Properties of conditional probability

For a fixed F with P(F) ≠ 0, conditional probabilities given F behave like ordinary probabilities on the smaller sample space F.

Property 1: P(S|F) = P(F|F) = 1.

Property 2: for a union under the condition F, add P(A|F) and P(B|F) and subtract P((A ∩ B)|F). If A and B are disjoint, the subtracted term is zero, so P((A ∪ B)|F) is just the sum.

Property 3: P(E′|F) = 1 − P(E|F). The complement is taken of the event, never of the condition: P(E|F′) is not 1 − P(E|F).

Also 0 ≤ P(E|F) ≤ 1.

A school has 1000 students, 430 of them girls, and 10% of the girls are in class XII. For a student picked at random, P(class XII | girl) = 0.043 ÷ 0.43 = 0.1.

A die is thrown three times. Let A be 'the third throw shows 4' and B be 'the first two throws show 6, then 5'. B has 6 of the 216 outcomes, and A ∩ B is the single outcome (6, 5, 4), so P(A|B) = 1/6.

A die is thrown twice and the sum is 6. That leaves five outcomes: (1, 5), (5, 1), (2, 4), (4, 2) and (3, 3). Two contain a 4, so P(4 appears at least once | sum 6) = 2/5.

Outcomes need not be equally likely. Toss a coin; on a head toss it again, on a tail throw a die. HH and HT have probability 1/4 each, and (T, 1), …, (T, 6) have 1/12 each.

In that experiment, with F: at least one tail, P(F) = 1/4 + 6/12 = 3/4. E: the die shows more than 4 has P(E ∩ F) = 2/12 = 1/6. So P(E|F) = (1/6) ÷ (3/4) = 2/9. Counting outcomes (2 of 7) would be wrong here.