Probability – important questions for Boards 2027

CBSE Class 12 Maths

1. VSA · 1 mark

   Given P(A|F)=0.6, P(B|F)=0.5 and P(A∩B|F)=0.2, evaluate P((A∪B)|F).

   Answer

   P((A ∪ B)|F) = P(A|F) + P(B|F) − P((A ∩ B)|F) = 0.6 + 0.5 − 0.2 ∴ P((A ∪ B)|F) = 0.9

2. SA · 1.5 marks

   Given P(A)=0.6, P(B)=0.4 and P(A∩B)=0.2, find P(A∪B).

   Answer

   P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.4 − 0.2 ∴ P(A ∪ B) = 0.8

3. SA · 2 marks

   Let P(F)=1/3 for an event F. Evaluate P(S|F) and P(F|F).

   Answer

   P(E|F) = P(E ∩ F) / P(F) P(S|F) = P(S ∩ F) / P(F) = P(F) / P(F) ∴ P(S|F) = 1 P(F|F) = P(F ∩ F) / P(F) = P(F) / P(F) ∴ P(F|F) = 1

4. MA · 3 marks

   A machine is correctly set up with probability 0.8 and incorrectly set up with probability 0.2. The probability that one item is acceptable is 0.9 for a correct setup and 0.4 for an incorrect setup. Assuming the two items’ outcomes are independent given the setup, find the probability that the setup was correct given that two items are acceptable.

   Answer

   Let C be the event of a correct setup and A the event that both items are acceptable. P(C) = 0.8, P(C′) = 0.2 P(A|C) = 0.9² = 0.81 P(A|C′) = 0.4² = 0.16 P(C|A) = P(C)P(A|C) / [P(C)P(A|C) + P(C′)P(A|C′)] = (0.8 × 0.81) / [(0.8 × 0.81) + (0.2 × 0.16)] = 0.648 / 0.68 ∴ P(C|A) = 81/85

5. LA · 4 marks

   The probability of a strike is 0.65. The probabilities that a construction job is completed on time during a strike and when there is no strike are 0.32 and 0.80, respectively. Using the theorem of total probability, calculate the probability that the job is completed on time.

   Answer

   Let A be the event that the job is completed on time, and B the event that there is a strike. P(B) = 0.65, P(B′) = 1 − 0.65 = 0.35 P(A|B) = 0.32, P(A|B′) = 0.80 P(A) = P(B)P(A|B) + P(B′)P(A|B′) = 0.65 × 0.32 + 0.35 × 0.80 = 0.208 + 0.28 ∴ P(A) = 0.488

6. LA · 5 marks

   Let P(E|F)=3/8, where P(F)>0. Prove that P(E′|F)=1−P(E|F), and evaluate P(E′|F).

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7. MCQ · 1 mark

   If A and B are disjoint events and P(A|F)=0.3, P(B|F)=0.4, which is P((A∪B)|F)? (A) 0.12 (B) 0.3 (C) 0.4 (D) 0.7

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8. SA · 2 marks

   Given P(F)=0.4, P(A∩F)=0.12, P(B∩F)=0.08 and P(A∩B∩F)=0.04, evaluate P((A∪B)|F).

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9. MA · 3 marks

   Each entry of a 2 × 2 determinant is chosen independently and is equally likely to be 0 or 1. Find the probability that the determinant is positive.

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10. LA · 4 marks

    A cause is selected from hypotheses E₁, E₂, E₃, which form a partition of S. An event A is then observed. Use Bayes’ theorem to obtain a formula for P(E₂|A), and state what the partition condition implies about how many of E₁, E₂, E₃ occur.

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