Sampling with/without replacement, tree diagrams, and Venn diagrams

A bag contains 7 counters: 4 orange and 3 purple. Two counters are drawn at random, one after another, without replacement. What is $P(\text{first orange, then purple})$?

• $\frac{2}{7}$

  Correct. $P(O_1) = \frac{4}{7}$, and once an orange counter is gone, 3 purple remain out of 6 total: $P(P_2 \mid O_1) = \frac{3}{6}$. Multiplying along the path: $\frac{4}{7} \times \frac{3}{6} = \frac{12}{42} = \frac{2}{7}$.

• B$\frac{12}{49}$

  This is $\frac{4}{7} \times \frac{3}{7}$ — the second fraction's denominator was left at 7, as though the first counter had been put back. This is exactly the mechanism a real examiner report documents on a without-replacement question spanning several draws: keeping the denominator fixed instead of letting it shrink by one at every draw.

• C$\frac{4}{21}$

  This computes $\frac{4}{7} \times \frac{2}{6}$, reducing the purple count from 3 to 2 as though a purple counter had been the one removed. It was an orange counter that came out first — the purple count stays at 3, and it is the total and the orange count that fall.

• D$\frac{1}{2}$

  This is only the second branch's own probability, $P(P_2 \mid O_1)$, on its own. A path through a tree diagram is a sequence of two events happening one after the other, and the product law says its probability is the product of both branches, not just the last one.

Traps tested: Denominator kept constant across repeated draws · Wrong colours count reduced · Second branch alone not multiplied

On a Venn diagram, $P(C) = 0.6$, $P(D) = 0.5$ and $P(C \cap D) = 0.2$. What is $P(C \mid D)$?

• $0.4$

  Correct. $P(C \mid D) = \frac{P(C \cap D)}{P(D)} = \frac{0.2}{0.5} = 0.4$ — restrict attention to the $D$ circle (denominator 0.5) and ask what share of it is also inside $C$ (numerator 0.2).

• B$0.6$

  This assumes $C$ and $D$ are independent and uses $P(C \mid D) = P(C)$ without checking — and here the assumption is wrong: $P(C) \times P(D) = 0.6 \times 0.5 = 0.3$, which does not equal $P(C \cap D) = 0.2$, so they are not independent. A real examiner report names exactly this shortcut as a pattern across the whole paper, not a one-off: candidates assume independence "when an appropriate conditional probability should be used instead."

• C$\frac{1}{3}$

  This divides the other way round, $\frac{P(C \cap D)}{P(C)} = \frac{0.2}{0.6}$, which gives $P(D \mid C)$ instead of $P(C \mid D)$. The event named after the vertical bar is the one to divide by.

• D$0.7$

  This adds $P(C \cap D) + P(D) = 0.2 + 0.5$. A conditional probability is a ratio — what fraction of $D$'s probability also lies in $C$ — not a sum of the two given values.

Traps tested: Independence assumed instead of tested · Conditioned on the wrong event · Added instead of divided

A Venn diagram for a class of 60 students and events $A$ (studies French) and $B$ (studies Spanish) shows 18 students in $A$ only, 22 in $B$ only, and 12 in both. How many students are outside both circles, and how should that be shown on the diagram?

• 8, written explicitly in the region outside both circles.

  Correct. $18 + 22 + 12 = 52$, leaving $60 - 52 = 8$ students who study neither — and that 8 has to be written on the diagram itself, in the region outside both circles, not left blank.

• B0, so nothing needs to be written there.

  The arithmetic gives 8, not 0 — three named regions here only account for 52 of the 60 students. More generally, even a genuinely empty region has to carry a written 0: a real examiner report specifically records candidates losing marks for leaving it blank, stating plainly that blank spaces are not assumed to be 0s in Venn diagrams.

• C8, but it doesn't need to be shown since it can be worked out from the other three.

  The value 8 is correct, but the same examiner-report finding applies regardless of whether it CAN be worked out — the diagram is marked on what it shows, not on what a reader could reconstruct from it.

• D42, since only the overlap should be subtracted from the total.

  This is $60 - 18$, subtracting only the $A$-only region and leaving out both the $B$-only region and the overlap. All three named regions — $A$ only, $B$ only, and both — have to come off the total, not just one of them.

Traps tested: Venn blank region assumed zero · Region value not required on diagram · Only one region subtracted

A bag contains 5 counters: 2 red and 3 green. Two counters are drawn at random, WITH replacement. What is $P(\text{both red})$?

• $\frac{4}{25}$

  Correct. With replacement, the bag is identical for both draws: $P(R_1) = \frac{2}{5}$ and $P(R_2) = \frac{2}{5}$ again, so $P(\text{both red}) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}$.

• B$\frac{1}{10}$

  This is $\frac{2}{5} \times \frac{1}{4}$ — the without-replacement answer, reducing both the red count and the total for the second draw. The question states the counter is replaced, so the second draw sees exactly the same bag as the first.

• C$\frac{2}{5}$

  This is only the probability of the first counter being red. "Both red" is two events happening in sequence, and the product law needs both branches multiplied, not just the first one stated on its own.

• D$\frac{1}{5}$

  This is $\frac{2}{5} \times \frac{2}{4}$ — reducing the total for the second draw as though a counter had been removed, while the question says it was replaced. With replacement, nothing about the bag changes between draws, so the second fraction should be identical to the first, not partly adjusted.

Traps tested: With replacement treated as without · Second draw not considered · Denominator reduced numerator not

A spinner is divided into 5 equal sections: 3 blue and 2 yellow. It is spun twice. What is $P(\text{exactly one yellow})$?

• $\frac{12}{25}$

  Correct. Two different paths give exactly one yellow — blue-then-yellow and yellow-then-blue — and a spinner resets itself every spin, so both spins use the same fractions, $P(\text{blue})=\frac{3}{5}$ and $P(\text{yellow})=\frac{2}{5}$. Each path: $\frac{3}{5}\times\frac{2}{5} = \frac{6}{25}$. Both paths count as "exactly one yellow" and cannot both happen on the same two spins, so the sum law adds them: $\frac{6}{25}+\frac{6}{25}=\frac{12}{25}$.

• B$\frac{6}{25}$

  This is only one of the two paths that give exactly one yellow — blue-then-yellow, say — leaving out yellow-then-blue. Both orders count as "exactly one yellow", and the sum law needs both added, not just one.

• C$\frac{4}{25}$

  This is $P(\text{both yellow}) = \frac{2}{5}\times\frac{2}{5}$, a different event from "exactly one yellow". Both yellow is one single path; exactly one yellow is two.

• D$\frac{3}{5}$

  This is just $P(\text{blue})$ on a single spin, with no second spin considered at all. The question asks about two spins together.

Traps tested: Only one order counted · Wrong event computed · Single spin probability given