Q.A coin is known to be biased so that it lands Heads 70% of the time and Tails 30% of the time. Can the two outcomes $\{H, T\}$ of a single toss of this coin be treated as equally likely? Explain what this means for applying the classical definition of probability directly by counting outcomes.

Equally likely requires every outcome to have the SAME chance — compare the two given percentages.

Since $P(H)=0.7 \neq P(T)=0.3$, the two outcomes do not have the same chance, so they are not equally likely; the classical formula $n(E)/n(S)$ cannot be applied here by simply counting outcomes.

By the definition in §4, outcomes are equally likely only when EVERY outcome has exactly the SAME chance of occurring as every other outcome.

Here, the coin is explicitly stated to land Heads 70% of the time and Tails only 30% of the time — these are two clearly DIFFERENT chances, so the outcomes $H$ and $T$ are not equally likely.

Consequence for the classical formula: the classical definition $P(E) = n(E)/n(S)$ (§5) is valid ONLY under the equally-likely assumption. If we were to apply it naively here by simply counting outcomes — treating $n(S)=2$ and each outcome as '1 favourable out of 2' — we would wrongly conclude $P(H)=P(T)=\tfrac12$, which directly contradicts the coin's actual known bias.

The correct probabilities for this biased coin must instead be assigned directly from the known bias itself: $P(H) = 0.7$ and $P(T) = 0.3$ (note these still satisfy $P(H)+P(T)=1$, consistent with $S$ being the sure event) — they are simply not derived by counting equally-weighted outcomes.

A common misconception is assuming EVERY coin toss automatically has a 50-50 chance regardless of context — the equally-likely assumption is a property of the specific experiment (a genuinely FAIR coin), not an automatic law of coin-tossing; a problem that explicitly states a bias must never be treated as equally likely.