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Probability Foundations

Model outcomes, complements, and conditional information.

Statistics is the math of data. You collect numbers, describe what they show, and then decide how much of the pattern could just be chance.

Probability Foundations: the central idea

Probability assigns a number from $0$ to $1$ to a clearly defined event. Rules for complements, unions, intersections, and conditioning prevent double counting.

Words you need

Sample space
The sample space is the complete list of every outcome an experiment can produce, such as the 36 ordered pairs you get from rolling two dice.
Event
An event is any collection of outcomes from the sample space that you want to measure, such as 'the two dice add to 7'.
Complement
The complement of an event is every outcome in the sample space where that event does not happen, and its probability is 1 minus the probability of the event.
Conditional probability
A conditional probability is the chance of one event given that another event has already happened, which shrinks the sample space to only the outcomes where the known event is true.
Independent events
Two events are independent when knowing that one happened does not change the probability of the other, which is exactly the case for two separate coin flips.
Mutually exclusive events
Two events are mutually exclusive when they cannot both happen on the same trial, so their overlap is empty and their probabilities can simply be added.

What to know before this lesson

Understand sets, fractions, counting outcomes, and the meanings of AND, OR, NOT, and GIVEN.

If one of those prerequisites is uncertain, use the Statistics subject guide to locate the earlier concept before memorizing a procedure.

Probability foundations: a worked example

Follow the mathematical structure
If $P(A)=0.3$, then $P(A^c)=1-0.3=0.7$.

Every step, with the arithmetic

  1. Step 1 - Describe the experiment and sample spacea bag holds 3 red marbles and 5 blue marbles, so there are 3 + 5 = 8 marbles and you draw two without putting the first one back
  2. Step 2 - Probability the first marble is redP(first red) = 3 / 8 = 0.375
  3. Step 3 - Use the complement for 'not red'P(first not red) = 1 - 3/8 = 5/8 = 0.625
  4. Step 4 - Condition on the first drawone red is gone, so 7 marbles remain with 2 red: P(second red given first red) = 2 / 7
  5. Step 5 - Multiply for 'both red'P(both red) = (3/8) x (2/7) = 6/56 = 3/28, which is about 0.107
  6. Step 6 - Do the same for 'both blue'P(both blue) = (5/8) x (4/7) = 20/56 = 5/14, which is about 0.357
  7. Step 7 - Use the complement for 'at least one red'P(at least one red) = 1 - P(both blue) = 1 - 20/56 = 36/56 = 9/14, which is about 0.643
  8. Step 8 - Check that the pieces add to 1P(exactly one red) = 2 x (3/8) x (5/7) = 30/56, and 6/56 + 30/56 + 20/56 = 56/56 = 1

If $P(A)=0.3$, the complement contains every outcome outside $A$, so $P(A^c)=1-0.3=0.7$.

The core probability rules, each with a worked number so you can see the rule fire

The core probability rules, each with a worked number so you can see the rule fire
RuleFormulaUse it whenWorked example
ComplementP(not A) = 1 - P(A)It is easier to count what you do not wantP(rain) = 0.3, so P(no rain) = 1 - 0.3 = 0.7
General additionP(A or B) = P(A) + P(B) - P(A and B)Two events can both happen at onceheart or face card: 13/52 + 12/52 - 3/52 = 22/52
Addition for disjoint eventsP(A or B) = P(A) + P(B)The two events cannot both happendice sum 7 or 11: 6/36 + 2/36 = 8/36 = 2/9
Multiplication for independent eventsP(A and B) = P(A) x P(B)The first result does not change the secondtwo coin heads: (1/2) x (1/2) = 1/4
General multiplicationP(A and B) = P(A) x P(B given A)The first result changes what is lefttwo red marbles: (3/8) x (2/7) = 3/28
Conditional probabilityP(B given A) = P(A and B) / P(A)You already know A happened(3/28) / (3/8) = 24/84 = 2/7

The step-by-step method for probability foundations

  1. Define the experiment, sample space, and event before selecting a rule.
  2. Decide whether outcomes are equally likely and whether events are disjoint, independent, or conditional.
  3. Calculate with exact fractions when possible and interpret the result as a long-run proportion, not a guarantee.

How to check your answer

Check that probabilities stay between $0$ and $1$ and that probabilities of an exhaustive disjoint partition add to $1$.

Verify that probabilities stay between 0 and 1, measures of spread are nonnegative, and numerical conclusions match the shape, units, and direction visible in the data.

A mistake that changes the mathematics

Mutually exclusive does not mean independent. If one nonempty disjoint event occurs, the other cannot occur.

Pause before continuing

Explain why the tempting step is invalid, then write the condition or definition that prevents it. This turns the error into a rule you can recognize in a new problem.

Where you will actually use this

Weather forecasts

A 70% chance of rain is a long-run statement: across many days that the forecaster labeled 70%, it rained on about 70 of every 100 of them, which is why one dry Tuesday does not prove the forecast wrong.

Medical screening

Conditional probability explains why a test that is 99% accurate can still give mostly false alarms for a rare disease, because the huge healthy group produces more false positives than the tiny sick group produces true ones.

Insurance pricing

An insurer sets your premium from the probability of a claim times its cost, so a 1-in-200 chance of a $40,000 claim costs the company about $200 a year before expenses.

How probability foundations connects to the rest of statistics

Try a transfer problem

A card is drawn from a standard deck. Find the probability it is a heart or a face card and subtract the overlap correctly.

Show the worked answer

A standard deck has 52 cards. There are 13 hearts, so P(heart) = 13/52. There are 12 face cards, the jack, queen, and king in each of the four suits, so P(face) = 12/52. The two events overlap, because the jack, queen, and king of hearts are counted in both lists: P(heart and face) = 3/52. Adding 13/52 + 12/52 = 25/52 would count those three cards twice, so use the general addition rule and subtract the overlap once: P(heart or face) = 13/52 + 12/52 - 3/52 = 22/52 = 11/26, which is about 0.423. You can confirm it by counting the winning cards directly: 13 hearts plus the 9 face cards in the other three suits gives 13 + 9 = 22 cards.

Work without copying the example. When finished, use the relevant focused calculator or formula reference to check the setup and result, then correct the first line where your reasoning changed. When the method feels reliable, move to statistics practice questions.

Questions about probability foundations

What is the difference between mutually exclusive and independent?

Mutually exclusive means the two events cannot both happen, like rolling a 2 and rolling a 5 on one die. Independent means one happening tells you nothing about the other, like a 2 on the first die and a 5 on the second. Mutually exclusive events with nonzero probability are never independent, because learning that one happened drops the other to a probability of 0.

What does a probability of 0.7 actually mean?

It is a statement about the long run, not about the next trial. If the same situation repeated 1,000 times, the event would show up on roughly 700 of them. A single trial either happens or does not, so 0.7 can never be checked against one outcome.

If a coin lands heads five times in a row, is tails more likely next?

No. The coin has no memory, so the sixth flip is still 1/2 heads and 1/2 tails. Believing otherwise is the gambler's fallacy. What is true is that a run of five heads was unlikely to be predicted in advance, since P(five heads) = (1/2)^5 = 1/32.

How do I know whether to add or multiply?

Look for the connecting word. 'Or' asks how many ways something can happen, so you add the probabilities and subtract any overlap. 'And' asks for two things to line up together, so you multiply, and you use P(B given A) for the second factor whenever the first event changed what was left.

Stuck on a problem?

Stuck on a probability foundations problem?

Paste your own question, or send the transfer problem above. You get the method, the answer, and a check you can repeat yourself.