1. The problem asks to find probabilities of numbers of bacteria in 1 mg of liquid, given that the mean number (\(\lambda\)) is 2 (assumed from provided probabilities). We use the Poisson distribution formula: $$P(X = k) = e^{-\lambda} \frac{\lambda^k}{k!}$$ where \(\lambda=2\) and \(k=0,1,2,3,4,5\). 2. Calculate each probability step-by-step: - \(P(X=0) = e^{-2} \frac{2^0}{0!} = e^{-2} = 0.1353\) - \(P(X=1) = e^{-2} \frac{2^1}{1!} = 0.2707\) - \(P(X=2) = e^{-2} \frac{2^2}{2!} = 0.2707\) - \(P(X=3) = e^{-2} \frac{2^3}{3!} = 0.1804\) - \(P(X=4) = e^{-2} \frac{2^4}{4!} = 0.0902\) - \(P(X=5) = e^{-2} \frac{2^5}{5!} = 0.0361\) 3. Verify the probabilities sum approximately to 1 for completeness: $$\sum_{k=0}^5 P(X=k) = 0.1353 + 0.2707 + 0.2707 + 0.1804 + 0.0902 + 0.0361 = 0.9834$$ which is close to 1, considering higher \(k\) values have small probabilities. 4. Final answer: probabilities for each \(k\) as above represent the chance of finding exactly \(k\) bacteria in 1 mg of liquid when mean number is 2.