A box contains 20 items of which 4 are defective.
Joshua draws one item after another with replacement until he gets a defective one.
What is the probability that the number of trials before Joshua gets a defective item is 3?
(Answer:$\frac{4}{625}$)
I thought it was a geometric distribution.
My answer: $P(X=x) = {\theta}(1-\theta)^{x-1}$
$\theta = \frac{4}{20} = \frac{1}{5}$
$P(X=3) = (\frac{1}{5})(\frac{4}{5})^{3-1} = (\frac{1}{5})(\frac{4}{5})^2 = \frac{16}{125}$
Am I using the wrong distribution?
Actually $\frac4{625}$ is the probability that Joshua needs four trials to get his first non-defective item, not the probability that he needs three trials to get his first defective item (hence, your textbook is making two mistakes to solve the same (elementary) exercise...).