# Typical urn conditional probability question

I'm studying the book Introduction to Mathematical Statistics and its Applications by Larsen and Marx, third edition. Here my paraphrasing of question 2.4.11, and I'm puzzled by the answer they give:

An urn contains a chip that's white and another chip equally likely to be white or black. A first chip is picked out, its color recorded, and is put back in the urn. A second chip is picked out again and its color is recorded. Given that the first chip picked out is white, what's the probability that the second chip is white?

Book answer: $5/6$

Here's what I've come up with.

$P(\textrm{chip with unknown color is white})= P(\textrm{chip with unknown color is black}) = 1/2$

$P(\textrm{picking a white chip from the urn}) = P(\textrm{white chip is picked}) + P(\textrm{unknown chip is white}) \cdot P(\textrm{unknown chip is picked})$

$= 1/2 + 1/2 \cdot 1/2 = 3/4$

$P(\textrm{first chip is white}) = 3/4$

$P(\textrm{first chip is white and second chip is white}) = 3/4 \cdot 3/4$

$P(\textrm{second chip is white | first chip is white}) = P(\textrm{first chip is white and second chip is white}) / P(\textrm{first chip is white})$

$= \frac{3/4 \cdot 3/4}{3/4} = 3/4$

What's wrong with my reasoning?

Here is one correct way of carrying out the calculation by using your your basic strategy. If the extra ball was white, then the probability of white and then white is $1$. If the extra ball was black, the probability of white and then white is $(1/2)(1/2)$.
So the probability that the first ball is white and the second is white is $$(1/2)(1)+(1/2)(1/2)(1/2).$$
This is $5/8$. Continue!