Conditional probability and Bayes' theorem: Class 12 practice
P(A|B) = P(A ∩ B)/P(B). Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Basic probability
- Fractions
The idea behind conditional probability and bayes' theorem
P(A|B) = P(A ∩ B)/P(B). Bayes' theorem reverses a condition: P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ) / Σ P(Eⱼ)P(A|Eⱼ).
Step-by-step method
- Name the cases E₁, E₂ and the observed event A.
- Write P(Eᵢ) and P(A|Eᵢ).
- Total probability: P(A) = Σ P(Eᵢ)P(A|Eᵢ).
- Divide the wanted case by P(A).
Worked example 1 (Basic)
Bag I has 2 red and 3 black balls; Bag II has 5 red and 4 black. A bag is chosen at random and one ball drawn. Find the probability it is red.
- P(red) = ½ × 2/5 + ½ × 5/9
- = 43/90
- Answer: 43/90
Worked example 2 (Standard)
Bag I has 4 red and 1 black balls; Bag II has 3 red and 4 black. A bag is chosen at random and a red ball is drawn. Find the probability it came from Bag I.
- P(E₁)P(R|E₁) = ½ × 4/5
- P(R) = ½ × 4/5 + ½ × 3/7
- P(E₁|R) = 28/43
- Answer: 28/43
Worked example 3 (Challenge)
Bag I has 2 red and 1 black balls; Bag II has 4 red and 3 black. A bag is chosen at random and a red ball is drawn. Find the probability it came from Bag I.
- P(E₁)P(R|E₁) = ½ × 2/3
- P(R) = ½ × 2/3 + ½ × 4/7
- P(E₁|R) = 7/13
- Answer: 7/13
Common mistakes and misconceptions
- Confusing P(A|E) with P(E|A)
- Forgetting the prior P(Eᵢ)
Practice questions
- (Basic) Bag I has 2 red and 3 black balls; Bag II has 1 red and 1 black. A bag is chosen at random and one ball drawn. Find the probability it is red.
- (Basic) Bag I has 3 red and 4 black balls; Bag II has 4 red and 3 black. A bag is chosen at random and one ball drawn. Find the probability it is red.
- (Standard) Bag I has 1 red and 3 black balls; Bag II has 2 red and 3 black. A bag is chosen at random and a red ball is drawn. Find the probability it came from Bag I.
- (Standard) Bag I has 1 red and 4 black balls; Bag II has 3 red and 3 black. A bag is chosen at random and a red ball is drawn. Find the probability it came from Bag I.
- (Challenge) Bag I has 4 red and 4 black balls; Bag II has 1 red and 3 black. A bag is chosen at random and a red ball is drawn. Find the probability it came from Bag I.
- (Challenge) Bag I has 1 red and 1 black balls; Bag II has 3 red and 1 black. A bag is chosen at random and a red ball is drawn. Find the probability it came from Bag I.
Answer key with full working
- Q1: P(red) = ½ × 2/5 + ½ × 1/2 → = 9/20 → Answer: 9/20
- Q2: P(red) = ½ × 3/7 + ½ × 4/7 → = 1/2 → Answer: 1/2
- Q3: P(E₁)P(R|E₁) = ½ × 1/4 → P(R) = ½ × 1/4 + ½ × 2/5 → P(E₁|R) = 5/13 → Answer: 5/13
- Q4: P(E₁)P(R|E₁) = ½ × 1/5 → P(R) = ½ × 1/5 + ½ × 3/6 → P(E₁|R) = 2/7 → Answer: 2/7
- Q5: P(E₁)P(R|E₁) = ½ × 4/8 → P(R) = ½ × 4/8 + ½ × 1/4 → P(E₁|R) = 2/3 → Answer: 2/3
- Q6: P(E₁)P(R|E₁) = ½ × 1/2 → P(R) = ½ × 1/2 + ½ × 3/4 → P(E₁|R) = 2/5 → Answer: 2/5
Syllabus note
This skill is listed under "Probability" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is conditional probability and bayes' theorem in?
"Probability", Class 12 Maths (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.