    1. Chapter 13 Class 12 Probability
2. Serial order wise
3. Ex 13.2

Transcript

Ex 13.2, 14 Probability of solving specific problem independently by A and B are 1﷮2﷯ and 1﷮3﷯ respectively. If both try to solve the problem independently, find the Probability that (i) the problem is solved. Two events A & B are independent if P(A ∩ B) = P(A) . P(B) Given, P(A) = 1﷮2﷯ & P(B) = 1﷮3﷯ Probability that the problem is solved = Probability that A solves the problem or B solves the Problem = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Since A & B are independent, P(A ∩ B) = P(A) . P(B) = 1﷮2﷯ × 1﷮3﷯ = 1﷮6﷯ Now, P(Problem is solved) = P(A) + P(B) – P(A ∩ B) = 1﷮2﷯ + 1﷮3﷯ – 1﷮6﷯ = 3﷮6﷯ + 2﷮6﷯ – 1﷮6﷯ = 4﷮6﷯ = 𝟐﷮𝟑﷯ Ex 13.2, 14 Probability of solving specific problem independently by A and B are 1﷮2﷯ and 1﷮3﷯ respectively. If both try to solve the problem independently, find the Probability that (ii) exactly one of them solves the problem. Two events A & B are independent if P(A ∩ B) = P(A) . P(B) Given, P(A) = 1﷮2﷯ & P(B) = 1﷮3﷯ Probability that exactly one of them solves the problem = Probability that only A solves + Probability that only B solves P(exactly one of them solves) = P(A alone) + P(B alone) = P(A ∩ B’) + P(B ∩ A’) = (P(A) – P(A ∩ B)) + (P(B) – P(B ∩ A)) = P(A) + P(B) – 2P(A ∩ B) = 1﷮2﷯ + 1﷮3﷯ – 2 × 1﷮6﷯ = 1﷮2﷯ + 1﷮3﷯ – 1﷮3﷯ = 1﷮2﷯

Ex 13.2 