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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 You are here
Ex 13.2, 7 Important
Ex 13.2, 11 (i)
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Example 17 Important
Example 18 Important
Example 20 Important
Example 21 Important
Ex 13.3, 2 Important
Ex 13.3, 4 Important
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Ex 13.3, 10 Important
Ex 13.3, 12 Important
Ex 13.3, 13 (MCQ) Important
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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
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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
Example 11 An unbiased die is thrown twice. Let the event A be ‘odd number on the first throw’ and B the event ‘odd number on the second throw’. Check the independence of the events A and B. Two events A & B are independent if P(A ∩ B) = P(A) . P(B) An unbiased die is thrown twice S = Let us define two events as A : Odd number on the First throw B : Odd number on the Second throw vA : Odd number on First throw A : { (1, 1), (1, 2), ………., (1, 6) (3, 1), (3, 2), ………., (3, 6) (5, 1), (5, 2), ………., (5, 6) } P(A) = 𝟏𝟖/𝟑𝟔 = 𝟏/𝟐 B : Odd number on Second throw B : { (1, 1), (2, 1), ………., (6, 1) (1, 3), (2, 3), ………., (6, 3) (1, 5), (2, 5), ………., (6, 5) } P(A) = 𝟏𝟖/𝟑𝟔 = 𝟏/𝟐 B : Odd number on Second throw B : { (1, 1), (2, 1), ………., (6, 1) (1, 3), (2, 3), ………., (6, 3) (1, 5), (2, 5), ………., (6, 5) } P(A) = 𝟏𝟖/𝟑𝟔 = 𝟏/𝟐 B : Odd number on Second throw B : { (1, 1), (2, 1), ………., (6, 1) (1, 3), (2, 3), ………., (6, 3) (1, 5), (2, 5), ………., (6, 5) } P(A) = 𝟏𝟖/𝟑𝟔 = 𝟏/𝟐 A ∩ B = Odd number on the First & Second throw = { (1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)} So, P(A ∩ B) = 9/36 = 1/4 Now, P(A) . P(B) = 1/2 × 1/2 = 1/4 Since P(A ∩ B) = P(A) . P(B), Therefoare, A and B are independent events