# Ex 13.2, 14 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 15, 2020 by Teachoo

Last updated at Feb. 15, 2020 by Teachoo

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.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

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Chapter 13 Class 12 Probability (Term 2)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.