Conditional probability by counting

When all outcomes are equally likely, you can find P(E|F) by counting, without the formula:

P(E|F) = (number of outcomes in E ∩ F) ÷ (number of outcomes in F)

In words: throw away every outcome that is not in F. What is left is the new sample space. Count how many of those are in E.

Example: a die is thrown and the number is even. What is the chance it is more than 3? The new sample space is {2, 4, 6}. Of these, 4 and 6 are more than 3. So P = 2/3 (without the information it would be 1/2).

Example: three coins are tossed, and at least two show heads. What is the chance all three are heads? Outcomes with at least two heads: HHT, HTH, THH, HHH. Only one is all heads, so P = 1/4.

Read the condition carefully. 'The first child is a boy' and 'at least one child is a boy' leave different sample spaces, so they give different answers.

List the outcomes when the numbers are small: it avoids mistakes and shows the examiner your reasoning.

If the outcomes are not equally likely (for example, a tree whose branches have different chances), counting does not work. Use the formula with the actual probabilities.