Mathematics 9709/52 — October/November 2025

Walkthrough

We are looking at the number of throws until the first head. The probability of a head is $0.4$, so the probability of a tail is $1-0.4=0.6$. For the first head to appear on the 5th throw, the first four throws must all be tails and the 5th throw must be a head. Because the throws are independent, we multiply the probabilities: $(0.6)^4(0.4)$. We compute $(0.6)^4=0.1296$ and then multiply by $0.4$ to get $0.05184$. The same result in fraction form is $\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)=\frac{162}{3125}$.

Key Takeaways

This part uses the geometric distribution. For a probability $p$ of success on each trial, the probability that the first success occurs on the $k$th trial is $(1-p)^{k-1}p$. Here $p=0.4$ and $k=5$.

Common Mistakes

• Using $0.4^4$ instead of $0.6^4$ for the four tails.
• Forgetting to multiply by the final $0.4$ for the head on the 5th throw.
• Treating the throws as dependent when they are independent.

Things to Be Careful About

• The probability of a tail is $0.6$, not $0.4$.
• The answer $0.05184$ is exact; the mark scheme also accepts $\frac{162}{3125}$.
• No rounding is required for this part.

Walkthrough

"After the 6th throw" means that the first head has not occurred during the first six throws. Therefore throws 1, 2, 3, 4, 5 and 6 must all be tails. Since each tail has probability $0.6$, the probability is $(0.6)^6$. This is also the geometric distribution tail probability $P(X>6)=(1-p)^6$. Evaluating gives $0.6^6=0.046656$, which is $\frac{729}{15625}$ in fraction form. A useful check is to subtract from 1 the probability that the first head appears on one of the first six throws; this gives the same result.

Key Takeaways

For a geometric distribution, $P(X>k)=(1-p)^k$. The phrase "after the $k$th throw" is exactly the event that the first $k$ throws are all failures.

Common Mistakes

• Interpreting "after the 6th throw" as "on the 6th throw" or "before the 6th throw".
• Using $0.4^6$ instead of $0.6^6$.
• Trying to calculate an infinite geometric series when the direct tail formula is simpler.

Things to Be Careful About

• "After the 6th throw" means the first head is on the 7th throw or later, so the first six throws must all be tails.
• The exact value is $0.046656$; if rounding, use at least 3 significant figures, so $0.0467$ is acceptable.
• The mark scheme accepts the complement method, but the direct tail formula is sufficient.

Walkthrough

The random variable $X$ counts how many draws are needed until the first green marble appears. Since there are 4 non-green marbles (3 red and 1 blue) and 2 green marbles, the first green marble must appear by the 5th draw. So the possible values are $X = 1, 2, 3, 4, 5$.

For each value, multiply the probability of drawing a non-green marble on each of the first $x-1$ draws, then a green marble on the $x$th draw. Because the marble is not replaced, the denominator decreases after each draw.

For example, for $X=3$, the first two draws must both be non-green and the third must be green:

After finding all five probabilities, check that they sum to 1. This confirms that the table is a valid probability distribution.

Key Takeaways

This question tests how to build a probability distribution table for a discrete random variable. It also tests sequential probability calculations without replacement, where each draw changes the composition of the bag.

Common Mistakes

• Using replacement, so the denominator stays 6 for every draw. This gives incorrect probabilities.
• Forgetting that the maximum value of $X$ is 5, not 6, because there are only 4 non-green marbles.
• Not linking each probability to the correct $x$ value in the table. The mark scheme requires the probabilities to be linked to the correct $x$ values.
• The mark scheme gives special-credit alternatives for tables starting at $X=0$; these do not score full marks.

Things to Be Careful About

• The probabilities must sum to 1.
• If decimals are used, give at least 3 significant figures.
• Unsimplified fractions are accepted in the table, so it is safe to leave probabilities as $\frac{5}{15}$, $\frac{4}{15}$, etc.

Walkthrough

Use the distribution table from part (a). First calculate $E(X)$ by multiplying each value of $X$ by its probability and adding the results:

Next calculate $E(X^2)$ by multiplying each squared value of $X$ by its probability:

Finally use the variance formula:

Key Takeaways

This question tests the definitions of expectation and variance for a discrete random variable. The key formula is $\text{Var}(X)=E(X^2)-[E(X)]^2$, so both $E(X)$ and $E(X^2)$ must be computed from the distribution table.

Common Mistakes

• Omitting the subtraction of $[E(X)]^2$ after computing $E(X^2)$.
• Using the incorrect formula $\text{Var}(X)=E(X^2)-E(X)$.
• Rounding $E(X)$ too early, which can change the final variance.

Things to Be Careful About

• The mark scheme allows follow-through from the student's table if the probabilities sum to 1; for the variance formula, it allows follow-through with at least 4 probabilities.
• If a special-case table starting at $X=0$ is used, the variance calculation can still earn method marks but not the final accuracy mark.
• Use exact fractions to avoid rounding errors. The final answer is $\frac{14}{9}$, which is approximately 1.56.

Walkthrough

The phrase "within 5 cm of the mean" means heights from $187.4 - 5 = 182.4$ cm to $187.4 + 5 = 192.4$ cm. Since the distribution is normal, we standardise these boundaries using $z = \frac{x - \mu}{\sigma}$. The lower boundary gives $z = \frac{-5}{6.4} = -0.78125$ and the upper boundary gives $z = 0.78125$. Because the normal curve is symmetric, the probability of $Z$ lying between $-0.78125$ and $0.78125$ is $2\Phi(0.7813) - 1$. Using tables, $\Phi(0.7813) \approx 0.7826$, so the probability is $0.5652$. Finally, the expected number of members is $124 \times 0.5652 \approx 70.08$, so we round to 70.

Key Takeaways

This question combines standardisation, reading a normal table, symmetry of the normal distribution, and converting a probability into an expected count. In a normal model, the expected frequency is simply the probability multiplied by the total number of observations.

Common Mistakes

• Forgetting to standardise both boundaries.
• Using the probability for only one tail instead of the central region.
• Reading $\Phi(z)$ incorrectly from the table.
• Giving the final answer as 70.08 instead of an integer number of members.

Things to Be Careful About

• The $z$-values are $\pm 0.78125$; using 0.78 or 0.7813 is acceptable as long as the final probability is in the range $0.565$ to $0.5655$.
• The expected number must be a whole number; 70.08 rounds to 70.
• The mark scheme requires the final answer to be an integer, with 70 or 71 expected depending on the table values used.

Walkthrough

We are told 23% of Junior members are shorter than 170.3 cm, so $\text{P}(X < 170.3) = 0.23$. Since 170.3 is below the mean 172.7, the corresponding $z$-value must be negative. From the standard normal table, the $z$-value with lower-tail probability 0.23 is $-0.739$. Substitute into $z = \frac{x - \mu}{\sigma}$: $\frac{170.3 - 172.7}{\sigma} = -0.739$. This gives $\frac{-2.4}{\sigma} = -0.739$, so $\sigma = \frac{2.4}{0.739} \approx 3.25$ cm.

Key Takeaways

This is an inverse normal problem: instead of finding a probability from a $z$-value, we find a $z$-value from a probability and then solve for the unknown parameter. Recognising whether the $z$-value should be positive or negative is essential.

Common Mistakes

• Using 0.77 as if it were the $z$-value instead of the upper-tail probability.
• Forgetting the negative sign on $z$, which can lead to an inconsistent equation.
• Using $\sigma^2$ or $\sqrt{\sigma}$ in the standardisation formula.
• Rearranging with inconsistent signs, for example writing $\frac{172.7 - 170.3}{\sigma} = -0.739$ and still obtaining a positive $\sigma$.

Things to Be Careful About

• The standard normal table gives $\Phi(z)$, the area to the left of $z$. For a lower-tail probability of 0.23, $z$ is negative.
• The mark scheme accepts $0.7385 < z < 0.7395$, so using $z = 0.739$ is appropriate.
• The final answer should satisfy $3.245 \leq \sigma \leq 3.25$; a common acceptable value is 3.25.
• Give $\sigma$ as a decimal, not as an improper fraction.

Walkthrough

We are choosing one card from each pack, so the two choices are independent. The probability that both cards show buses is the product of the probability that Gio's card is a bus and the probability that Ivy's card is a bus.

Gio has 18 cards, 6 of which are buses, so $\text{P}(\text{Gio bus}) = \frac{6}{18}$. Ivy has $x$ cards, $x - 12$ of which are buses, so $\text{P}(\text{Ivy bus}) = \frac{x-12}{x}$. Therefore $\text{P}(\text{BB}) = \frac{6}{18} \times \frac{x-12}{x}$.

Similarly, Gio has 10 cars out of 18, and Ivy has 9 cars out of $x$, so $\text{P}(\text{CC}) = \frac{10}{18} \times \frac{9}{x}$.

The problem states that $\text{P}(\text{BB}) = 2 \times \text{P}(\text{CC})$. Substituting the expressions gives an equation we can solve.

Simplify the fractions: $\frac{6}{18} = \frac{1}{3}$ and $2 \times \frac{10}{18} \times \frac{9}{x} = \frac{10}{x}$. So the equation becomes $\frac{x-12}{3x} = \frac{10}{x}$.

Cross-multiplying: $x(x-12) = 30x$, which expands to $x^2 - 12x = 30x$, then $x^2 - 42x = 0$. Factorising gives $x(x-42) = 0$, so $x = 0$ or $x = 42$.

We reject $x = 0$ because Ivy must have a positive number of bus cards ($x - 12 > 0$). Hence $x = 42$.

Key Takeaways

This question tests the multiplication law for independent events — the probability that two independent events both occur is the product of their individual probabilities. It also tests forming and solving a quadratic equation from a probability statement. When a quadratic gives two solutions, check whether both make sense in the context of the problem.

Common Mistakes

• Forgetting to multiply the car probability by 2 — the equation must reflect that the bus probability is twice the car probability.
• Losing the factor of $x$ in the denominator when cross-multiplying, which leads to a wrong linear equation.
• Accepting $x = 0$ without checking that it is impossible (Ivy cannot have a negative number of bus cards).

Things to Be Careful About

• Ivy's total number of cards is $x$, so the denominator for Ivy's probabilities is always $x$.
• The number of bus cards in Ivy's pack is $x - 12$, which must be positive, so $x > 12$. This eliminates $x = 0$.
• The mark scheme requires the equation to be formed correctly before solving; show the unsimplified equation to earn the method marks.

Walkthrough

Once we know $x = 42$, Ivy has $42 - 12 = 30$ bus cards, 9 car cards, and 3 train cards.

The probability that both cards show the same type of vehicle is the sum of three separate probabilities: both buses, both cars, and both trains. These three outcomes are mutually exclusive — they cannot happen at the same time — so we add them.

$\text{P}(\text{BB}) = \frac{6}{18} \times \frac{30}{42} = \frac{10}{42}$.

$\text{P}(\text{CC}) = \frac{10}{18} \times \frac{9}{42} = \frac{5}{42}$.

$\text{P}(\text{TT}) = \frac{2}{18} \times \frac{3}{42} = \frac{1}{126}$.

Adding: $\frac{10}{42} + \frac{5}{42} + \frac{1}{126}$. Convert to a common denominator of 126: $\frac{30}{126} + \frac{15}{126} + \frac{1}{126} = \frac{46}{126} = \frac{23}{63} \approx 0.365$.

Key Takeaways

This part applies the addition law for mutually exclusive events. Each matching outcome (BB, CC, TT) is computed using the multiplication law, and since the outcomes are mutually exclusive, their probabilities are added. It also reinforces substituting a found value back into probability expressions.

Common Mistakes

• Forgetting the train-train case — the question asks for the same type of vehicle, which includes trains, not just buses and cars.
• Not converting fractions to a common denominator before adding.
• Using the wrong number of bus cards for Ivy (using $x - 12$ without substituting $x = 42$, or using 42 instead of 30).

Things to Be Careful About

• The three matching outcomes are mutually exclusive, so they are added, not multiplied.
• The answer should be given as a fraction or a decimal accurate to at least 3 significant figures ($0.365$).
• Keep the fractions unsimplified until the end to make the addition clear.

Walkthrough

When drawing a histogram for grouped data with unequal class widths, the area of each bar must be proportional to the frequency. This means the height of each bar must represent the frequency density, not the raw frequency.

First, we find the continuous class boundaries. Since the times are recorded correct to the nearest minute, the upper boundary of one class and the lower boundary of the next class meet at the halfway point (e.g., 10.5 between 1-10 and 11-20). We subtract 0.5 from the lower limit and add 0.5 to the upper limit to get the boundaries: 0.5, 10.5, 20.5, 25.5, 30.5, 50.5.

Next, we calculate the class width for each interval by subtracting the lower boundary from the upper boundary: 10, 10, 5, 5, and 20.

Then, we calculate the frequency density for each class by dividing the frequency by the class width: $12/10 = 1.2$, $38/10 = 3.8$, $68/5 = 13.6$, $76/5 = 15.2$, and $46/20 = 2.3$.

Finally, we draw the histogram on the provided grid. The x-axis represents time in minutes from 0.5 to 50.5, and the y-axis represents frequency density. We draw five adjacent bars with the calculated widths and heights. The 'daylight rule' must be applied, meaning there should be a gap between the top of the bar and the top of the grid if the bar does not reach it, ensuring the full height of the bar is visible.

Key Takeaways

• Histograms with unequal class widths require frequency density on the y-axis to ensure area represents frequency.
• Continuous class boundaries are found by taking the midpoint between the upper limit of one class and the lower limit of the next.
• The area of each bar (width $\times$ height) equals the frequency of that class.

Common Mistakes

• Using frequency instead of frequency density for the y-axis heights.
• Using the class limits (1, 10, 11, 20, etc.) as boundaries instead of the continuous boundaries (0.5, 10.5, 10.5, 20.5, etc.).
• Forgetting to apply the daylight rule when drawing the bars, which can obscure the exact height of the bar.
• Using an incorrect scale on the y-axis that does not accommodate the maximum frequency density (15.2).

Things to Be Careful About

• Ensure the x-axis scale covers the full continuous range from 0.5 to 50.5.
• The y-axis must start at 0 and have a linear scale that can clearly show values up to at least 15.2.
• Bars must be drawn adjacent to each other with no gaps between them, as the data is continuous.
• Labels on the axes must clearly state 'frequency density' and 'time (minutes)' or equivalent.

Walkthrough

When data is grouped into intervals, we do not know the exact values, so we estimate the mean by assuming all observations in a class fall exactly at the class midpoint.

First, we calculate the midpoint for each time interval by adding the lower and upper bounds and dividing by 2. For example, the midpoint of 1–10 is $(1+10)/2 = 5.5$. We do this for all five classes to get 5.5, 15.5, 23, 28, and 40.5.

Next, we multiply each midpoint by the corresponding frequency to get the estimated total time for each class. For the first class, this is $5.5 \times 12 = 66$. We calculate this for all classes and sum them up to get the total estimated time for all 240 competitors: $66 + 589 + 1564 + 2128 + 1863 = 6210$.

Finally, we divide this total estimated time by the total number of competitors (240) to find the average time per competitor: $6210 / 240 = 25.875$ minutes.

Key Takeaways

• The estimated mean of grouped data uses class midpoints as representative values.
• The formula is $\bar{x} = \frac{\sum f x}{\sum f}$, where $x$ is the midpoint and $f$ is the frequency.
• This is an estimate because it assumes uniform distribution within each class interval.

Common Mistakes

• Using the lower or upper class limit instead of the midpoint.
• Forgetting to multiply by the frequency and just averaging the midpoints.
• Arithmetic errors when calculating the products $f \times x$ or their sum.
• Using the wrong total frequency (e.g., summing the frequencies incorrectly instead of using the given 240).

Things to Be Careful About

• Ensure midpoints are calculated correctly, especially for the last class (31–50), where the midpoint is $(31+50)/2 = 40.5$, not 40 or 50.
• The question asks for an estimate, so the answer is inherently approximate; providing the exact decimal 25.875 or the fraction $207/8$ is acceptable.

Walkthrough

We have 10 people, and for each person there are two categories for the parts of the week: Saturday/Sunday or one of the other five days. Since each day is equally likely, the probability that a particular person's next birthday is on one of the other five days is $\frac{5}{7}$.

The people are independent, so instead of adding probabilities we multiply: the probability that all 10 fall in the 'not Saturday/Sunday' category is $\frac{5}{7} \times \frac{5}{7} \times \dots$ ten times, namely $\left(\frac{5}{7}\right)^{10}$.

This is exactly the binomial probability $P(X=0)$ when there are 10 independent trials and the probability of success on each trial is $\frac{5}{7}$.

Key Takeaways

This question tests the multiplication law for independent events and the simplest single use of a binomial model. It also requires converting the wording 'none' into the event $X=0$.

Common Mistakes

• Using $\frac{2}{7}$ as the probability of success. Here success is 'not Saturday or Sunday', so the probability is $\frac{5}{7}$.
• Treating the people as if they were not independent and adding probabilities instead of multiplying.
• Writing $\left(\frac{5}{7}\right)^{10}$ but then evaluating a different power such as $\left(\frac{1}{7}\right)^{10}$ or $10 \times \frac{5}{7}$.

Things to Be Careful About

The probability is less than 1 and very small, about 0.0346. In exact form the answer may be written as the fraction $\frac{9765625}{282475249}$. Since the mark scheme accepts either the exact fraction or the decimal 0.0346, either is acceptable.

Walkthrough

We first identify the random variable. There are 10 fixed trials (the 10 people), each trial is independent, and each has the same probability $\frac{1}{7}$ of being a Wednesday birthday. This is a binomial situation.

'Fewer than 3' means 0, 1 or 2. These are separate cases, so we calculate each binomial probability separately and add them.

The binomial probability formula gives each individual probability. For example, when exactly 2 people have Wednesday birthdays, we choose which 2 of the 10 people they are ($\binom{10}{2}$), and multiply by the two Wednesday probabilities $\left(\frac{1}{7}\right)^2$ and the eight non-Wednesday probabilities $\left(\frac{6}{7}\right)^8$.

Repeating this for 1 and 0, then adding the three results, gives approximately 0.838.

Key Takeaways

A binomial experiment must have a fixed number of independent trials, two outcomes and a constant probability of success. The phrase 'fewer than 3' is cumulative and requires adding $P(X=0)$, $P(X=1)$ and $P(X=2)$.

Common Mistakes

• Computing only $P(X=2)$ or only $P(X=1)$ instead of adding all three required probabilities.
• Using $p = \frac{6}{7}$ for a Wednesday birthday instead of $\frac{1}{7}$.
• Forgetting the binomial coefficients, especially $\binom{10}{2}=45$ and $\binom{10}{1}=10$.
• Reading 'fewer than 3' as '3 or fewer', which would include $X=3$.
• Writing a correct unsimplified expression but then pressing the calculator incorrectly; the method mark requires a valid binomial term and the accuracy mark requires the full un-simplified sum.

Things to Be Careful About

The mark scheme awards the method mark for correctly using one binomial term of the form $\binom{10}{r}\left(\frac{1}{7}\right)^r\left(\frac{6}{7}\right)^{10-r}$. To gain the accuracy mark, the full un-simplified sum should be written before evaluating. The final answer should be given to about three decimal places; the accepted range is $0.838$.

Walkthrough

We have $n=392$ independent people, each with probability $\frac{1}{7}$ of next birthday on Friday. This is binomial, but calculating $\sum_{r=66}^{392}$ directly is not practical. Because $n$ is large and both $np=56$ and $n(1-p)=336$ are comfortably greater than 5, the normal approximation is suitable.

First find the mean and variance of the approximating normal distribution: $\mu = np = 56$ and $\sigma^2 = np(1-p) = 48$.

The discontinuity is handled by a continuity correction. For a discrete variable, $X>65$ means $X=66,67,\dots$. In the continuous normal model we take the lower boundary of the first included integer, $65.5$, and standardise:

The normal table gives the area to the left of $z$, $\Phi(1.3712)\approx0.9149$. Since we need the area to the right, subtract from 1 to get $0.0851$.

Key Takeaways

This is the normal approximation to the binomial. The three essential steps are: compute $\mu$ and $\sigma^2$ from the binomial parameters, apply the correct continuity correction, and standardise using $\frac{x-\mu}{\sigma}$ before using normal tables.

Common Mistakes

• Using $65$ instead of $65.5$ in the standardisation formula because no continuity correction has been applied.
• Using $64.5$ for $X>65$; that boundary would include the possibility $X=65$, which is not wanted.
• Believing the variance is $\sqrt{48}$ and dividing by $48$; the standard deviation is $\sqrt{48}$.
• Forgetting that normal tables give the cumulative probability to the left, so for 'more than 65' the final step is $1-\Phi(z)$.
• Not checking that the approximation is suitable; here $np=56$ and $n(1-p)=336$ are both large.

Things to Be Careful About

The direction of the continuity correction is important: for $X>65$, values start at 66, so the lower boundary is 65.5; for $X\ge65$, values start at 65, so the lower boundary would be 64.5. The final answer should be a probability less than 0.5; the accepted range is $0.0851$ (about $0.085$ to $0.0852$).

Walkthrough

Start with the 10 letters of ZOOLOGICAL. Since all three Os are identical, when we insist they are together we collapse them into one block [OOO]; there is no extra 3! factor inside the block because the Os are indistinguishable. Now there are 8 units to arrange: the OOO block, the two identical Ls, and the five distinct letters Z, G, I, C, A. The identical Ls mean we divide by 2!, giving 8!/2! arrangements with the Os together.

This total includes the arrangements where the two Ls are adjacent, which we must exclude. To count those, collapse the Ls into a [LL] block as well. There are now 7 blocks: [OOO], [LL], and the five distinct letters. Because both blocks have no internal distinct arrangements, the count is 7!.

Finally subtract the forbidden arrangements from the total. This is the principle of inclusion-exclusion for not next to each other: total allowed = total with Os together - total with Os together and Ls together.

Key Takeaways

• Repeated identical objects are handled by dividing by the factorial of their count.
• Treating required-adjacent letters as a single block reduces the problem to arranging fewer objects.
• Not next to each other is often best counted as total minus the cases where they are next to each other.
• Identical objects inside a block do not create extra arrangements.

Common Mistakes

• Using 8! instead of 8!/2! for the two identical Ls.
• Forgetting to subtract the arrangements where the two Ls are together.
• Treating the three Os as distinct inside the block and multiplying by 3!, which is wrong because they are identical.
• In the subtracted case, counting 8 units instead of 7 units after forming the [LL] block.

Things to Be Careful About

• The OOO block has no internal permutations, and the LL block also has no internal permutations.
• The units in the second count are [OOO], [LL], and five distinct letters, so there are exactly 7 units.
• The answer must be an integer; check that 8!/2! - 7! = 20160 - 5040 = 15120.

Walkthrough

For exactly 5 letters between the two Ls, the positions of the Ls must differ by 6: if one L is in position $p$, the other is in position $p+6$. In a 10-letter word the valid position pairs are (1,7), (2,8), (3,9), (4,10), so there are 4 possibilities. The order of the Ls does not matter because they are identical.

After placing the Ls, the remaining 8 positions are filled with the other 8 letters: O, O, O, Z, G, I, C, A. Three of these are identical, so there are $8!/3!$ arrangements of those letters. Multiply by the 4 ways of choosing the L positions.

Key Takeaways

• A fixed gap between two identical letters corresponds to a fixed difference in their position numbers.
• Count position choices first, then fill the remaining slots.
• Repeated letters again require division by the appropriate factorial.

Common Mistakes

• Using 5 possible position pairs instead of 4. The pairs are determined by a difference of 6, not 5.
• Forgetting to divide 8! by 3! for the three identical Os.
• Considering the two Ls as distinct and doubling the count; they are identical.

Things to Be Careful About

• Exactly 5 letters between means the position numbers differ by 6, so there are four choices: (1,7), (2,8), (3,9), (4,10).
• The same remaining letters are used for every choice, so the multiplier is always $8!/3!$.
• Final total is $4 \times 6720 = 26880$.

Walkthrough

We need the probability that two letters chosen from ZOOLOGICAL are different. Since the choice is unordered, total outcomes are combinations of 10 letters choose 2. Then count outcomes where the letters are the same: both Os or both Ls. There are 3 Os, so ${}^{3}C_2 = 3$; there are 2 Ls, so ${}^{2}C_2 = 1$. Therefore the probability of the same letter is $4/45$. The complement gives the probability that they are different.

Key Takeaways

• Use combinations for unordered selection.
• The complement probability is often easier: different = 1 - same.
• Letters of the same type are indistinguishable, so count by choosing among the copies.

Common Mistakes

• Forgetting that there are two Ls, so a pair of Ls is another way to get the same letter.
• Using ordered selections such as 10 × 9 without matching the denominator, which can lead to double-counting errors.
• Writing a final answer without any working; the mark scheme awards only SCB1 for a correct final answer with no method.

Things to Be Careful About

• There are exactly 3 Os and 2 Ls; the other 5 letters are all distinct.
• Total unordered pairs is ${}^{10}C_2 = 45$.
• The complement gives $1 - 4/45 = 41/45$, which is about 0.911.
• If using a product rule, remember $P(OO)=3/10 \times 2/9$ and $P(LL)=2/10 \times 1/9$.