Dave E. answered 15d
33 years as a high school and college teacher
I don't know what level this is but there is more going on here than that last answer is considering. Your answer is the "Textbook" answer that everyone assumes. But it is not the only possibility.
Remember, Slimy/Smooth frogs can be SSww or Ssww. You can't tell them apart.
They made it hard by telling you the ratios with different denominators. If you adjust everything to be out of 12 you get 6:3:2:1.
They are then asking you to cross two frogs from the F1 generation that are Slimy / Smooth. The answer above from the previous tutor made the same mistake you did. In fact, they didn't even notice that they gave the same answer as you did.
You (and they) assumed that you were crossing Ssww / Ssww. However that isn't the only possibility
If you randomly grab any Slimy/Smooth frog, you might get two that are Ssww. But you could also grab one or both that are SSww. If you do the Punnett square, 50% of the Slimy/smooth should be SSww. So if you grab two random slimy/smooth frogs, odds are you will get at least one that is actually SSww. Possibly both. So that means that you could really be crossing any of the following:
SSww x SSww = 100% Slimy/smooth
Ssww x SSww= 100% Slimy/smooth
Ssww x Ssww = 75% Slimy/Smooth and 25% Dry Smooth
Only that last one will give you 3/4 Slimy/Smooth to 1/4 Dry/Smooth
The other two would result in 100% Slimy/Smooth
Since 2 of the 3 possible random Slimy/Smooth crosses give all Slimy/smooth, the odds are you would get more than 3/4 Slimy/Smooth and less than 1/4 Dry/Smooth.
If that is what they meant, this is a seriously nasty question. If not, I don't know what they took off points for.