Conditional probability, and independence vs. mutually exclusive

Events $C$ and $D$ satisfy $P(C) = 0.6$, $P(D) = 0.5$, and $P(C \cap D) = 0.3$. Are $C$ and $D$ independent, mutually exclusive, both, or neither?

• Independent, and therefore not mutually exclusive — $P(C) \times P(D) = 0.6 \times 0.5 = 0.3 = P(C \cap D)$

  Correct. The product-rule test passes exactly, so $C$ and $D$ are independent — and since $P(C \cap D) = 0.3 \neq 0$, they are not mutually exclusive, consistent with the mechanism proved earlier: independent events with positive probabilities can never be mutually exclusive.

• BMutually exclusive, since $C$ and $D$ are clearly two separate, distinct events

  'Distinct events' is not the same as 'mutually exclusive' — mutually exclusive specifically means they share no outcomes, i.e. $P(C \cap D) = 0$. Here $P(C \cap D) = 0.3$, given directly in the question, so they plainly do overlap.

• CBoth independent and mutually exclusive

  Impossible whenever both probabilities are positive, by the same argument proved in the mechanism block: mutually exclusive would force $P(C \cap D) = 0$, but it's given as $0.3$. Only one of the two properties can hold here, and the calculation shows it's independence.

• DNeither — $0.3$ doesn't obviously relate to $0.6$ and $0.5$

  $0.3$ is exactly $0.6 \times 0.5$ — the relationship is there, it just needs the multiplication carried out rather than eyeballed. This is the whole reason the independence test exists: the connection between the numbers is rarely visible without actually computing it.

Traps tested: Distinct events assumed mutually exclusive · Mutually exclusive and independent treated as compatible · Independence check not attempted

In a class, $P(\text{studies Chemistry}) = 0.5$, $P(\text{studies Biology}) = 0.6$, and $P(\text{studies both}) = 0.2$. Find $P(\text{Biology} \mid \text{NOT Chemistry})$.

• $0.8$ — from $P(B \cap C') = P(B) - P(B \cap C) = 0.6 - 0.2 = 0.4$ and $P(C') = 1 - 0.5 = 0.5$, so $P(B \mid C') = 0.4 \div 0.5$

  Correct. $P(B \cap C')$ is the students who study Biology but NOT Chemistry — everyone who studies Biology, minus the ones who also study Chemistry — and $P(C')$ is the complement rule applied to the conditioning event. Both intermediate steps have to be found before the conditional probability formula can be applied at all.

• B$0.6$ — since 'not Chemistry' shouldn't change anything about Biology

  This assumes the answer rather than testing it — exactly the habit this lesson is built to break. $P(\text{Biology}) = 0.6$ unconditionally, but conditioning on NOT studying Chemistry does change the figure here, to $0.8$: among the non-Chemistry students specifically, Biology is more common than in the class as a whole, because the overlap between the two subjects has been removed from the comparison.

• C$0.4$ — using $P(B \cap C')$ alone, without dividing by $P(C')$

  $0.4$ is a correct intermediate value — but it's $P(B \cap C')$, the probability of studying Biology and not Chemistry out of the WHOLE class, not the conditional probability restricted to non-Chemistry students. The division by $P(C') = 0.5$ is the step that's missing.

• D$\dfrac{1}{3}$ — from $P(C \cap B) \div P(B) = 0.2 \div 0.6$

  This computes $P(\text{Chemistry} \mid \text{Biology})$ — the wrong pair of events conditioned the wrong way round, not $P(\text{Biology} \mid \text{NOT Chemistry})$. Both the event being conditioned ON and which event is complemented need to match the question exactly.

Traps tested: Conditioning assumed to have no effect · Conditioning step omitted · Conditioning event reversed

Events $E$ and $F$ satisfy $P(E) = 0.2$, $P(F) = 0.3$, $P(E \cap F) = 0.06$. A student writes: 'E and F are independent, because their probabilities are both small, so they can't really affect each other.' Is the student's final answer correct, and is the reasoning acceptable?

• The answer is correct ($0.2 \times 0.3 = 0.06 = P(E \cap F)$, so they ARE independent) — but the reasoning given would not earn the method mark, because 'small probabilities' was never actually tested

  Correct on both counts. The independence test happens to pass here, so the conclusion is right — but a mark scheme rewards the calculation, not a correct guess about it. Swap in different small-but-not-multiplying-to-match probabilities (say $P(E)=0.2$, $P(F)=0.3$, $P(E \cap F)=0.1$) and the same reasoning ('both small, so independent') would produce the wrong answer, because it was never actually connected to the numbers.

• BThe answer is correct and the reasoning is acceptable, since it correctly identifies that small probabilities tend to indicate independence

  There is no general rule connecting the SIZE of two probabilities to whether they're independent — a large $P(E)$ and small $P(F)$ can just as easily be independent, or a small $P(E)$ and small $P(F)$ can just as easily be dependent. The only valid test is the specific equality $P(E \cap F) = P(E) \times P(F)$, checked with these exact numbers.

• CThe answer is wrong — $E$ and $F$ are not independent, because $0.06$ is a small number and small intersections usually mean dependent events

  $0.06$ being 'small' says nothing on its own; it has to be compared to $P(E) \times P(F) = 0.06$ specifically, and it matches exactly. The size of $P(E \cap F)$ in isolation, without that comparison, cannot answer an independence question.

• DThere isn't enough information to check independence without knowing more about what $E$ and $F$ actually represent

  The independence test needs only the three numbers given — $P(E)$, $P(F)$, and $P(E \cap F)$ — and nothing about what the events physically represent. That's precisely what makes the test objective rather than a matter of judgement: two events can be independent (or not) regardless of how intuitively 'related' they seem.

Traps tested: Probability magnitude substituted for independence test · Intersection magnitude substituted for independence test · Independence treated as unanswerable without context