Multiplication theorem
To draw two red balls one after the other from a bag, the first draw must be red and then the second draw must also be red. The chance of the second depends on what is left in the bag, so the two chances are multiplied.
The probability that both A and B occur is P(A ∩ B) = P(A) · P(B|A). For three events in a row, P(A ∩ B ∩ C) = P(A) · P(B|A) · P(C|A ∩ B). Without replacement the second chance changes; with replacement it does not.
Formula: P(A ∩ B) = P(A) · P(B|A) = P(B) · P(A|B)
Rule 1: Without replacement — Reduce the total and the favourable count after each draw.
- Easy: Two aces from a pack without replacement: 4/52 × 3/51 = 1/221.
- Medium: A bag has 5 red and 3 blue balls. Two red without replacement: 5/8 × 4/7 = 5/14.
- Exam-type: Three kings without replacement: 4/52 × 3/51 × 2/50 = 1/5525.
Rule 2: With replacement and orders — The total stays the same.
- Easy: With replacement, two red from the same bag: (5/8)² = 25/64.
- Medium: Red then blue without replacement: 5/8 × 3/7 = 15/56.
- Exam-type: One red and one blue in any order, without replacement: 2 × 15/56 = 15/28.
Write the first probability, then the second with the changed bag. Multiply along the path.
Using the same total for the second draw when the first ball is not put back.
Chance, then chance in the changed bag, then multiply.
Practice questions
- 1If P(A) = 0.5 and P(B|A) = 0.4, find P(A ∩ B).
- 2Two cards are drawn without replacement. Find the probability that both are aces.
- 3A bag has 5 red and 3 blue balls. Two are drawn without replacement. Find the probability that both are red.
- 4The same bag, with replacement: find the probability that both are red.
- 5The same bag, without replacement: find the probability of red first and blue second.
- 6Three cards are drawn without replacement. Find the probability that all are kings.
- 7An urn has 4 white and 6 black balls. Two are drawn without replacement. Find the probability that both are white.
- 8In the bag with 5 red and 3 blue balls, find the probability of one red and one blue in any order (without replacement).
Show answers (try them yourself first)
- 10.5 × 0.4 = 0.2
- 24/52 × 3/51 = 1/221
- 35/8 × 4/7 = 5/14
- 4(5/8)² = 25/64
- 55/8 × 3/7 = 15/56
- 64/52 × 3/51 × 2/50 = 1/5525
- 74/10 × 3/9 = 2/15
- 82 × 15/56 = 15/28
Independent events
A coin toss in Pune and a coin toss in Delhi do not affect each other. The result of one tells nothing about the other. Such events are independent.
Events A and B are independent if P(A ∩ B) = P(A) · P(B), which means P(B|A) = P(B). If A and B are independent, then so are A and B′, A′ and B, and A′ and B′. Mutually exclusive events with non-zero probabilities are never independent.
Formula: P(A ∩ B) = P(A) · P(B), P(A ∪ B) = 1 − P(A′) · P(B′) for independent A and B
Rule 1: Test the product rule — Compare P(A ∩ B) with P(A) · P(B).
- Easy: Two coins: P(head and head) = ½ × ½ = ¼.
- Medium: A die: A = {2, 4, 6} and B = {1, 2, 3, 4}. P(A ∩ B) = 1/3 = ½ × 2/3, so A and B are independent.
- Exam-type: If P(A) = 0.3 and P(B) = 0.4 are independent, then P(A ∪ B) = 0.3 + 0.4 − 0.12 = 0.58.
Rule 2: Repeated trials — Multiply the probabilities of each trial.
- Easy: A die: A = {1, 2} and B = {3, 4} cannot occur together, so P(A ∩ B) = 0, but P(A) P(B) = 1/9. They are not independent.
- Medium: A coin and a die: P(head and 6) = ½ × 1/6 = 1/12.
- Exam-type: A shooter hits with probability 0.6 each time. P(no hit in 3 shots) = 0.4³ = 0.064, so P(at least one hit) = 0.936.
Independent does not mean mutually exclusive. Exclusive events have zero overlap, independent events keep the same chance.
Saying that two events are independent because they cannot occur together.
Independent: multiply. Exclusive: add.
Practice questions
- 1If P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2, are A and B independent?
- 2If A and B are independent with P(A) = 0.3 and P(B) = 0.4, find P(A ∩ B).
- 3For the same events, find P(A ∪ B).
- 4A die is thrown. Are A = {2, 4, 6} and B = {1, 2, 3, 4} independent?
- 5A die is thrown. Are A = {1, 2} and B = {3, 4} independent?
- 6A coin and a die are thrown. Find P(head and 6).
- 7A shooter hits with probability 0.6 each time, independently. Find the probability that all 3 shots miss.
- 8For the same shooter, find the probability that at least one of 3 shots hits.
Show answers (try them yourself first)
- 1Yes, since 0.5 × 0.4 = 0.2
- 20.3 × 0.4 = 0.12
- 30.3 + 0.4 − 0.12 = 0.58
- 4Yes, since P(A ∩ B) = 1/3 = ½ × 2/3
- 5No, since P(A ∩ B) = 0 but P(A) P(B) = 1/9
- 6½ × 1/6 = 1/12
- 70.4 × 0.4 × 0.4 = 0.064
- 81 − 0.064 = 0.936