Select numbers a and b between 0 and 1

independently and at random, and let c be their sum. Let A, B, and C be the results when a, b, and c, respectively, are rounded to the nearest integer. What is the probability that A + B = C?

To solve this problem, we need to consider all the possible combinations of values for integers A, B, and C that satisfy the condition A + B = C. Let's break it down step by step:

Step 1: Define the range of possible values for a and b
We are given that a and b are random numbers between 0 and 1. Since they are inclusive, the range of possible values for a and b is [0, 1].

Step 2: Calculate the range of possible values for c
Since c is the sum of a and b, the range of possible values for c is the sum of all possible values of a and b. In this case, it is the range [0 + 0, 1 + 1] which is [0, 2].

Step 3: Define the rounding conditions
Let's consider the possible rounding scenarios:
- If the decimal part of a is less than 0.5, it rounds down to 0; otherwise, it rounds up to 1.
- Similarly, if the decimal part of b is less than 0.5, it rounds down to 0; otherwise, it rounds up to 1.
- For c, the rounding will be similar to the rounding of the sum of a and b.

Step 4: Calculate the range of possible values for A, B, and C
Considering the rounding conditions, the possible values for A, B, and C are integers that range from 0 to 2.

Step 5: Determine the combinations that satisfy A + B = C
To find the combinations that satisfy A + B = C, we can systematically evaluate all the possible combinations of A, B, and C. There are only 9 possible combinations to check:

(A=0, B=0, C=0)
(A=0, B=0, C=1)
(A=0, B=1, C=1)
(A=0, B=1, C=2)
(A=1, B=0, C=1)
(A=1, B=0, C=2)
(A=1, B=1, C=2)
(A=1, B=1, C=2)
(A=2, B=0, C=2)

Step 6: Calculate the probability
To calculate the probability, we need to determine how many of these combinations satisfy A + B = C. In this case, there are 5 combinations out of the total 9 that satisfy the condition.

So, the probability that A + B = C is 5/9.

If a and b are decimals between 0 and 1, then

A and B could be either 0 or 1
The prob of a rounded up to 1 is 1/2 and
the prob of a becoming 0 is 1/2

forming a table:
A B -----C
0 0 ----- 0 e.g. .2+.1 = .3 --> 0
0 0 ----- 1 e.g. .4 + .4 = .8 ---> 1 ...... FALSE
0 1 ----- 1 e.g. .4 + .9 = 1.3 ---> 1
1 0 ----- 1 e.g. .9+.4 = 1.3 --> 1
1 1 ----- 1 e.g. .6 + .6 = 1.2 ---> 1 ... FALSE
1 1 ----- 2 e.g. .9+.9 = 1.8 ---> 2

I see 4 cases out of the 6 where A+B = C
so prob = 4/6 = 2/3