Question bank — Independent events — multiplication rule

False — that describes mutually exclusive events (no overlap in the Venn picture). Independent events usually can co-occur; in fact if both are possible, $P(A\cap B)=P(A)P(B)>0$, so they happen together sometimes.

True — this equation is the definition/test of independence, and it is symmetric, so it also guarantees $P(A\mid B)=P(A)$ and $P(B\mid A)=P(B)$.

False — they are the strongest form of dependence: if $A$ happens, $B$ is forced not to happen, so knowing $A$ changes $B$'s chance from $P(B)$ down to $0$.

False — removing the first card changes what's left, so $P(\text{2nd}\mid \text{1st})\ne P(\text{2nd})$; you must use the General Multiplication Rule with a conditional probability.

True — since $P(A^c\cap B^c)=1-P(A\cup B)=1-(P(A)+P(B)-P(A)P(B))=(1-P(A))(1-P(B))=P(A^c)P(B^c)$, which is exactly the independence test for the complements.

False — summing double-counts the overlap and can exceed $1$; for independent events use $1-{\prod }_i(1-P(A_i))$ via the Complement Rule.

False (except in degenerate cases) — it requires $P(A)=P(A{)}^2$, which forces $P(A)=0$ or $P(A)=1$; any genuinely uncertain event ($0<P(A)<1$) is dependent on itself.

True — but note the essential caveat: both conditionals must be defined, i.e. $P(A)>0$ and $P(B)>0$; when either probability is $0$ the conditional is undefined and this symmetric implication doesn't apply.

True — independence is a numerical property $P(A\cap B)=P(A)P(B)$, so the very same labelled events can be independent under one probability assignment and dependent under another.

False — independence is not transitive; pairwise links between $A$–$B$ and $B$–$C$ tell you nothing guaranteed about $A$–$C$.

False — mutual independence is stronger; it also requires $P(A\cap B\cap C)=P(A)P(B)P(C)$, which can fail even when all pairs check out.

The error is multiplying two unconditional probabilities. The bag changed, so the second factor must be $P(R_2\mid R_1)=\frac{3}{9}$, giving $\frac{4}{10}\cdot \frac{3}{9}\approx 0.133$.

"Can't both happen" means $P(A\cap B)=0$, which is exclusivity, not independence. It actually makes them dependent, since $A$ occurring guarantees $B$ does not.

The gambler's fallacy: independence means past flips leave the next flip's probability exactly $\frac{1}{2}$; the coin has no memory, so nothing is "due".

Independence is the wrong tool here; addition without subtracting $P(A\cap B)$ is only valid for mutually exclusive events. For independent events, $P(A\cup B)=P(A)+P(B)-P(A)P(B)$.

Adding overlaps double-counts and this only approximates. The clean method is the complement: $1-0.9^3=0.271$.

This confuses "no information transfer" with "no overlap". Look at the Venn figure above: independent events typically overlap (their intersection area equals $P(A)P(B)$); a zero overlap would make them exclusive and hence dependent.

The second is the General Multiplication Rule — always true by definition of conditional probability. The first is a special case obtained only after replacing $P(A\mid B)$ with $P(A)$, a swap independence permits.

"At least one" splits into many messy cases (exactly one, exactly two, …), but its opposite — "none occur" — is a single clean product $\prod (1-P(A_i))$, so $1-\prod (1-P(A_i))$ is far easier.

Replacing the drawn item resets the bag to its original composition, so the second draw faces identical odds; the first draw leaves no trace, which is exactly the meaning of independence.

Each probability is at most $1$, so multiplying shrinks the result — and it should: requiring both events is harder than requiring either alone, so "both" deserves a smaller chance.

The formula always produces a number, but that number is only correct when independence holds. Skipping the check silently gives wrong answers (as in the without-replacement bag).

Independence means one event carries zero information about the other; exclusivity means one event carries maximal information — it tells you the other definitely did not happen.

Yes — $P(A\cap B)\le P(A)=0$, so $P(A\cap B)=0=0\cdot P(B)=P(A)P(B)$; an impossible event is trivially independent of anything.

Yes — a certain event satisfies $P(A\cap B)=P(B)=1\cdot P(B)=P(A)P(B)$, so knowing $B$ never changes $A$'s (already certain) status.

Only in degenerate cases — you'd need $P(A\cap B)=0$ and $P(A\cap B)=P(A)P(B)$ simultaneously, forcing $P(A)=0$ or $P(B)=0$; with both positive it's impossible.

Yes — physically separate devices share no mechanism, so $P(\text{H}\cap 6)=\frac{1}{2}\cdot \frac{1}{6}=\frac{1}{12}$ matches the product, confirming independence.

Only in special numeric cases — covering everything doesn't force dependence, but you must still check $P(A\cap B)=P(A)P(B)$; e.g. $A$ and $A^c$ with $0<P(A)<1$ are not independent since $P(A\cap A^c)=0\ne P(A)P(A^c)$.