Given: A dice is thrown twice. And each time number appearing on it is recorded. 

To Find: Describe the given events. 

Explanation: when the dice is thrown twice then the number of sample spaces are 6 2 = 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 are: 

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)} 

Therefore, 

(AՍB) = {(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)} 

(AՌB) = {Փ} 

(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)} 

(AՌC) = {(1, 1)(1, 3)(3, 1)} 

Hence, (AՌB) = Փ and (AՌC)≠Փ, A and B are mutually exclusive, but A and C are not.

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,

(AՍB) = {(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)}

(AՌB) = {Փ}

(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)}

(AՌC) = {(1, 1), (1, 3), (3, 1)}

∴ (AՌB) = Փ and (AՌC) ≠ Փ, A and B are mutually exclusive, but A and C are not.