Three defective electric toothbrushes were accidentally shipped to a drugstore by fresh
and white products along with 17 nondefective ones.
i what is the probability the first two electric toothbrushes sold will returned to the
drugstore because they are defective?
ii what is the probability the first two electric toothbrushes sold will not be defective?

 When you are sampling from a small finite lot, the hypergeometric distribution applies. The binomial is a poor approximation in this case. 
The general equation for the hypergeometric where aCx means the number of combinations of a items selected x at-a-time.: 
P(x) =[(aCx)(N-aCn-x)]/NCn 
Where 
N is the lot size = 20 
a is the number of defectives in the lot = 3. 
x is the number of defectives in the sample. 
n is the sample size = 2. 
A. The probability that the first item is defective is 
P(x=1) = [(3C1)(17C1)]/(20C2) 
= (3)(17)/190 = 0.268 
The probability that the second item is defective is 
P(x = 1) = [(2C1)(17C1)]/(19C2) = (2)(17)/171 = 0.199. 
So the total probability is (0.268)(0.199) = 0.0532 
B. The probability that the first item is good is: 
P(x = 0) = (3C0)(16C2)]/20C2 = (1)(120)/190 = 0.632 
The probability that the second item is defective is 
P(x = 0) =[(3C0)(16C2)]/19C2 
= (1)(120)/171 = 0.670. 
The total probability is 0.632(0.670) = 0.4234

I’m confused this is what i’m doing in class but the question is confusing

Yup correct because it right