Three red and eight blue balls sit in an urn, and two are drawn with replacement. In how many of the outcomes are both balls blue?

1. Only the blue balls matter for this count. 2. There are 8 ways to draw a blue ball on the first draw. 3. The ball is replaced, so there are again 8 ways on the second draw. 4. Multiply the two counts: 8 times 8. 5. 8 times 8 is 64. 6. So 64 of the outcomes have blue on both draws. 7. Answering 24 counts one blue and one red, which is 8 times 3. 8. Answering 16 adds the two counts of 8 instead of multiplying them. _Source: OpenStax High School Statistics (CC BY 4.0), Ch 2 "Descriptive Statistics" and Ch 3 "Probability Topics", section 3.5 Tree and Venn Diagrams_