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Chapter 13 Class 12 Probability
Example 6
Ex 13.1, 10 (a) Important
Ex 13.1, 12 Important
Example 11 Important
Ex 13.2, 7 Important
Ex 13.2, 11 (i)
Ex 13.2, 14 Important You are here
Example 17 Important
Example 18 Important
Example 20 Important
Example 21 Important
Ex 13.3, 2 Important
Ex 13.3, 4 Important
Ex 13.3, 8 Important
Ex 13.3, 10 Important
Ex 13.3, 12 Important
Ex 13.3, 13 (MCQ) Important
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Important Deleted for CBSE Board 2024 Exams
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Question 15 Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams
Example 23 Important
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Misc 7 Important
Misc 10 Important
Chapter 13 Class 12 Probability
Last updated at May 29, 2023 by Teachoo
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