Fighting Fish

the number of ways of selecting non-harming pairs are ${5 \choose 3} + {5 \choose 2}*{4 \choose 1} = 50$

Total number of ways of selecting 3 fishes from 10 fishes are = ${10 \choose 3} = 120$

so the probability os selection of non-harming pairs = $\frac{50}{120}$ $*100$ $= 41.66667$ % $= 42$ %(approx.)

There are only 2 scenarios where no fish will fight each other: 1. They are all gold fish. & 2. There are 2 gold fish and 1 Siamese fighting fish (it will only bite other fighting fish). The piranha can't be let out; otherwise, it will eat other fish.

The total number of ways we can pick 3 fish out of 10= $\frac{10!}{3!7!}$ = 120

The number of ways for scenario 1 (picking 3 out of 5 gold fish) = $\frac{5!}{3!2!}$ = 10

The number of ways for scenario 2 (picking 2 out of 5 gold fish & 1 out of 4 fighting fish) = $\frac{5!}{3!2!}$ $\frac{4!}{3!1!}$ = 40

Therefore, the probability that no fish will be injured or dead = $\frac{10+40}{120}$ = $\frac{50}{120}$ ≈ $\boxed{42}$ %