On average 15% of tomatoes are defective. In a batch of 20, the binomial probability of exactly 3 defective tomatoes is:

1. Use $P(X=x)=\binom{n}{x}p^{x}q^{n-x}$ with $n=20$, $p=0.15$, $q=0.85$. 2. $\binom{20}{3}=1140$. 3. $P(X=3)=1140\times(0.15)^3\times(0.85)^{17}$. 4. This evaluates to $0.2428$. 5. $0.0415$ comes from substituting $n=6$ instead of $n=20$; $0.2240$ is the Poisson value for the same count. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 1: Probability_