Acomputer generates two random integers in the range 1 to 10, inclusive, where eachvalue in the range 1 to 10 is equally likely to be generated. what is the probability thatthe second randomly generated integer has a value that is (strictly) greater than the first?

0.45 or 45%

Explanation:

For the first integer:

propability=

For the second number (strictly greater), we consider every case possible

if the first integer is:

10, then 0 numbers would satisfy the criteria

9, then 1 number would satisfy the criteria

8, then 2 numbers would satisfy the criteria

7, then 3 numbers would satisfy the criteria

6, then 4 numbers would satisfy the criteria

5, then 5 numbers would satisfy the criteria

4, then 6 numbers would satisfy the criteria

3, then 7 numbers would satisfy the criteria

2, then 8 numbers would satisfy the criteria

1, then 9 numbers would satisfy the criteria

So, the probability for each case si calculated dividing the numbers that would satisfy the criteria by the total:

10, (0/10) we consider it 0

9, (1/10) = 0.1

8, (2/10) = 0.2

7, (3/10) = 0.3

6, (4/10) = 0.4

5, (5/10) = 0.5

4, (6/10) = 0.6

3, (7/10) = 0.7

2, (8/10) = 0.8

1, (9/10) = 0.9

finally, add all the probabilities for all possible cases, and divide it by the number of cases (10)

answer; i believe its a ( bond );

b and c ( ip address and port number)