Is P(at least one six) just n × 1/6?

Here is a question that tests this misconception head-on, in the spirit of JEE Advanced 2021 Paper 1 Q5:

The misconception answer adds a $\tfrac{1}{6}$ per throw: $7 \times \tfrac{1}{6} = \tfrac{7}{6} \approx 1.17$. But a probability can never exceed 1, so the additive rule has refuted itself. The correct move is the complement — the one clean failure is no six on any throw:

About 72%, comfortably below 1 — the complement always gives a sensible probability exactly where the sum cannot.

Adding feels like the way to combine "chances of a six", and for two or three throws the result ($\tfrac{2}{6}$, $\tfrac{3}{6}$) looks plausible enough to pass unchallenged. The rule only visibly breaks when stretched far enough to cross 1, by which point the student has long trusted it. Worse, the additive rule has no built-in cap, so a student can produce an impossible probability without noticing.

Fischbein's work on primary intuitions in probability, and Tversky and Kahneman's on disjunction-event misjudgement, both find that students under-use the complement rule even when it is the only tractable route. Being told the multiplication rule rarely dislodges the addition habit. What dislodges it is pushing the student's own rule to $\tfrac{7}{6}$ and letting them see it produce an impossible number — then offering $1 - P(\text{none})$ as the repair.

The complement of "at least one" is "none". There is exactly one way to get none — every trial fails — so its probability is a clean product across independent trials, and $P(\text{at least one}) = 1 - P(\text{none})$. For $n$ independent trials each succeeding with probability $p$:

This never climbs past 1, because $(1-p)^n$ always lies between 0 and 1. Adding the per-try probabilities is right only when the events are mutually exclusive, that is when they cannot both occur — which repeated trials are not. So whenever you read "at least one", flip to the complement instead of summing.

Take a fair die thrown twice, and find $P(\text{at least one six})$ both ways.

1. The additive attempt. $\tfrac{1}{6} + \tfrac{1}{6} = \tfrac{2}{6} = \tfrac{12}{36}$. It looks harmless at two throws, so the rule survives — but watch it at seven throws, where $7 \times \tfrac{1}{6} = \tfrac{7}{6} > 1$, which is impossible.
2. The one failure path. The single way to get no six at all is "not a six" on both throws. On one throw $P(\text{not a six}) = \tfrac{5}{6}$.
3. Multiply across independent throws. $P(\text{no six at all}) = \left(\tfrac{5}{6}\right)^2 = \tfrac{25}{36}$.
4. Subtract from 1.
   $$P(\text{at least one six}) = 1 - \tfrac{25}{36} = \tfrac{11}{36} \approx 0.31.$$
5. Compare. The complement gives $\tfrac{11}{36}$, while the additive rule gave $\tfrac{12}{36}$ — already too big by $\tfrac{1}{36}$ at two throws, exactly the double-counted "both sixes" outcome. The gap only widens, until the sum breaks past 1.

The same framework settles JEE Advanced 2020 Paper 2 Q2: to make $P(\text{at least one hit}) \ge 0.95$ when each missile hits with probability $0.75$, solve $1 - (0.25)^n \ge 0.95$, i.e. $(0.25)^n \le 0.05$. Since $(0.25)^2 = 0.0625$ and $(0.25)^3 = 0.0156$, the minimum is $n = 3$ missiles — found through the complement, never a sum of per-missile chances.