Drawing twice with replacement from an urn of three red and eight blue balls, 88 of the 121 outcomes have blue first. What is the probability of red second given blue first?

1. Conditioning on blue first reduces the sample space from 121 outcomes to 88. 2. Those 88 split into 24 blue-then-red outcomes and 64 blue-then-blue outcomes. 3. The conditional probability is therefore 24/88. 4. That fraction simplifies to 3/11. 5. Getting the same 3/11 as the plain probability of red confirms that drawing with replacement makes the draws independent. 6. The value 3/10 would apply without replacement, where only ten balls remain for the second draw. _Source: OpenStax Introductory Business Statistics (CC BY 4.0), Ch 3 "Probability Topics", section 3.4 Contingency Tables and Probability Trees_