If 1 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 2, 3, or 4. Similarly, if 2 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 1, 3, or 4. The same holds true for the remaining numbers too.
Thus, the sample space of this experiment is given by
S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

Given: There are 4 slips in the box and mixed thoroughly. 

To Find: Describe the given events . 

Explanation: Here, Four slips of paper 1, 2, 3, and 4 are put in a box. 

If Two slips are drawn from it one after the other without replacement. 

Then The sample space for the experiment is: 

S = {(1, 2)(1, 3), (1, 4)(2, 1)(2, 3)(2, 4)(3, 1)(3, 2)(3, 4)(4, 1)(4, 2)(4, 3)} 

(i) A = number on the first slip is larger than the one on the second slip, 

So, The sample space for A is: 

{(2, 1)(3, 1)(3, 2)(4, 1)(4, 2)(4, 3)} 

(ii) B = number on the second slip is greater than 2 

So, The sample space for B is 

{(1, 3)(2, 3)(1, 4)(2, 4)(3, 4)(4, 3)} 

(iii) C = sum of the numbers on the two slips in 6 or 7 

The sample space for C is 

{(2, 4)(3, 4)(4, 2)(4, 3)} 

(iv) D = number on the second slip is two times the number on the first slip The sample space if: 

{(1, 2)(2, 4)} 

Now, 

We can see, AՌD = Փ.

Therefore, A and D are mutually exclusive events 

Hence, A and D are mutually exclusive events.

In number `1` is drawn in first slip, then, possible outcomes will be,
`(1,2),(1,3),(1,4)`

In number `2` is drawn in first slip, then, possible outcomes will be,
`(2,1),(2,3),(2,4)`

In number `3` is drawn in first slip, then, possible outcomes will be,
`(3,1),(3,2),(3,4)`

In number `4` is drawn in first slip, then, possible outcomes will be,
`(4,1),(4,2),(4,3)`

So, sample space will be,
`(1,2),(1,3),(1,4),“(2,1),(2,3),(2,4),“(3,1),(3,2),(3,4)`,`(4,1),(4,2),(4,3)`