Trial vs Outcome

In statistics a trial is not a court case and not a whole clinical study. It is one repetition of a chance process, and it ends with exactly one outcome from the sample space. Flip a coin once and you have run one trial, landing heads or tails. The problem defines what one trial is, and writing that definition down is the first move, because every count that follows is measured in those units.

Flip a fair coin 20 times and you have run 20 trials. If 12 come up heads, the relative frequency is $12/20 = 0.60$ while the probability of heads is still 0.5 and the expected count was $20(0.5) = 10$. Trials are called independent when the result of one leaves the probabilities for the next unchanged, which fair coin flips satisfy.

Here is the error that quietly ruins simulations. To estimate the chance that a family with three children has at least one girl, one trial is three births, not one birth. "I flipped the coin 100 times, so I ran 100 trials" is wrong twice over: 100 flips supply 33 complete trials with one flip left dangling, and dividing by 100 understates every proportion. A trial runs until the question has a single yes or no answer, so a trial that takes three flips consumes three flips. The exact answer here is $1 - (0.5)^3 = 0.875$, and listing the 8 equally likely birth orders shows 7 of them contain a girl.

The binomial setting demands a fixed number of trials that are independent and share one probability of success. Sampling without replacement breaks the last two: pull chips from a bag and each trial changes what the next one faces, which is why the ten percent condition exists.

Trials are the counting unit of simulation, topic 2.3, Estimating Probabilities Using Simulation.

An outcome is one complete result of a single trial and cannot be broken into smaller results. An event is any set of outcomes, so events are assembled out of outcomes and never the reverse, and an event's probability is the total probability of the outcomes inside it. What counts as an outcome is fixed by what you record on each trial, and that choice does most of the damage below.

Roll two fair dice and record both faces. The sample space holds $6 \times 6 = 36$ ordered outcomes, each carrying probability $1/36 \approx 0.0278$ by symmetry. Six of them total 7, running from $(1,6)$ to $(6,1)$, so $P(\text{sum of } 7) = 6/36 \approx 0.1667$. Exactly one totals 2, so $P(\text{sum of } 2) = 1/36 \approx 0.0278$.

Now the misreading: there are 11 possible sums, 2 through 12, so each has probability $1/11 \approx 0.0909$. Those 11 sums are a legitimate way to describe a roll, but they are events rather than outcomes, and they are events of very different sizes. A sum of 7 collects six outcomes while a sum of 2 collects one, so the first is six times as likely as the second, not equal to it. Counting only produces a probability across items that carry equal probability, and grouping outcomes into categories is the quickest way to lose that.

Outcomes are not required to be equally likely, and nothing in the definition promises they will be. A loaded die still has six outcomes; you add their stated probabilities rather than counting them. What does always hold across a finite sample space is that the outcome probabilities total 1.

Topic 2.4, Introduction to Probability, defines the sample space as the set of all possible nonoverlapping outcomes. Nonoverlapping is the load-bearing word. If two items on your list could both be true of the same trial, you have written down events, and the counting shortcut no longer applies to them.