Probability and Combined Events

Updated 7 October 2026

Probability is the one branch of school mathematics where a correct calculation regularly contradicts a strong intuition — and where the intuition is usually the thing that is wrong. That makes it unusually worth doing carefully. The arithmetic is easy; the difficulty is always in stating precisely what event you are asking about.

What does a probability actually measure?

A probability measures how likely something is, as a number from 0, impossible, to 1, certain. When every outcome is equally likely, it is the number of outcomes where the event happens divided by the total number. A fair die has six faces, and three are even, so the probability of an even number is $\frac{3}{6}=\frac{1}{2}$3/6 = 1/2. See The Probability Scale, which places words such as unlikely and even chance on the same line. The fraction works because each equally likely outcome carries the same share of the certainty. The steps change when outcomes are not equally likely, as with a drawing pin landing point up: then the probability is estimated from trials, as successes over attempts. The chance that an event does not happen is 1 minus the chance that it does, so the chance of not rolling a six is $1-\frac{1}{6}=\frac{5}{6}$1 − 1/6 = 5/6. The probabilities of all the outcomes must add to 1: $\frac{1}{2}+\frac{1}{2}=1$1/2 + 1/2 = 1.

Getting a number requires listing what could happen. Sample Space is the discipline of writing out every outcome, and it is the step people skip and then regret: the sample space for two dice has 36 entries, not 11, because a 3 and a 4 is a different outcome from a 4 and a 3.

Single-Event Probability then divides — favorable outcomes over total outcomes — with the condition that makes the division legitimate: the outcomes must be equally likely. Complementary Events turns a question about what does not happen into one about what does.

Probability from experiment rather than theory

Experimental Probability covers outcomes that cannot be counted.

The Law of Large Numbers says that estimate converges: the more trials, the closer the relative frequency settles to the true probability.

Expected Counts from a Chance runs it the other way: multiply the probability by the number of trials to predict how many times something should occur. Twenty rolls of a die should give roughly three or four sixes, and the gap between that and what happened is exactly the quantity every statistical test later measures.

How do you handle two events at once?

To handle two events at once, draw them so that every outcome has a place. A Venn diagram shows two events as overlapping circles inside a box, splitting the outcomes into four regions: A only, B only, both, and neither. See Naming the Regions of a Venn Diagram. In a class of 30, suppose 12 play soccer, 10 play tennis, and 4 play both. Then 8 play soccer only, 6 play tennis only, and 30 − 8 − 6 − 4 = 12 play neither. The probability that a student picked at random plays a sport is $\frac{18}{30}=\frac{3}{5}$18/30 = 3/5. The regions never overlap and together cover everyone, so each student is counted exactly once. Adding the two totals counts the middle twice, which is why 12 + 10 = 22 overstates the 18 who play; subtract the overlap once. Confirm the diagram by adding the four regions back to the class: 8 + 6 + 4 + 12 = 30.

Possibility Diagrams is the grid version for two-stage experiments, with the outcomes of one stage along the top and the other down the side. Two dice become a 6 × 6 grid where you can literally count the cells summing to 7, and Chance from Areas extends the same idea to continuous cases, where probability becomes a ratio of areas rather than a count.

Mutually Exclusive Events is the condition under which you may simply add: if two events cannot both happen, the chance of one or the other is the sum. When they can overlap, The Addition Rule for Overlapping Events subtracts the overlap once: P(A or B) = P(A) + P(B) − P(A and B).

Independent Events is the condition under which you may simply multiply: if one event's outcome does not affect the other's chance, the probability of both is the product. Two coin tosses are independent; two cards drawn without replacement are not.

How do tree diagrams work?

A tree diagram shows a sequence of events, with a set of branches for each stage and a probability on every branch. Multiply along a path for the chance of that sequence, and add the paths that answer the question. A bag holds 2 red and 3 blue counters; one is taken, replaced, and another taken. Both red has probability $\frac{2}{5}\times \frac{2}{5}=\frac{4}{25}$2/5 × 2/5 = 4/25. One of each color happens two ways, so add both paths: $\frac{2}{5}\times \frac{3}{5}+\frac{3}{5}\times \frac{2}{5}=\frac{12}{25}$2/5 × 3/5 + 3/5 × 2/5 = 12/25. See Tree Diagrams. Multiplying works because the second stage happens in a fraction of the cases the first leaves. Adding works because different paths cannot happen together. If the counter is not replaced, the second branches change: after a red, 1 red is left among 4 counters, so both red becomes $\frac{2}{5}\times \frac{1}{4}=\frac{1}{10}$2/5 × 1/4 = 1/10. The branches from each point add to 1, and so do all the paths: $\frac{4}{25}+\frac{12}{25}+\frac{9}{25}=1$4/25 + 12/25 + 9/25 = 1.

The Chance of At Least One is the first serious payoff. "At least one six in three rolls" spreads across seven paths; "no six at all" is one path. $1-{\left(\frac{5}{6}\right)}^3\approx 0.42$1 − (5/6)³ ≈ 0.42. Doing it the long way is not wrong, it is just an invitation to lose a case.

Then the branches start changing. Without Replacement is the situation where taking one item alters what is left, so the second layer's fractions differ from the first — and differ per branch. Multiplying Along a Tree Without Replacement works the full calculation.

A bag holds 5 red and 3 blue. $P\left(\text{both red}\right)=\frac{5}{8}\times \frac{4}{7}=\frac{20}{56}$P(both red) = 5/8 × 4/7 = 20/56. The denominator dropped because a ball is gone; the numerator dropped because the ball that went was red.

What is conditional probability?

Conditional probability is the probability of one event given that another has happened, so the sample space shrinks to the outcomes where the given event occurred. Conditional Probability shows that shrinking, and Conditional Probability Notation writes it: $P\left(A|B\right)=P\left(A\cap B\right)$P(A | B) = P(A ∩ B) / P(B). Roll a fair die and suppose you are told the result is even. Only 2, 4 and 6 remain, so the chance of a 4 given an even result is $\frac{1}{3}$1/3, not $\frac{1}{6}$1/6. With the formula, $\frac{1}{6}÷\frac{3}{6}=\frac{1}{3}$1/6 ÷ 3/6 = 1/3. Dividing rescales the smaller sample space so that its probabilities again add to 1. Order matters: the chance of an even result given a 4 is 1, since every 4 is even, while the chance of a 4 given an even result is $\frac{1}{3}$1/3. In a tree diagram, every branch in the second stage is already a conditional probability. The three outcomes left must share all of the probability: $\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$1/3 + 1/3 + 1/3 = 1.

Read that formula as a restriction rather than a division. The numerator counts outcomes where both happened; the denominator counts every outcome in the new, smaller world. On a two-way table it is reading down a single column and ignoring the rest, and the charts and displays guide covers the frequency tree.

Independence gets its real definition here: A and B are independent exactly when P(A | B) = P(A) — knowing B changes nothing.

Why do the famous problems catch everybody?

Famous probability puzzles catch people because each hides a condition the intuitive answer ignores. In the Monty Hall problem there are three doors and one prize. You pick a door, the host opens another door with no prize, and you may switch. Switching wins two times in three. Your first pick is right with probability $\frac{1}{3}$1/3, and the host's move does not change that, so the other closed door carries the remaining $1-\frac{1}{3}=\frac{2}{3}$1 − 1/3 = 2/3. The hidden condition is that the host knows where the prize is and never opens that door; a host opening a door at random would leave an even chance. Listing cases makes it plain: with the prize behind door 1, picking each door in turn, switching loses once and wins twice. The birthday paradox hides a count: 23 people form 23 × 22 ÷ 2 = 253 pairs, and each pair could share a birthday. The Monty Hall shares cover the whole prize: $\frac{1}{3}+\frac{2}{3}=1$1/3 + 2/3 = 1.

The Monty Hall Problem: the host knows where the prize is, and that knowledge moves the two-thirds onto the unopened door.

The Birthday Paradox: in a room of 23 people it is more likely than not that two share a birthday. The instinct compares you against 22 others; the event compares every pair. Computing it as 1 minus the chance of all different — $\frac{365}{365}\times \frac{364}{365}\times \frac{363}{365}\times \dots$365/365 × 364/365 × 363/365 × … — makes the collapse visible.

Why a Positive Test Can Still Mean Healthy is the one with consequences outside a classroom. Take a disease affecting 1 in 1,000 and a test that is 99% accurate both ways, and run 100,000 people through it: about 99 of the 100 who have it test positive, and about 999 of the 99,900 who do not also test positive. A positive result is correct about 9% of the time.

Reversing a Tree: the Base-Rate Answer is the general method behind it. Build the tree in the order events happen, then read it backwards: the answer to "given the test was positive, what is the chance of illness" is the one relevant path divided by every path ending in a positive. That is Bayes' theorem, arrived at by counting rather than by formula.

The mistakes worth naming

• Adding probabilities of events that can both happen. Subtract the overlap, or you have counted it twice.
• Multiplying when the events are not independent. Without replacement, the second fraction must change.
• Listing outcomes that are not equally likely. Two dice have 36 equally likely outcomes; their 11 possible totals are not equally likely.
• Believing a run is "due". Independent trials have no memory. The proportion evens out by dilution, not by correction.
• Swapping P(A | B) with P(B | A). This is the base-rate error, and it is the most consequential mistake in the topic.
• Enumerating "at least one" case by case. Use the complement; the long way loses a branch.

Where this leads next

Counting outcomes stops being feasible once the sample space is large, and the fix is to count without listing — permutations, combinations and the set notation that organizes them, in the sets, counting and distributions guide, which also turns repeated trials into the binomial distribution. Turning an observed frequency into a claim about a population is the sampling and inference guide. And the fraction arithmetic every tree diagram runs on is in the fractions guide.