Get live Maths 1-on-1 Classs - Class 6 to 12

Examples

Example 1

Example 2

Example 3

Example 4

Example 5 Important

Example 6

Example 7 Important

Example 8

Example 9 Important

Example 10

Example 11 Important You are here

Example 12 Important

Example 13 Important

Example 14 Important

Example 15 Important

Example 16

Example 17 Important

Example 18 Important

Example 19 Important

Example 20 Important

Example 21 Important

Example 22

Example 23

Example 24 Important

Example 25 Important

Example 26 Important

Example 27

Example 28 Important Deleted for CBSE Board 2023 Exams

Example 29 Important

Example 30 Important Deleted for CBSE Board 2023 Exams

Example 31 Important Deleted for CBSE Board 2023 Exams

Example 32 Important Deleted for CBSE Board 2023 Exams

Example 33 Important

Example 34 Deleted for CBSE Board 2023 Exams

Example 35

Example 36 Important

Example 37 Important

Chapter 13 Class 12 Probability

Serial order wise

Last updated at March 16, 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