Option1: no one of the 3 students has a dark brown eye color.

Option2: at least one person has a dark brown eye color.

Option1 + option2 contains all the possibilities so P(options1)+p(option2)=1.

Usually, the easier way to understand is to write all the options

for simple writing:

1 - dark brown eye

0 - other eyes color

Following all options:

Option, student-1, student-2, student-3

**1,** 0,0,0

**2**, 1,0,0

**3**, 0,1,0

**4**, 0,0,1

**5**, 1,1,0

**6**, 0,1,1

**7**, 1,0,1

**8**, 1,1,1

The question is: What is the probability that at least one person has a dark brown eye color?

So the is the probability that one of the following options will happen: 2,3,4,5,6,7,8

You calculated only option **2**. you need to add options 3,4,5,6,7,8

You may do it the hard way p(2)+p(3)+...p(8)

Or you may use the hint, as you know that p(1)+p(2)+...+p(8)=1.

I suggest that you will try to calculate on both ways to improve your intuition

Cheers.

PS, you may look also the binomial distribution, saving calculations using combinatorics,