Conditional probability
If you are told that the card drawn from a pack is a face card, the possibilities shrink from 52 cards to 12. The chance of a king is now found inside these 12 cards, not inside 52.
The probability of A given that B has already occurred is P(A|B) = P(A∩B)/P(B), where P(B) is not zero. The sample space shrinks to B, and we count only the favourable outcomes of A within B.
Formula: P(A|B) = P(A ∩ B) / P(B), P(B) ≠ 0
Rule 1: Use the formula — Divide the probability of both by the probability of the given event.
- Easy: If P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.3, then P(A|B) = 0.3/0.5 = 0.6 and P(B|A) = 0.3/0.6 = 0.5.
- Medium: P(king | face card) = 4/12 = 1/3.
- Exam-type: Two coins are tossed: P(both heads | at least one head) = (1/4)/(3/4) = 1/3.
Rule 2: Shrink the sample space — Count outcomes inside the given event.
- Easy: A die is thrown: P(2 | even) = 1/3.
- Medium: Two dice: P(sum 8 | first die is 3) = 1/6, since only (3, 5) works.
- Exam-type: Two dice: P(sum 8 | first die even) = 3/18 = 1/6, with (2, 6), (4, 4), (6, 2).
In P(A|B) the denominator is P(B), the event that is given. P(A|B) and P(B|A) are generally different.
Dividing by P(A) instead of P(B).
Divide by the given event.
Practice questions
- 1If P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.3, find P(A|B).
- 2With the same values, find P(B|A).
- 3A die is thrown. Find P(2 | the number is even).
- 4Two coins are tossed. Find P(both heads | at least one head).
- 5Two dice are thrown. Find P(sum 8 | first die shows 3).
- 6Two dice are thrown. Find P(sum 8 | first die is even).
- 7A card is drawn from a pack. Find P(king | face card).
- 8If P(A|B) = 0.4 and P(B) = 0.5, find P(A ∩ B).
Show answers (try them yourself first)
- 10.3/0.5 = 0.6
- 20.3/0.6 = 0.5
- 31/3
- 41/3
- 51/6
- 63/18 = 1/6
- 74/12 = 1/3
- 80.4 × 0.5 = 0.2
Properties of conditional probability
Inside the shrunken sample space all the usual rules of probability still hold. Conditional probabilities behave like ordinary probabilities, only the universe is smaller.
For a given event F with P(F) not zero: P(S|F) = 1; P(A′|F) = 1 − P(A|F); and P(A ∪ B | F) = P(A|F) + P(B|F) − P(A ∩ B | F).
Formula: P(A′|F) = 1 − P(A|F), P(A ∪ B | F) = P(A|F) + P(B|F) − P(A ∩ B | F)
Rule 1: Same rules as ordinary probability — Keep the same given event F.
- Easy: If P(A|B) = 0.3, then P(A′|B) = 0.7.
- Medium: If P(A|F) = 0.4, P(B|F) = 0.5 and P(A ∩ B | F) = 0.2, then P(A ∪ B | F) = 0.7.
- Exam-type: Two dice: P(sum 7 | at least one die shows 6) = 2/11, with (1, 6) and (6, 1) among the 11 outcomes.
Rule 2: Use inside a smaller universe — Count favourable and total outcomes inside F.
- Easy: P(heart | red card) = 13/26 = 1/2.
- Medium: P(ace | black card) = 2/26 = 1/13.
- Exam-type: If P(A|F) = 0.6 and P(B|F) = 0.3, and A and B cannot occur together, then P(A ∪ B | F) = 0.9.
Only the first place (before the bar) changes. The given event after the bar stays the same.
Writing P(A′|B) = 1 − P(A|B′), which changes the given event.
The bar condition stays, the rest follows the usual rules.
Practice questions
- 1If P(A|B) = 0.3, find P(A′|B).
- 2If P(A|F) = 0.4, P(B|F) = 0.5 and P(A ∩ B | F) = 0.2, find P(A ∪ B | F).
- 3What is P(S|E) for any event E with P(E) not zero?
- 4Two dice are thrown. Find P(sum 7 | at least one die shows 6).
- 5If P(A|F) = 2/5, find P(A′|F).
- 6A card is drawn from a pack. Find P(heart | the card is red).
- 7A card is drawn from a pack. Find P(ace | the card is black).
- 8If P(A|F) = 0.6, P(B|F) = 0.3 and A, B cannot both occur, find P(A ∪ B | F).
Show answers (try them yourself first)
- 11 − 0.3 = 0.7
- 20.4 + 0.5 − 0.2 = 0.7
- 31
- 42/11
- 53/5
- 613/26 = 1/2
- 72/26 = 1/13
- 80.6 + 0.3 = 0.9