Refer attached graphical representation for more understanding..

The solution makes sense only once you assume that there is no fish that has all three colors.
If we assume there is no fish with all three colors then g = 0

Following things are given to us..
a + d + g + f = 3 (3 Purple fish)
b + d + g + e = 5 (5 Red fish)
c + e + g + f = 10 (10 Yellow Fish)
it is given that f = 0 and d = 2.
We have to find e.

a + b + c + d + e + f + g = 15
a + b + c + d + e = 15 (since g=f=0) …………………. (1)

a + d + g + f = 3
c = 3 (again because g=f=0) ……………………(2)

c + e + g + f = 10
c + e = 10 (g=f=0)…………………. (3)

subtract (2) and (3) from (1)
(1) – (2) – (3)
(a + b + c + d + e) – (a + d) – (c + e) = 15 – 3 – 10
Which gives us b = 2

Now,
b + d + g + e = 5
b + d + e = 5 (g =0)
2 + 2 + e = 5 (b=d=2)
e = 1

e represents red and yellow, which is our answer..

HTH..

Rather than draw, I just think of these problems this way:

Total = # groupA + # groupB + # groupc - overlap AB - Overlap AC - Overlap BC - Overlap ABC - Overlap ABC +# in none of the groups.

This is just an expansion of the equation for if you have two groups:
total = #A + #B - OverlapAB + Neither

The reason you have to subtract ABC twice is that it gets counted 3 times and you need to only count it once (to understand that, look at the diagram above). Now that I believe the theory, I just apply this formula.

So, for this problem:
15 = 3 + 5 + 10 - 2 - x - 0 - 0 + 0
15 = 16 - x
1 = x
Answer = B

-Carrie

Thank you both! I have a much better understanding now.