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First, calculate the total number of ways to choose 3 fish from the 13 fish in the tank:
Next, calculate the number of ways to select exactly 2 salmon and 1 non-salmon. There are 6 salmon, so the number of ways to choose 2 salmon is:
There are 7 non-salmon fish (4 trout + 3 bass), so the number of ways to choose 1 non-salmon is:
The probability of selecting exactly 2 salmon is then:
Simplifying the fraction (divide numerator and denominator by their greatest common divisor, 1):
We want the probability that exactly 3 out of 4 selected samples have at least high organic matter. This is a binomial probability:
We want the number of ways to assign exactly 3 H, 3 M, and 2 L:
The probability of any one such assignment (e.g., assigning H to 3 zones, M to 3, L to 2) is:
This fraction is already in simplest form (560 and 6561 share no common factors).
