I have been doing the following example, and there is something I do not understand. It goes like this:
From an urn with four white balls and six black balls, one ball is drawn repeatedly with replacement until a white ball is obtained. Let X be the number of drawings resulting in a black ball. Compute the probability $P(X\geq 3)$.
Now, according to the solutions, the way to solve this is to calculate the following: $\sum_{k=3}^∞ ((\frac{6}{10})^k\cdot\frac{4}{10}) $
My question: Why cannot I solve it like this: $P(X\geq 3)=1-(P(X=0)+P(X=1)+P(X=2))$?
Thank you