The method you suggested is the simplest. It's very easy to count the number of series 3 consecutive integers, especially since you don't need to do permutations. There are a total of 8 such series (1,2,3), (2,3,4)...(8,9,10).

1 - 8/10C3 = 1 - 1/15 = 14/15 E
_________________

Giving +1 kudos is a better way of saying 'Thank You'.

Went by counting the possible 3 consecutive set = 8
8/10C3 = 8/120
1/15
15/15-1/15 = 14/15

Responding to a pm:

Wy do you restrict the first choice to 8 options?, the way I understand this problem you could pick any of the numbres 1 to 10 and then the next two would have constraints.

For example if my first pick is 10, my second 8 and my last pick is 9, wouldn't that set comply with being a set of consecutive numbers?

The point is 'does the order of picking numbers matter'. If only the end result is important i.e. you get three numbers which are consecutive, then the answer is different but if the order of selecting numbers is important too i.e. after 8, I must pick 9 and then 10, then the answer is different. The question would have been clearer if they had mentioned that three numbers are picked simultaneously. Then the order doesn't matter. When they say that the numbers are picked one after the other, it gets you thinking if the order of selection is important too. Anyway, the intention of the original question is what you suggested. I offered a different spin on it.

In fact, there are two different spins on the question.

When I restrict the first number to one of 1-8, I am assuming that they want the probability of 3 consecutive numbers picked in increasing order.

If you want to find the probability of 3 consecutive number picked in either increasing or decreasing order (say you pick 8, 9, 10 OR you pick 8, 7, 6 in that order), you do not need to have this restriction but you do need different cases.
If you pick 1/2/9/10, the next two numbers can be chosen in only 1 way each.
If you pick any other number, you can choose the next number in 2 ways and the third number in one way.
Convert this logic into math and find out what you get.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

I just thought of another way to figure out the problem:


1.) Select any number: P1 10/10
2.) Select any number next to the first one: P2 2/9
3.) Select and number next to the first or second chosen one: P3 2/8

P1 * P2 * P3 = 1/18

1 - 1/18 = 17/18 .... this is closest to E. The only problem is that I can also chose numbers 1 and 10 which don't have two neighboring numbers.

Taking this into account will make it too difficult and long-lasting (at least for me)