Question -
Answer -
Given: A dice is thrown twice. And each time number appearing on it is recorded.
When the dice is thrown twice then the number of sample spaces are 62 = 36
Now,
The possibility both odd numbers are:
A = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
Since, possibility of both even numbers is:
B = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
And, possible outcome of sum of the numbers is less than 6.
C = {(1, 1)(1, 2)(1, 3)(1, 4)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(4, 1)}
Hence,
(A╒НB) = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5) (2, 2)(2, 4)(2, 6)(4, 2)(4, 4)(4, 6)(6, 2)(6, 4)(6, 6)}
(A╒МB) = {╒У}
(AUC) = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5) (1, 2)(1, 4)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(4, 1)}
(A╒МC) = {(1, 1), (1, 3), (3, 1)}
тИ┤ (A╒МB) = ╒У and (A╒МC) тЙа ╒У, A and B are mutually exclusive, but A and C are not.