Ex 13.2, 14 - Chapter 13 Class 12 Probability (Term 2)
Last updated at Feb. 15, 2020 by Teachoo
Independent events
Ex 13.2, 6
Ex 13.2, 10 Important
Ex 13.2, 5
Example 10
Example 11 Important
Example 12 Important
Ex 13.2, 15 (i)
Ex 13.2, 8
Ex 13.2, 7 Important
Ex 13.2, 11 (i)
Ex 13.2, 4
Ex 13.2, 13 Important
Ex 13.2, 14 Important You are here
Ex 13.2, 18 (MCQ) Important
Example 13 Important
Example 14 Important
Independent events
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.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. Probability that exactly one of them solves the problem = Probability that only A solvesa + Probability that only B solves Therefore, P(exactly one of them solves) = P(A alone solves) + P(B alone solves) = 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