Counting and conditional probability

Count outcomes that the question distinguishes

Before using a formula, describe one outcome. An unordered committee, an assignment of different jobs and an ordered sequence of draws distinguish different things. A probability ratio also needs an experiment that makes the counted underlying outcomes equally likely.

The multiplication principle combines successive choices: four shirt choices and three trouser choices give 4 × 3 = 12 outfits. The addition principle combines disjoint alternatives: five bus options or two train options give seven choices if every option belongs to exactly one group. Overlapping cases need their shared outcomes removed from a simple sum.

For r ordered selections without replacement from n distinct objects,

If order does not matter,

because each unordered selection occurs in r! different orders. Here 0 ≤ r ≤ n are integers and 0! = 1. Choosing two of five pupils gives ten pairs; assigning first and second places gives twenty ordered results.

Restrictions change the structure before they change the arithmetic. Five distinct books have 5! = 120 linear orders. If two specified books must be together, treat them as one block: four units have 4! orders and the pair has two internal orders, giving 48. If they must be apart, the complete complement gives 120-48 = 72.

Use event regions to organise the sample space

Suppose one pupil is chosen uniformly from a group of 25. Of these pupils, 18 take Art, 12 take Music and 8 take both. Let A mean Art and M mean Music. The intersection belongs to both event totals, so Art only has 18-8 = 10 pupils and Music only has 12-8 = 4.

The same information in a two-way table makes both the row and column reference groups visible.

The union contains 18 + 12-8 = 22 pupils, leaving three in neither event. For overlapping events,

The intersection is subtracted once because adding both totals counted it twice. A complement gives P((A∪ M)') = 1-22/25 = 3/25. In general, P(E') = 1-P(E) for an event E.

A condition changes the reference group

Conditional probability restricts attention to the stated event:

Given Music in this group, the denominator is the 12 Music pupils and the numerator is the eight who also take Art, so P(A ∣ M) = 8/12 = 2/3. Reversing the condition gives P(M ∣ A) = 8/18 = 4/9. The intersection is the same, but the reference group changes.

Independence and mutual exclusion make different claims

Events E and F are independent when P(E∩ F) = P(E)P(F). If P(F) > 0, this is equivalent to P(E ∣ F) = P(E). Mutual exclusion means their intersection is empty. Two mutually exclusive events with positive probabilities cannot be independent: their joint probability is zero while the product of their probabilities is positive.

Overlapping events may be independent or dependent. Overlap alone does not decide the product test, and the direction of a condition matters when comparing conditional probabilities.

A tree records conditional stages

A bag contains three red and two blue balls. Two balls are drawn uniformly one after the other without replacement. The second draw uses the remaining four balls, and its colour counts depend on the first draw.

Multiply along a route using its conditional branch probabilities. For example, P(R₁∩ R₂) = (3/5)(2/4) = 3/10. This is the general multiplication rule P(E∩ F) = P(E)P(F ∣ E), not an assumption that the draws are independent.

The four colour routes and their probabilities are:

They sum to one, but the four colour routes are not equally likely. Add disjoint routes for an event such as exactly one red: the red-blue and blue-red probabilities add to 3/5. Multiplying down a path and adding alternative paths perform different jobs.