Chapter 6 — Data Handling and Probability

Answer: A

Explanation

Step 1 — Sum all values.

$$\text{Sum} = 4 + 7 + 3 + 9 + 2 + 5 = 30$$

Step 2 — Divide by the count.

$$\bar{x} = \frac{30}{6} = 5$$

Why the distractors are wrong:

B) 4.5 — common error from dividing by 7 instead of 6 (the number of values, not the highest index).

C) 6 — results from including a phantom extra value or miscounting.

D) 3.5 — this is the minimum value divided by 1; a computation shortcut gone wrong.

Takeaway: Count the number of values carefully before dividing. With six data points you divide by 6, not by any other number.

Answer: C

Explanation

Step 1 — Sort the data in ascending order.

$$1,\ 2,\ 3,\ 5,\ 7,\ 8,\ 9$$

Step 2 — Locate the middle position.

With $n = 7$ values, the median is at position $\dfrac{7+1}{2} = 4$. The 4th value is $\mathbf{5}$.

Why the distractors are wrong:

A) 3 — the 3rd value, not the middle one.

B) 7 — the 5th value; the error of counting from the end rather than finding the exact middle.

D) 6 — this would be the mean of 7 and 5, which is the average of the two middle values in an even-sized set. There are 7 values here, so there is one exact middle value.

Takeaway: Always sort first, then find position $\frac{n+1}{2}$ for odd $n$. Never find the median of unsorted data.

Answer: B

Explanation

Step 1 — Count the frequency of each value.

Step 2 — Identify the highest frequency.

The value 2 appears 3 times — more than any other value. The mode is $\mathbf{2}$.

Why the distractors are wrong:

A) 4 — appears twice, less than 2.

C) 7 — also appears twice.

D) 4 and 2 — a data set is bimodal only when two values share the highest frequency. Here 2 has a strictly higher frequency (3 vs 2), so there is one mode only.

Takeaway: A data set has a unique mode only when one value appears strictly more often than all others. If two values are tied for most frequent, the set is bimodal.

Answer: D

Explanation

Step 1 — Compute the new mean.

$$\bar{x}_\text{new} = \frac{5+5+6+7+7+50}{6} = \frac{80}{6} \approx 13.3$$

The mean jumped from 6 to about 13.3 — an increase of over 7 units.

Step 2 — Find the new median.

Sorted set: $5,\ 5,\ 6,\ 7,\ 7,\ 50$. With $n=6$, the median is the average of the 3rd and 4th values:

$$\text{Median}_\text{new} = \frac{6+7}{2} = 6.5$$

The median increased by only 0.5 — far less than the mean.

Why the distractors are wrong:

A) Adding 50 certainly changes the mean — it inflates the sum enormously.

B) The median barely moved (6 to 6.5); it is the mean that increased dramatically.

C) The mean increased by ~7.3, the median by 0.5 — very different amounts.

Takeaway: The mean is sensitive to outliers because it uses every value in its calculation. The median is resistant because only the middle position matters.

Answer: A

Explanation

Step 1 — Calculate $fx$ for each row.

Step 2 — Divide $\sum fx$ by $\sum f$.

$$\bar{x} = \frac{\sum fx}{\sum f} = \frac{410}{20} = 20.5$$

Why the others are wrong: B (22), D (21) and C (19.5) are each one bookkeeping slip away — a mis-multiplied row or a wrong column total. The table gives you two checkpoints: $\sum f = 20$ must match the number of data points, and $\sum fx = 410$. Verify both before dividing.

(Also resist averaging the four midpoints. That ignores the frequencies and only works if every class is equally full.)

Takeaway: Grouped mean $= \frac{\sum fx}{\sum f}$ using midpoints. It is an estimate — the exact mean was lost when the data was grouped — and the two column totals are your only defence against slips.

Section 6.2 — Measures of Spread

The range is easy to calculate but fragile — one extreme value changes it completely. The IQR is more robust because it focuses on the middle 50 % of the data. The five-number summary (minimum, $Q_1$, median, $Q_3$, maximum) provides a compact description of the entire distribution and is displayed graphically as a box-and-whisker plot. The outer fences $Q_1 - 1.5 \times \text{IQR}$ and $Q_3 + 1.5 \times \text{IQR}$ define the outlier boundaries.

Answer: B

Explanation

$$\text{Range} = \text{maximum} - \text{minimum} = 23 - 4 = 19$$

Why the others are wrong: the data are unsorted, and every wrong option is a scanning error —

• D (15) is $19 - 4$: the maximum misidentified as 19, with the true maximum 23 overlooked further down the list.
• A (12) is $23 - 11$: the minimum misidentified as 11 (4 hides in the middle).
• C (17) is a subtraction slip on the correct pair.

Takeaway: Range = max − min $= 23 - 4 = 19$. With unsorted data, underline the max and min before subtracting — both scanning errors above are one glance from happening. And remember the range says nothing about the shape in between.

Answer: C

Explanation

Step 1 — Split into lower and upper halves.

With $n = 8$ values, split after the 4th:

• Lower half: $\{3,\ 5,\ 7,\ 9\}$
• Upper half: $\{11,\ 13,\ 15,\ 17\}$

Step 2 — Find $Q_1$ and $Q_3$.

$$Q_1 = \frac{5+7}{2} = 6 \qquad Q_3 = \frac{13+15}{2} = 14$$

Step 3 — IQR.

$$\text{IQR} = Q_3 - Q_1 = 14 - 6 = 8$$

Why the distractors are wrong:

A) 14 — this is $Q_3$, not the IQR.

B) 6 — this is $Q_1$, not the IQR.

D) 10 — results from using incorrect quartile positions (e.g., 2nd and 7th values).

Takeaway: Split the data at the median position, then find the median of each half. For an even number of values, each half contains $n/2$ values.

Answer: D

Explanation

Min and Max: 2 and 25.

Median ($n=10$): average the 5th and 6th values, $\frac{10+12}{2} = 11$.

Split into halves (the median belongs to neither): lower $\{2,5,7,8,10\}$ → $Q_1 = 7$; upper $\{12,15,18,20,25\}$ → $Q_3 = 18$

Summary: $2,\ 7,\ 11,\ 18,\ 25$

Why the others are wrong: all three agree on Min, Median and Max — the whole fight is over how to split the halves.

• A averages pairs, using a different quartile convention. With five values per half, each quartile is a single datum, never an average.
• B counted the quartile positions from the wrong ends.
• C splits the lower half wrongly but gets $Q_3$ right — mixed conventions in one answer, always a red flag.

Takeaway: For even $n$, each half has $n/2$ values and the quartiles are their medians. Write both halves out — quartile errors are nearly always splitting errors.

Answer: A

Explanation

By definition, $Q_1$ is the value below which 25 % of the data falls, and $Q_3$ is the value below which 75 % falls. Therefore the interval $[Q_1,\ Q_3]$ contains the middle $75\% - 25\% = 50\%$ of the data.

Why the distractors are wrong:

B) 25 % — this is the proportion in each whisker segment (from Min to $Q_1$, or from $Q_3$ to Max).

C) 75 % — this is the proportion below $Q_3$, not between the two quartiles.

D) 100 % — only the full range (Min to Max) contains all the data.

Takeaway: The box in a box-and-whisker plot always represents 50 % of the data. Each whisker represents 25 %.

Answer: B

Explanation

Three short lines:

1. IQR $= 20 - 8 = 12$
2. Upper fence $= Q_3 + 1.5 \times \text{IQR} = 20 + 18 = 38$
3. Compare: $40 > 38$, so 40 is an outlier.

Why the others are wrong:

• A gets IQR right but reverses the final comparison. Slow down at the finish line.
• C (IQR 14) counted 8 to 20 inclusively — a fencepost error. IQR is plain subtraction.
• D (IQR 10) is an arithmetic slip whose "outlier" verdict is only accidentally right — the NBT often pairs a wrong intermediate with a right-sounding conclusion.

Takeaway: Write all three lines. Each one is checkable on its own, and a right conclusion from a wrong intermediate still earns nothing.

Section 6.3 — Data Representation

Histograms, ogives, and scatter plots are the three graphical displays most likely to appear in NBT data questions. A histogram shows the shape of a distribution; an ogive (cumulative frequency curve) lets you estimate medians and percentiles; a scatter plot shows the relationship (correlation) between two variables.

Answer: C

Explanation

The frequencies rise from 3 to a peak of 12 in the middle class $[20;30)$, then fall back to 4 — a roughly symmetric bell shape. The rising and falling sides are nearly mirror images of each other.

Why the distractors are wrong:

A) Positively skewed — a positively skewed distribution has a long tail to the right (small peak at low values, trailing off slowly at high values). That is not the case here.

B) Uniform — a uniform distribution has roughly equal frequencies across all classes.

D) Negatively skewed — would peak near the high end and trail off to the left.

Takeaway: Visualise the bar heights: if they form a hill centred in the data range, the distribution is symmetric. A long tail to the right means positive skew; a long tail to the left means negative skew.

Answer: D

Explanation

Median position: $\frac{50}{2} = $ the 25th value.

Which class holds it? 15 values by the 40 boundary, 30 by the 60 boundary — so the 25th sits in $[40;60)$.

Interpolate: $$40 + \frac{25-15}{30-15} \times 20 = 40 + \tfrac{10}{15}(20) \approx 53$$

Why the others are wrong:

• A (40) is the class's lower boundary — the interpolation never started.
• B (50) is both the class midpoint and an echo of "$n=50$" — doubly baited, and it skips the arithmetic.
• C (45) mangled the fraction. Build it in words: how far into this class's 15 values does the 25th sit? Ten of fifteen.

Takeaway: Find the class straddling $\frac{n}{2}$, then: boundary $+ \frac{\text{values still needed}}{\text{values in the class}} \times \text{class width}$.

Answer: A

Explanation

Correlation measures the tendency for two variables to move together. If there is no pattern — points spread randomly — neither variable can predict the other. This is called no correlation (or zero correlation).

Why the distractors are wrong:

B) Strong positive correlation means as $x$ increases, $y$ clearly tends to increase — a visible upward trend.

C) Weak negative means a faint downward trend is visible.

D) Perfect positive correlation means all points lie exactly on an upward straight line.

Takeaway: The key question for any scatter plot is: "Is there a trend?" If yes, classify the direction (positive/negative) and strength (strong/weak). If no visible trend, the answer is no correlation.

Answer: B

Explanation

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{14 - 4}{10 - 0} = \frac{10}{10} = 1$$

The equation of the line is $y = x + 4$.

Why the others are wrong: A (0.5), D (1.5) and C (2) all come from misreading one coordinate — halving, off-by-five, or doubling the rise or run. The defence is mechanical: write $\Delta y = 14 - 4 = 10$ and $\Delta x = 10 - 0 = 10$ on separate lines before dividing. A gradient never comes out of a single glance at two points.

Takeaway: $m = \frac{\Delta y}{\Delta x} = \frac{10}{10} = 1$: $y$ rises one unit per unit of $x$, and the intercept 4 is the prediction at $x = 0$ — so the line is $y = x + 4$. Compute the two differences explicitly, every time.

Answer: C

Explanation

Standard deviation measures spread around the mean. Adding or subtracting a constant shifts the entire data set by the same amount — the mean shifts, but the distances between values stay the same, so $\sigma$ is unchanged.

Multiplying every value by a constant $k$ scales all distances. If the original values are $x_i$, the new values are $kx_i$. The deviations from the new mean $(kx_i - k\bar{x}) = k(x_i - \bar{x})$ are all scaled by $k$. Therefore the new standard deviation is $|k|\sigma$.

Multiplying by 2 doubles $\sigma$.

Why D is wrong: You cannot add a constant to the mean without changing the individual values. The mean is derived from the values, not set independently.

Takeaway: Adding or subtracting a constant — no change to $\sigma$. Multiplying or dividing by a constant $k$ — $\sigma$ is multiplied by $|k|$.

Section 6.4 — Basic Probability

The probability of event $A$ is $P(A) = \dfrac{n(A)}{n(S)}$, where $n(A)$ is the number of outcomes in $A$ and $n(S)$ is the total number of equally likely outcomes in the sample space $S$. Probability values always lie in $[0,1]$.

Answer: D

Explanation

Sample space: $S = \{1,2,3,4,5,6\}$, so $n(S) = 6$.

Even numbers: $A = \{2,4,6\}$, so $n(A) = 3$.

$$P(A) = \frac{3}{6} = \frac{1}{2}$$

Why the others are wrong:

• B ($\frac{1}{6}$) is the probability of one specific face — "even" admits three faces, not one.
• A ($\frac{1}{3}$) counts only two even faces (2 and 4, with 6 forgotten) — or divides the three categories {odd, even, …} carelessly.
• C ($\frac{2}{3}$) sweeps four faces into "even" — an odd face smuggled in. List the set: $\{2, 4, 6\}$, exactly three.

Takeaway: $P = \frac{n(A)}{n(S)}$: write out the favourable set before counting it. For a fair die, three even faces out of six gives $\frac12$ — and explicit listing is what keeps 3 from becoming 2 or 4 under time pressure.

Answer: A

Explanation

For any event $A$ and its complement $A'$:

$$P(A) + P(A') = 1 \implies P(A') = 1 - 0.3 = 0.7$$

Why the others are wrong:

• B (0.3) hands back $P(A)$ unchanged — as if an event and its complement were equally likely by default. They are only equal at 0.5.
• C (0.4) subtracts 0.3 twice ($1 - 0.3 - 0.3$) — a "remove it from both sides" reflex with no rule behind it.
• D (1.3) adds the two — but no probability can exceed 1, which makes D self-refuting the moment you look at it. Options outside $[0, 1]$ are free eliminations.

Takeaway: $P(A) + P(A') = 1$, always: complementary events are exhaustive and mutually exclusive. And scan the options for impossible probabilities first — examiners include them to reward students who know the boundaries.

Answer: B

Explanation

There are 2 red kings in a deck: the king of hearts and the king of diamonds.

$$P(\text{red king}) = \frac{2}{52} = \frac{1}{26}$$

Why the distractors are wrong:

A) $\frac{1}{52}$ would be the probability of a specific single card, e.g. the king of hearts only.

C) $\frac{1}{13}$ is the probability of any king (4 kings out of 52).

D) $\frac{2}{13}$ is the probability of any red king or any other compound event with 8 favourable outcomes.

Takeaway: Count the favourable outcomes carefully. "Red king" is more specific than "any king" (4 outcomes) but less specific than "king of hearts" (1 outcome).

Answer: C

Explanation

Mutually exclusive means $A$ and $B$ cannot occur simultaneously, so $P(A \cap B) = 0$.

$$P(A \cup B) = P(A) + P(B) = 0.25 + 0.35 = 0.60$$

Why the others are wrong:

• A (0.0875) is $0.25 \times 0.35$ — the product rule, which belongs to independent events. Mutually exclusive events are the opposite of independent: if one happens, the other is impossible.
• B (0.25) echoes $P(A)$ alone — the "or" was never processed.
• D (1.0) treats "A or B" as covering everything, but the two events only account for 60 % of the probability; the remaining 0.40 belongs to "neither".

Takeaway: Mutually exclusive ⇒ $P(A \text{ or } B) = P(A) + P(B)$, nothing to subtract — and nothing to multiply. "Mutually exclusive" and "independent" are different (in fact incompatible) conditions; identify which one the question grants you before choosing a rule.

Answer: D

Explanation

$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0.1 = 0.6$$

The intersection is subtracted because those outcomes are counted twice if you simply add $P(A)$ and $P(B)$.

Why the others are wrong:

• A (0.7) adds without subtracting — valid only for mutually exclusive events, and the question hands you a non-zero overlap precisely so that this shortcut fails.
• B (0.12) multiplies $0.4 \times 0.3$ — importing the independence product rule where nothing says "independent" (and indeed $P(A \cap B) = 0.1 \neq 0.12$, so these events are not independent).
• C (0.1) echoes the intersection itself — an ingredient posing as the meal.

Takeaway: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, always; the overlap is counted twice in the sum, so subtract it once. When the intersection is given, use it — its value even tells you whether the events are independent ($0.1$ vs $0.4 \times 0.3 = 0.12$: they aren't).

Answer: A

Explanation

For independent events, the occurrence of one does not affect the probability of the other:

$$P(A \cap B) = P(A) \times P(B) = 0.4 \times 0.5 = 0.2$$

Why the others are wrong:

• B (0.9) adds the probabilities — the union-style move, and even then only correct for mutually exclusive events. "And" questions multiply.
• C (0.45) averages the two probabilities — a compromise with no probabilistic meaning.
• D (0.5) echoes $P(B)$ — as if "and" meant "whichever is larger".

Takeaway: Independent: $P(A \cap B) = P(A) \times P(B) = 0.2$. Mutually exclusive: $P(A \cap B) = 0$. Two completely different conditions — an event pair cannot be both (unless a probability is 0). Translate the English first: and ⇒ intersection ⇒ multiply (under independence); or ⇒ union ⇒ add-then-subtract.

Section 6.5 — Conditional Probability, Tree Diagrams and Contingency Tables

When events are not independent, the probability of the second event depends on what happened first. Conditional probability captures this: $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$. Tree diagrams display multi-stage experiments visually; the probability along each branch is a conditional probability. Two-way tables (contingency tables) organise counts for two categorical variables and allow direct reading of conditional probabilities.

Answer: B

Explanation

$$P(\text{1st red}) = \frac{3}{12} = \frac{1}{4}$$

After drawing one red ball, 2 red balls remain in a bag of 11:

$$P(\text{2nd red} \mid \text{1st red}) = \frac{2}{11}$$

$$P(\text{both red}) = \frac{1}{4} \times \frac{2}{11} = \frac{2}{44} = \frac{1}{22}$$

Why the others are wrong:

• A ($\frac{1}{16}$) is the with-replacement answer: $\left(\frac{3}{12}\right)^2$ — the bag never updated after the first draw.
• C ($\frac{1}{44}$) updates too hard: $\frac{1}{4} \times \frac{1}{11}$ leaves only one red behind, but removing one of three reds leaves two.
• D ($\frac{1}{12}$) matches no consistent method — a plausible-sized decoy. Both legitimate routes agree on B: sequentially $\frac14 \times \frac{2}{11} = \frac{1}{22}$, or by combinations $\frac{\binom{3}{2}}{\binom{12}{2}} = \frac{3}{66} = \frac{1}{22}$.

Takeaway: "Without replacement" shrinks both the favourable count and the total: numerator and denominator each drop by one for the second draw. When unsure, cross-check with the combinations route — two methods agreeing is proof against all four distractor styles.

Answer: C

Explanation

$$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.12}{0.4} = 0.3$$

Why the others are wrong:

• A (0.048) multiplies $0.12 \times 0.4$ — running the multiplication rule when the formula calls for division.
• B (0.52) adds them ($0.4 + 0.12$) — no rule anywhere adds an intersection to a marginal.
• D (0.12) echoes $P(A \cap B)$ unchanged — "given" ignored entirely. Conditioning must rescale: inside the world where $A$ happened (which has size 0.4), the slice where $B$ also happens (0.12) occupies $\frac{0.12}{0.4} = 30\%$.

Takeaway: $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$ — "given $A$" shrinks the sample space to $A$, and the division is that shrinkage. If your "conditional" probability equals the plain intersection, no conditioning has actually happened.

Answer: D

Explanation

Total students who play sport $= 30 + 15 = 45$.

$$P(\text{sport}) = \frac{45}{100} = 0.45$$

Why the others are wrong: each is a different piece of the table posing as the answer —

• A (0.30) is only the top-left cell (sporty maths students) — the 15 sporty non-maths students got dropped.
• C (0.35) is the bottom-right cell (plays neither... i.e. non-maths non-sport) — the wrong region entirely.
• B (0.55) is the complement: the "does not play sport" column total ($20 + 35$). Right technique, wrong column.

Takeaway: Probability from a table = (relevant column or row total) ÷ (grand total). Sum the full "plays sport" column first ($30 + 15 = 45$), and reserve subtotals-as-denominators for questions that say "given that".

Answer: A

Explanation

The condition "plays sport" restricts us to the 45 students in the sport column.

Of those 45, exactly 30 study maths.

$$P(\text{maths} \mid \text{sport}) = \frac{30}{45} = \frac{2}{3}$$

Why the others are wrong:

• B ($\frac{3}{10} = \frac{30}{100}$) divides by the grand total — the "given that" was never applied. That computes $P(\text{maths and sport})$, a different quantity.
• D ($\frac{3}{5} = \frac{30}{50}$) conditions on the wrong event: dividing by the maths row total gives $P(\text{sport} \mid \text{maths})$ — the reversed conditional. $P(A \mid B)$ and $P(B \mid A)$ are rarely equal, and the exam loves offering both.
• C ($\frac{1}{2}$) is a "half-ish" guess matching no cell-over-subtotal in the table.

Takeaway: "Given that X" ⇒ denominator = X's subtotal (here the 45 sport players). Say the fraction in words — "of the 45 who play sport, 30 study maths" — and the reversed conditional can't sneak in.

Answer: B

Explanation

$$P(\text{1st white}) = \frac{4}{10}$$

After removing one white ball, 9 balls remain (3 white + 6 black):

$$P(\text{2nd black} \mid \text{1st white}) = \frac{6}{9} = \frac{2}{3}$$

$$P(\text{white then black}) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90} = \frac{4}{15}$$

Why the others are wrong:

• A ($\frac{6}{25}$) is the with-replacement version: $\frac{4}{10} \times \frac{6}{10}$ — the second denominator never dropped to 9.
• C ($\frac{3}{10}$) drops the denominator too far ($\frac{4}{10} \times \frac{6}{8}$) — two balls removed after one draw.
• D ($\frac{2}{9}$) is $\frac{24}{90}$ mis-simplified ($\frac{20}{90}$ territory) — a cancellation slip at the very last step. Reduce fractions in one careful move: $\frac{24}{90} = \frac{4}{15}$.

Takeaway: Sequential dependent events: $P(A \cap B) = P(A) \times P(B \mid A)$, with the second fraction rebuilt from the post-draw bag (9 balls: 3 white, 6 black). One draw removes exactly one ball — no more, no fewer.

Section 6.6 — Counting Principle, Permutations and Combinations

The Fundamental Counting Principle: if task 1 can be done in $m$ ways and task 2 in $n$ ways, then both can be done in $m \times n$ ways. When order matters, use permutations: $P(n,r) = \dfrac{n!}{(n-r)!}$. When order does not matter, use combinations: $C(n,r) = \dbinom{n}{r} = \dfrac{n!}{r!(n-r)!}$.

Answer: C

Explanation

$$\text{Total outfits} = 3 \times 4 \times 2 = 24$$

Why the others are wrong:

• A (9) adds the choices ($3 + 4 + 2$). Adding counts the ways to pick one item of one kind; an outfit needs one of each, and each shirt pairs with every trouser–shoe combination.
• D (48) doubles the true count — the "2 pairs of shoes" counted as 4 individual shoes, or a stage multiplied twice.
• B (16) mixes adding and multiplying ($3 \times 4 + 4$) — a chain that changes rules midway.

Takeaway: Independent choices in sequence multiply: $3 \times 4 \times 2 = 24$. The test for multiplying is the word each — "for each shirt, every pair of trousers; for each of those, both pairs of shoes."

Answer: D

Explanation

$$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$

For the first position there are 5 choices, for the second there are 4 (one book is placed), and so on.

Why the others are wrong:

• A (25) is $5^2$ — a "five times something" guess with no counting model.
• B (60) is $5 \times 4 \times 3$ — the factor chain abandoned two books early. All five books need placing, so all five factors belong.
• C (100) is round-number bait; factorials are rarely round.

Takeaway: Arranging all $n$ distinct objects gives $n!$ — one shrinking factor per position, all the way down to 1. $5! = 120$ is worth instant recall (as are $4! = 24$ and $6! = 720$).

Answer: A

Explanation

MATHS has 5 distinct letters. We want to select 3 and arrange them in order:

$$P(5,3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = 5 \times 4 \times 3 = 60$$

Why the others are wrong:

• C (120) is $5!$ — arrangements of all five letters, but only three positions exist here.
• B (10) is $\binom{5}{3}$ — the combination count, which ignores order. Arrangements of letters are ordered (CAT ≠ ACT), so each trio counts $3! = 6$ times: $10 \times 6 = 60$.
• D (20) stops the factor chain early ($5 \times 4$) — two slots filled, one abandoned.

Takeaway: "Arrangements" ⇒ order matters ⇒ $P(n,r) = n \times (n-1) \times \cdots$, one factor per slot: $5 \times 4 \times 3 = 60$. Count your factors against your slots before moving on.

Answer: B

Explanation

$$\binom{8}{3} = \frac{8!}{3!\,5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$$

Why the others are wrong:

• A (336) is $P(8,3) = 8 \times 7 \times 6$ — ordered selections. A committee has no first, second or third member, so each group of three is counted $3! = 6$ times over.
• D (168) divides those 336 by 2 instead of by $3! = 6$ — a half-hearted correction that remembers order matters but forgets how many orderings a trio has.
• C (24) is $8 \times 3$ — the two visible numbers multiplied, with no counting structure behind them.

Takeaway: "Committee / group / selection" ⇒ order doesn't matter ⇒ $\binom{n}{r}$ = permutation count ÷ $r!$. The divisor is the factorial of the group size — a trio has six orderings, not two.

Section 6.7 — Venn Diagrams and Advanced Probability

Venn diagrams represent events as overlapping circles within a rectangle (the sample space). The overlap region is $A \cap B$; the combined shaded region is $A \cup B$. The region outside both circles is $(A \cup B)'$.

Answer: C

Explanation

$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.4 - 0.2 = 0.7$$

Why the others are wrong:

• B (0.9) adds without subtracting the overlap — the double-counted 0.2 left in place.
• A (0.8) subtracts only half the overlap — an arithmetic compromise with no rule behind it.
• D (0.5) echoes $P(A)$ — as if the union were just "the bigger event".

Takeaway: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.7$, always — mutually exclusive events are merely the special case where the subtracted term is 0. If your union isn't larger than both individual probabilities yet smaller than their sum, something is off; use that squeeze as a sanity check.

Answer: D

Explanation

"$A$ only" means $A \cap B'$, the part of $A$ that does not overlap with $B$:

$$P(A \cap B') = P(A) - P(A \cap B) = 0.5 - 0.2 = 0.3$$

Why the others are wrong:

• C (0.5) hands back $P(A)$ whole — but $P(A)$ includes the overlap where $B$ also happens; "A but not B" requires carving that 0.2 out.
• A (0.2) is the overlap itself — the piece to be removed, presented as the remainder.
• B (0.4) echoes $P(B)$ — the wrong circle altogether.

Takeaway: $P(A \text{ only}) = P(A) - P(A \cap B) = 0.3$. Sketch the two circles and shade the crescent — Venn questions are region-labelling exercises, and every wrong option here is a different mislabelled region.

Answer: A

Explanation

"At least one" is the complement of "neither":

$$P(\text{at least one}) = 1 - P(A' \cap B')$$

Since $A$ and $B$ are independent, so are $A'$ and $B'$:

$$P(A' \cap B') = P(A') \times P(B') = (1-0.6)(1-0.5) = 0.4 \times 0.5 = 0.2$$

$$P(\text{at least one}) = 1 - 0.2 = 0.8$$

Why the others are wrong:

• B (0.7) subtracts $P(\text{both})$ from 1 — the complement of the wrong event. "At least one" is the opposite of "neither".
• C (0.3) is $P(\text{both})$, answering a different question.
• D (0.6) echoes $P(A)$. "At least one" must be bigger than either single probability, so 0.6 is too small on sight.

Takeaway: $P(\text{at least one}) = 1 - P(\text{none})$. Say out loud which event you are complementing before subtracting from 1 — that sentence is exactly where B goes wrong.

Answer: B

Explanation

$$n(\text{soccer} \cup \text{hockey}) = 18 + 12 - 5 = 25$$

$$n(\text{neither}) = 30 - 25 = 5$$

Why the others are wrong:

• A (7) is $30 - 18 - 5$ — a subtraction chain that removes "both" as if it were a third separate group. The 5 who play both are already inside the 18 and inside the 12.
• C (3) and D (10) are the same story with different mis-handlings of the overlap (subtracting it twice, or not at all, plus slips). All three die at the checkpoint: students in at least one sport $= 18 + 12 - 5 = 25$, leaving exactly 5 outside.

Takeaway: Inclusion–exclusion first ($n(A \cup B) = 18 + 12 - 5 = 25$), subtract from the total second. Better yet, fill in the Venn regions — 13 soccer-only, 5 both, 7 hockey-only — and confirm they sum with "neither" to 30 ✓.

Answer: C

Explanation

Total outcomes: $6 \times 6 = 36$ ordered pairs.

Pairs summing to 7: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — six.

$$P = \frac{6}{36} = \frac16$$

Why the others are wrong:

• A ($\frac1{12}$) counted unordered pairs, halving the truth. $(1,6)$ and $(6,1)$ are different rolls.
• B ($\frac19$) missed two pairs. List systematically — one for each value of the first die — and six is unmissable.
• D ($\frac{7}{36}$) put the target sum in the numerator. The favourable count is 6.

Takeaway: Two dice always give 36 ordered outcomes. Sum 7 has exactly one pair per value of the first die — six ways, the most of any sum.

Mixed Practice Questions — M1 to M30

These questions cover all six chapters. No explanations are given. Work through each one using the methods from earlier chapters.

Answer: A

Answer: C

Answer: C

Answer: B

Answer: A

Answer: C

Answer: B

Answer: D

Answer: A

Answer: C

Answer: B

Answer: D

Answer: A

Answer: C

Answer: B

Answer: D

Answer: A

Answer: C

Answer: B

Answer: D

Answer: A

Answer: C

Answer: B

Answer: D

Answer: A

Answer: C

Answer: C

Answer: D

Answer: A

Answer: C