Two outcomes does not mean fifty-fifty

A random process generates results that are individually unpredictable but that follow a stable pattern over many repetitions. The sample space is the set of all possible outcomes, and an event is any collection of those outcomes.

Rolling two fair dice has a sample space of 36 ordered outcomes, from (1,1) to (6,6). "The sum is at least 10" is an event containing 6 of them: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6). Because the 36 outcomes are equally likely here, $P(\text{sum} \ge 10) = \frac{6}{36} = \frac{1}{6}$.

The counting must be done over the sample space that the random process actually generates. Rolling two dice has 36 outcomes, not 11: the eleven possible sums are events built from those outcomes, and they are not equally likely, since a sum of 7 collects six outcomes while a sum of 2 collects one. Choosing the wrong list to count over is what makes an answer wrong before any arithmetic starts.

Two rules hold for any probability model whatsoever:

1. $0 \le P(A) \le 1$ for every event $A$. A probability of 0 means the event never occurs in this model; a probability of 1 means it always does.
2. The probabilities of all the outcomes in the sample space sum to exactly 1.

These are working tools, not decoration. An answer of 1.4 or $-0.2$ is wrong on sight, no matter how the arithmetic got there, and a probability distribution whose listed values sum to 0.9 or 1.2 is not a distribution. Checking both is the cheapest error detection in the unit, and it catches the most common arithmetic slip in the next topic, where "or" probabilities are computed by adding overlapping pieces and quietly exceed 1.

A third statement follows immediately: the complement of $A$, written $A^c$ and read "not $A$", satisfies

$$P(A^c) = 1 - P(A),$$

because $A$ and "not $A$" together are the whole sample space.

When the outcomes of a sample space are equally likely, probability is a counting problem: $P(A) = \frac{\text{outcomes in } A}{\text{outcomes in the sample space}}$. Fair dice, a well-shuffled deck, and a properly executed simple random sample all justify that assumption.

Most situations do not. "Either it rains tomorrow or it does not, so the probability is 0.5" lists two outcomes and then assumes, with no support, that they are equally likely. "A game ends in a win, a loss, or a draw, so each is $\frac{1}{3}$" makes the same move. The number of outcomes tells you how many pieces the total 1 is divided into, never how big the pieces are.

When outcomes are not equally likely, the probabilities come from somewhere else: a long-run relative frequency observed in data, a simulation, or a model built from the structure of the process. Whatever the source, the assignment still has to satisfy both rules above.

"At least one" events are the standard use of the complement rule, because the complement of "at least one" is the single tidy event "none".

Suppose a quality check fails a part with probability 0.2, independently for each of 4 parts, and the question is the probability that at least one part fails. Listing the ways (exactly one, exactly two, exactly three, all four) means four calculations. The complement takes one:

$$P(\text{at least one fails}) = 1 - P(\text{none fails}) = 1 - (0.8)^4 = 1 - 0.4096 = 0.5904.$$

The error this replaces is adding: $4 \times 0.2 = 0.8$ treats the four ways to fail as if they could not overlap and inflates the answer. Push the same reasoning to ten parts and adding gives 2.0, which the 0-to-1 rule rejects on sight while the complement gives $1 - (0.8)^{10} \approx 0.893$.

Two wording traps ride along. "At least one" is not "exactly one", and its complement is "none", not "all". Reading the event out loud in the complement's language before computing is what keeps them apart.