Using it's concept, it is found that there is a 0.64 = 64% probability that the sum of two randomly chosen numbers from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is no greater than 10.
A probability is the <u>number of desired outcomes divided by the number of total outcomes</u>.
In this problem:
- Two numbers that can be repeated chosen from a set of 10 numbers, thus, in total, there are
outcomes.
For a sum <u>no greater than 10</u>, we have that:
- 0 can be added with all the 10 numbers, as can 1.
- 2 can be added with 9 of them, bar 9.
- 3 can be added with 8 of them.
- 4 can be added with 7 of them.
- 5 can be added with 6 of them.
- 6 can be added with 5 numbers, 7 with 4, 8 with 3, and 9 with 2.
Hence, the number of desired outcomes is:
![D = 10(2) + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 = 64](https://tex.z-dn.net/?f=D%20%3D%2010%282%29%20%2B%209%20%2B%208%20%2B%207%20%2B%206%20%2B%205%20%2B%204%20%2B%203%20%2B%202%20%3D%2064)
The probability is:
![p = \frac{D}{T} = \frac{64}{100} = 0.64](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7BD%7D%7BT%7D%20%3D%20%5Cfrac%7B64%7D%7B100%7D%20%3D%200.64)
0.64 = 64% probability that the sum of two randomly chosen numbers from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is no greater than 10.
A similar problem is given at brainly.com/question/25401798