Answer
$P(2)=0.251$
In about 251 of every 1000 5-gram samples, there will be exactly 2 insect fragments.
Work Step by Step
$P(x)=\frac{(λt)^x}{x!}e^{-λt}$
λ = 0.3 -> (0.3 insect fragment per gram)
t = 5 -> (5 grams)
$P(2)=\frac{(0.3\times5)^2}{2!}e^{-0.3\times5}=\frac{1.5^2}{2}e^{-1.5}=0.251$
