.

The number of ways to choose which 2 of the first 6 attempts are successes is $\binom{6}{2} = 15$. Each such arrangement has probability $(0.6)^2(0.4)^4$ for the first 6 attempts, because there are 2 successes and 4 failures. Then the 7th attempt must be a success, with probability $0.6$. Multiplying gives

Evaluating this gives $0.082944$, which is $0.0829$ to 3 significant figures.

Key Takeaways

This is a negative binomial / waiting-time problem: the $r$th success occurs on the $n$th trial. The probability is

Here $r = 3$, $n = 7$, $p = 0.6$.

Common Mistakes

• Forgetting the factor $\binom{6}{2}$; the successes can be in any 2 of the first 6 attempts, not just a specific order.
• Using $\binom{7}{3}$ instead of $\binom{6}{2}$. The 7th attempt is fixed as a success, so the choices only apply to the first 6 attempts.
• Adding probabilities instead of multiplying. This is an 'and' situation: exactly 2 successes in the first 6 attempts AND a success on the 7th.

Things to Be Careful About

• The mark scheme accepts $0.0829$ or $0.082944$ correct to at least 3 significant figures, or the exact fraction $\frac{1296}{15625}$.
• Make sure the powers are correct: $(0.6)^3$ for the three total successes and $(0.4)^4$ for the four failures.
• If the final answer is given without working, the mark scheme may still award a special B1 for the correct answer, but showing the method is safer.

Walkthrough

A histogram is the correct choice when the class widths of grouped data are not all the same. The defining property of a histogram is that the area of each bar equals (or is proportional to) the frequency in that class. So the height of each bar must be chosen so that
$\text{height} \times \text{class width} = \text{frequency} \quad\Rightarrow\quad \text{height} = \frac{\text{frequency}}{\text{class width}} = \text{frequency density}.$
Computing each height:

• $0 \leq t < 20$: width $20$, so $\tfrac{8}{20} = 0.4$.
• $20 \leq t < 30$: width $10$, so $\tfrac{23}{10} = 2.3$.
• $30 \leq t < 35$: width $5$, so $\tfrac{35}{5} = 7.0$.
• $35 \leq t < 40$: width $5$, so $\tfrac{52}{5} = 10.4$.
• $40 \leq t < 50$: width $10$, so $\tfrac{20}{10} = 2.0$.
• $50 \leq t < 70$: width $20$, so $\tfrac{12}{20} = 0.6$.

The class boundaries 0, 20, 30, 35, 40, 50, 70 are where the bars meet (no gaps between adjacent bars). Place a linear scale on each axis starting at 0, with at least three values shown on each axis so the scale is unambiguous. The horizontal axis is labelled "Time taken (mins)" and the vertical axis "Frequency density".

Key Takeaways

• A histogram is used for grouped continuous data, with area representing frequency.
• Frequency density = frequency ÷ class width is the bar height.
• Both axes must be on a linear scale starting at 0, with each axis labelled.

Common Mistakes

• Plotting the raw frequencies on the vertical axis instead of frequency density (this only works when all class widths are equal).
• Leaving gaps between bars (a histogram is continuous — bars meet at the class boundaries).
• Using a non-linear scale or starting the vertical axis above 0.
• Forgetting to label the vertical axis "Frequency density".

Things to Be Careful About

• The maximum frequency density is $10.4$, so choose a vertical scale that comfortably accommodates this (e.g. 0 to 11).
• The mark scheme gives B1 for a correct linear scale on each axis (with at least three values shown) and B1 for the correct axis labels; the M1 is for the frequency-density calculations and the A1 is for the correct bar heights.

Walkthrough

Because we only know that each time lies somewhere in a class interval, we cannot compute the exact mean. Instead we estimate it by assuming every observation in a class is at the class midpoint. The midpoint of a class is the average of its two boundaries, so for $30 \leq t < 35$ the midpoint is $(30+35)/2 = 32.5$.

The grouped-mean formula weighs each midpoint by the number of observations in that class:
$\bar{x} = \frac{\sum f m}{\sum f}.$

Computing each $f \times m$:

• $10 \times 8 = 80$
• $25 \times 23 = 575$
• $32.5 \times 35 = 1137.5$
• $37.5 \times 52 = 1950$
• $45 \times 20 = 900$
• $60 \times 12 = 720$

Sum $= 80 + 575 + 1137.5 + 1950 + 900 + 720 = 5362.5$.

Dividing by the total frequency $150$:
$\bar{x} = \frac{5362.5}{150} = 35.75 \text{ minutes}.$

Key Takeaways

• For grouped data, the midpoint of each class is the representative value used in the mean.
• The grouped-mean formula is $\bar{x} = \dfrac{\sum f m}{\sum f}$.
• The result is an estimate, because we have approximated every observation in a class by a single value.

Common Mistakes

• Using the lower boundary or upper boundary of each class instead of the midpoint.
• Forgetting to multiply each midpoint by its frequency (treating every class as if it had the same frequency).
• Dividing by the number of classes (6) instead of by the total frequency (150).

Things to Be Careful About

• The frequencies must sum to $150$: $8 + 23 + 35 + 52 + 20 + 12 = 150$ — check this before dividing.
• The two non-integer midpoints are $32.5$ and $37.5$; carry the decimals through the calculation.
• The mark scheme accepts $35.75$ or $35\tfrac{3}{4}$ or $35.8$, but not the fraction $\tfrac{143}{4}$.

Walkthrough

The lower quartile $Q_1$ is the value below which $25\%$ of the data lie, so it sits at position $\tfrac{n}{4}$ in the ordered list. With $n = 150$ we need the $37.5$th observation.

A cumulative-frequency table adds each frequency to the running total:

• up to $t = 20$: $8$
• up to $t = 30$: $8 + 23 = 31$
• up to $t = 35$: $31 + 35 = 66$
• up to $t = 40$: $66 + 52 = 118$
• up to $t = 50$: $118 + 20 = 138$
• up to $t = 70$: $138 + 12 = 150$

The cumulative count crosses $37.5$ between $31$ and $66$, so the $37.5$th observation must lie in the class $30 \leq t < 35$.

Key Takeaways

• The lower quartile is at position $n/4$ in the ordered data.
• A cumulative-frequency table is the quickest way to locate a quartile for grouped data.
• For grouped data we can identify the class containing the quartile, but not its exact value, without interpolation.

Common Mistakes

• Confusing the lower quartile with the lower boundary of the data (i.e. the smallest value).
• Using $n/2$ (the median position) by mistake instead of $n/4$.
• Forgetting to convert the frequencies to cumulative frequencies before searching.

Things to Be Careful About

• The position $37.5$ is a value between the cumulative counts $31$ and $66$, so the class boundary $35$ is excluded in the upper-bound style of writing; the answer is the closed-at-lower-bound interval $30 \leq t < 35$.
• The mark scheme accepts the answer as "$30 \leq t < 35$", the "3rd interval", or simply "$30 - 35$".

Walkthrough

We are told the masses of the small bags are normally distributed with mean $1.20\text{ kg}$ and standard deviation $0.16\text{ kg}$. We want the expected number of bags, out of 500, with mass greater than $1.26\text{ kg}$.

First, we standardise the boundary $1.26$ using the formula $z = \frac{x - \mu}{\sigma}$. This gives $z = \frac{1.26 - 1.20}{0.16} = 0.375$, telling us that $1.26$ kg is 0.375 standard deviations above the mean.

Next, we find the probability that a single bag exceeds this mass: $P(X > 1.26) = P(Z > 0.375)$. Since the standard normal table gives the area to the left, we use $1 - \Phi(0.375) = 1 - 0.6462 = 0.3538$.

Finally, we multiply this probability by the sample size: $500 \times 0.3538 = 176.9$. Since we are counting bags, we round to the nearest whole number, giving 177.

Key Takeaways

• Standardisation converts any normal variable to the standard normal $Z$ so we can use the tables.
• The probability that a single item exceeds a value, multiplied by the sample size, gives the expected count.
• Counts must be reported as whole numbers.

Common Mistakes

• Using the variance $\sigma^2$ instead of the standard deviation $\sigma$ in the standardisation formula.
• Forgetting to subtract the table value from 1 when the question asks for a "greater than" probability.
• Leaving the expected count as a decimal instead of rounding to a whole number of bags.

Things to Be Careful About

• The mark scheme requires the standardisation formula with $\sigma = 0.16$, not $\sigma^2$.
• The final answer must be a single positive integer; the mark scheme accepts 176 or 177.

Walkthrough

For the large bags, $X \sim N(2.50, \sigma^2)$, and we are told that 20% have a mass less than $2.40\text{ kg}$.

We start with the probability statement $P(X < 2.40) = 0.20$ and standardise it: $P(Z < \frac{2.40 - 2.50}{\sigma}) = 0.20$.

We need the z-value whose lower-tail probability is 0.20. From the standard normal table, this is $z = -0.842$ (negative because $2.40$ is below the mean $2.50$).

Setting $\frac{2.40 - 2.50}{\sigma} = -0.842$ and solving gives $\sigma = \frac{0.1}{0.842} = 0.1188 \approx 0.119$.

Key Takeaways

• To find an unknown parameter from a probability, work backwards: read the inverse z-value first, then solve the equation.
• A lower-tail probability below 0.5 corresponds to a negative z-value.

Common Mistakes

• Using 0.20 or 0.80 directly as the z-value instead of finding the actual z from the table.
• Dropping the negative sign on the z-value, which would give a negative $\sigma$.

Things to Be Careful About

• The z-value must be in the range $-0.842 \leq z < -0.8415$.
• The final value of $\sigma$ must lie in $0.1185 < \sigma \leq 0.119$.

Walkthrough

Each large bag independently has probability 0.2 of having mass less than $2.40\text{ kg}$ (from part (b)). So the number $X$ of bags, out of 80, with mass below $2.40$ kg follows a binomial distribution $X \sim B(80, 0.2)$.

Because $n = 80$ is large, we approximate the binomial with a normal distribution. The mean is $np = 80 \times 0.2 = 16$ and the variance is $np(1-p) = 80 \times 0.2 \times 0.8 = 12.8$.

We want $P(X < 22)$. Since the binomial is discrete and the normal is continuous, we apply a continuity correction: "fewer than 22" means $X \leq 21$, which becomes $21.5$ in the normal approximation.

Standardising: $z = \frac{21.5 - 16}{\sqrt{12.8}} = \frac{5.5}{3.5777} = 1.537$. Then $P(Z < 1.537) = \Phi(1.537) = 0.938$.

Key Takeaways

• The normal approximation to the binomial uses mean $np$ and variance $np(1-p)$.
• The continuity correction is essential when approximating a discrete distribution with a continuous one.
• "Fewer than 22" translates to $X \leq 21$, hence the boundary $21.5$.

Common Mistakes

• Forgetting the continuity correction entirely.
• Using 22 instead of 21.5 in the standardisation.
• Using the small-bag parameters (mean 1.20, SD 0.16) instead of the binomial mean and variance.

Things to Be Careful About

• The mean must be 16 and the variance 12.8.
• The final probability must lie in $0.9375 < p \leq 0.938$.

Walkthrough

We want the probability that Anil scores exactly 2 points in 3 games. A game gives 0, 1 or 2 points, so a total of 2 can happen in only two ways: one win and two losses, or two draws and one loss. For the first way, the probability of one particular order such as WLL is $0.5 \times 0.2 \times 0.2$. Because the win could be in any of the 3 games, there are 3 orders, so multiply by 3. For the second way, one particular order such as DDL has probability $0.3 \times 0.3 \times 0.2$, and again there are 3 orders. These two cases cannot happen at the same time, so add their probabilities. This gives $0.06 + 0.054 = 0.114$.

Key Takeaways

This question uses the multiplication law for independent events and the addition law for mutually exclusive events. It also shows the importance of counting arrangements when several results are identical.

Common Mistakes

• Forgetting to multiply by 3 for the possible positions of the win or the loss.
• Using 6 arrangements instead of 3, because two of the three results are identical.
• Adding the cases without checking they are mutually exclusive.
• Writing only probabilities such as $0.5 \times 0.2^2$ without identifying the outcomes WLL and DDL; the mark scheme requires outcomes to be linked to probabilities.

Things to Be Careful About

This is a 'show that' question, so the final value must be exactly 0.114 and the working must justify it. Use $^3C_1$ or multiply by 3 explicitly. Remember that the games are independent, so multiplying probabilities is valid.

Walkthrough

Start with the totals not given. $X=0$ only when all three games are losses: $0.2^3=0.008$. $X=1$ only when there is one draw and two losses; the draw can be in any of 3 positions, so $3 \times 0.3 \times 0.2^2 = 0.036$. $X=3$ can happen in two ways: one win, one draw and one loss in any order (6 orders), or three draws. So $6 \times 0.5 \times 0.3 \times 0.2 + 0.3^3 = 0.18 + 0.027 = 0.207$. $X=5$ happens with two wins and one draw: $3 \times 0.5^2 \times 0.3 = 0.225$. The remaining values are given. Check the probabilities sum to 1: $0.008 + 0.036 + 0.114 + 0.207 + 0.285 + 0.225 + 0.125 = 1$.

Key Takeaways

Completing a probability distribution requires enumerating all possible outcomes and using independence and arrangements. Always check that probabilities sum to 1.

Common Mistakes

• Missing the three-draws case when calculating $X=3$.
• Forgetting the 6 arrangements for one win, one draw and one loss.
• Using 3 instead of 6 when all three results are different.
• Not checking that the final probabilities sum to 1.

Things to Be Careful About

The table already contains 0.114, 0.207, 0.285 and 0.125. Only three values are missing: 0.008, 0.036 and 0.225. If using the sum-to-one check, these three must sum to 0.269. The mark scheme allows a special case if three additional probabilities that sum to 0.269 are identified.

Walkthrough

Once the distribution is complete, expectation is the sum of each value multiplied by its probability. Calculate $E(X) = 3.9$. To find variance, use $\text{Var}(X) = E(X^2) - [E(X)]^2$. Calculate $E(X^2)$ by summing $x^2 p(x)$, giving 17.04. Then subtract $3.9^2 = 15.21$ to get 1.83. Alternatively, since each game has expected points 1.3 and variance 0.61, and the three games are independent, $\text{Var}(X) = 3 \times 0.61 = 1.83$.

Key Takeaways

Variance can be found from the distribution using $E(X^2) - [E(X)]^2$. For independent games, variances add, so the linearity shortcut is also valid.

Common Mistakes

• Using $\text{Var}(X) = E(X^2) - E(X)$ instead of subtracting $[E(X)]^2$.
• Forgetting to square the $x$-values when computing $E(X^2)$.
• Using the probabilities incorrectly or not following through from an incorrect table; method marks can still be earned if the formula is applied to their probabilities.

Things to Be Careful About

If using the table, make sure the probabilities sum to 1. If using the shortcut, remember that adding variances requires independence, which is given in the question. The final answer should be 1.83; $\frac{183}{100}$ is also acceptable.

Walkthrough

The committee is an unordered selection, so each choice is counted with combinations, not permutations. Start by listing all possible numbers of men and women on a 5-member committee that meet both conditions.

Since there are 6 men, at least 2 men means the possible numbers of men are 2, 3, 4 or 5. But a 5-member committee can contain at most 4 women, so 5 men and 0 women would violate at least 1 woman. Thus only 4 men and 1 woman, 3 men and 2 women, or 2 men and 3 women work.

For each case, choose the men and the women independently. For example, for 4 men and 1 woman, the number is $\binom{6}{4} \times \binom{4}{1}$. Compute each product, then add the three results. The cases are mutually exclusive because each committee has a unique number of men, so adding the three counts gives the total number of valid committees.

Key Takeaways

A selection without order is counted with combinations. The choice of men and the choice of women are independent, so multiply within each case. When a problem gives conditions such as at least, enumerate every valid split and add the disjoint cases.

Common Mistakes

• Missing one of the three valid splits, especially 2 men and 3 women.
• Including 5 men and 0 women, which violates at least 1 woman.
• Multiplying the three case totals instead of adding them.
• Using permutations when the order of committee members does not matter.

Things to Be Careful About

• The product in each scenario should have the form $\binom{6}{a}\binom{4}{b}$ with $a+b=5$.
• Each valid committee is counted in exactly one case, so there is no double-counting.
• Show the individual products before adding; an unsupported total of 240 may not earn full method marks.

Walkthrough

The condition that all men stand together and all women stand together means the 10 people form two blocks on the line: one block containing the 6 men and one block containing the 4 women.

First arrange the two blocks. The men's block can be first or the women's block can be first, giving 2 orders. Then arrange the 6 men inside their block in $6!$ ways and the 4 women inside their block in $4!$ ways. These choices are independent, so multiply: $6! \times 4! \times 2$.

Key Takeaways

The block method is the standard tool for together restrictions. Arrange the blocks, then arrange within each block, then multiply. The order of the two blocks matters.

Common Mistakes

• Forgetting the factor 2 for the order of the two blocks.
• Using $10!$ as if there were no restriction.
• Arranging members inside blocks but not considering which block comes first.

Things to Be Careful About

• All men together and all women together means exactly two blocks, so the only block orders are men-women and women-men.
• To earn full marks, show $6! \times 4! \times 2$, not just the final value.

Walkthrough

Olly and Petra must both be on the front row and must be adjacent. The front row has 4 positions, so choose the 2 other people who stand in the front row from the remaining 8 members: $\binom{8}{2}$ ways.

Now arrange the front row. Treat Olly and Petra as a single block, so the front row consists of three objects: the OP block and the two chosen extra people. These three objects can be placed in $3!$ orders. Inside the block, Olly and Petra can be ordered in 2 ways, so the front row has $3! \times 2$ arrangements.

The other 6 people are in the back row and can be arranged in $6!$ ways. Since the front row choice, front row arrangement, and back row arrangement are independent, multiply all factors:

$\binom{8}{2} \times 3! \times 2 \times 6! = 241\,920$.

Key Takeaways

For an adjacency restriction, merge the adjacent pair into one block and arrange the reduced number of objects. Use combinations to choose unordered groups for a row, then arrange the row. Count all independent stages by multiplication.

Common Mistakes

• Forgetting the factor 2 for Olly and Petra swapping places.
• Placing Olly and Petra in the photograph without ensuring they are on the front row.
• Choosing the two extra front-row members with order $P(8,2)$ and then also arranging them, which can double-count.
• Forgetting to arrange the 6 people in the back row.

Things to Be Careful About

• The front row arrangement is $3! \times 2$, not $4!$; the block reduces the number of objects by one.
• The final answer is $241\,920$; the mark scheme allows several equivalent method routes.
• If a separate selection step is used, use a combination for choosing the two extras to avoid overcount.