What is conditional probability?
Sometimes we get extra news before we guess. The news is an event B that has already happened. Then every outcome outside B is impossible now. B becomes the new sample space (the list of all possible outcomes).
The chance of A given B is written P(A|B). Read the bar "|" as "given".
P(A|B) = P(A ∩ B) ÷ P(B), where P(B) ≠ 0.
With equally likely outcomes you can simply count: P(A|B) = n(A ∩ B) ÷ n(B).
Properties of conditional probability
- 0 ≤ P(A|B) ≤ 1, like any probability.
- P(S|B) = P(B|B) = 1: inside the new world, B is sure.
- P(A′|B) = 1 − P(A|B) (A′ means "not A").
- For any A and C: P(A ∪ C | B) = P(A|B) + P(C|B) − P(A ∩ C | B).
- P(A|B) and P(B|A) are usually different. The bottom of the fraction changes.
Multiplication theorem of probability
Multiply both sides of the definition by P(B). You get the multiplication rule:
P(A ∩ B) = P(B) · P(A|B) = P(A) · P(B|A)
Use it when things happen one after another, like drawing two cards without putting the first one back. First chance × chance of the second, given the first.
For three events: P(A ∩ B ∩ C) = P(A) · P(B|A) · P(C|A ∩ B).
Independent events
A and B are independent if knowing one tells you nothing about the other: P(A|B) = P(A) (and P(B|A) = P(B)).
Put this in the multiplication rule and you get the test: P(A ∩ B) = P(A) · P(B).
- Two different coins, two different dice, draws with replacement → independent.
- Draws without replacement → usually dependent.
- If A and B are independent, so are A and B′, A′ and B, A′ and B′.
- Three events are independent only if every pair works and P(A ∩ B ∩ C) = P(A)P(B)P(C).
Independent is not the same as mutually exclusive
Mutually exclusive means A and B cannot happen together: P(A ∩ B) = 0. Independent means they do not affect each other. If both have a non-zero chance, they cannot be both: if A happens, a mutually exclusive B becomes impossible, so the news clearly changed B.
Try it at home
Take a pack of 10 cards: 1 to 10. Guess the chance a card is more than 6 (it is 4/10). Now a friend peeks and says "it is even". Remove the odd cards. Only 2, 4, 6, 8, 10 are left, and 8 and 10 are more than 6: 2/5. Same number 0.4! So "more than 6" and "even" are independent here. Try "more than 5" and see the chance change.
Key formulas and definitions
- P(A|B) = P(A ∩ B) / P(B), P(B) ≠ 0
- P(A|B) = n(A ∩ B) / n(B) (equally likely outcomes)
- P(A′|B) = 1 − P(A|B)
- P(A ∩ B) = P(B)·P(A|B) = P(A)·P(B|A)
- P(A ∩ B ∩ C) = P(A)·P(B|A)·P(C|A ∩ B)
- Independent: P(A ∩ B) = P(A)·P(B)
Worked examples
1. P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.2. Find P(A|B) and P(B|A).
Step 1: P(A|B) = P(A ∩ B)/P(B) = 0.2/0.5 = 0.4. Step 2: P(B|A) = P(A ∩ B)/P(A) = 0.2/0.6 = 1/3. Note: the two answers are different.
2. Two fair dice are thrown. Find the chance that the sum is 8, given that the first die shows an even number.
Step 1: B = first die even → 3 × 6 = 18 outcomes. Step 2: A ∩ B = (2,6), (4,4), (6,2) → 3 outcomes. Step 3: P(A|B) = 3/18 = 1/6.
3. A family has two children. Given that at least one is a girl, what is the chance both are girls?
Step 1: All cases: GG, GB, BG, BB (equally likely). Step 2: B = at least one girl → GG, GB, BG (3 cases). Step 3: A ∩ B = GG (1 case). Step 4: P = 1/3 (not 1/2!).
4. A box has 5 red and 3 blue pens. Two pens are taken out one after another without putting back. Find the chance both are red.
Step 1: P(first red) = 5/8. Step 2: Now 4 red out of 7: P(second red | first red) = 4/7. Step 3: Multiplication rule: 5/8 × 4/7 = 20/56 = 5/14.
5. Three cards are drawn one by one without replacement from a pack of 52. Find the chance all three are kings.
Step 1: P(K1) = 4/52. Step 2: P(K2 | K1) = 3/51. Step 3: P(K3 | K1 ∩ K2) = 2/50. Step 4: Multiply: 4/52 × 3/51 × 2/50 = 24/132600 = 1/5525.
6. A die is thrown. E = "multiple of 3", F = "even". Are E and F independent?
Step 1: E = {3, 6}, P(E) = 1/3. F = {2, 4, 6}, P(F) = 1/2. Step 2: E ∩ F = {6}, P(E ∩ F) = 1/6. Step 3: P(E)·P(F) = 1/3 × 1/2 = 1/6 = P(E ∩ F). So yes, they are independent.
7. A solves a problem with chance 1/2 and B with chance 1/3, working alone (independent). Find the chance the problem gets solved.
Step 1: P(A fails) = 1/2, P(B fails) = 2/3. Step 2: Both fail (independent): 1/2 × 2/3 = 1/3. Step 3: Solved = 1 − 1/3 = 2/3.
8. P(A) = 0.3, P(B) = 0.4, and A, B are independent. Find P(A ∪ B) and P(A′ ∩ B′).
Step 1: P(A ∩ B) = 0.3 × 0.4 = 0.12. Step 2: P(A ∪ B) = 0.3 + 0.4 − 0.12 = 0.58. Step 3: P(A′ ∩ B′) = 1 − P(A ∪ B) = 0.42 (check: 0.7 × 0.6 = 0.42 ✓).
Common mistakes
- Mixing P(A|B) with P(B|A). The event after the bar is the one you already know; it goes in the denominator.
- Dividing by the full sample space after being told B. Given B, count only inside B.
- Thinking "independent" means "cannot happen together". That is mutually exclusive, a different idea.
- Using P(A)·P(B) for draws without replacement. The second draw depends on the first; use P(A)·P(B|A).