Chapter 13 - Counting and Probability - 13.3 Probability - 13.3 Asses Your Understanding - Page 866: 55

$0.936$

We know that $P(\text{an event happens})=1-P(\text{an event doesn't happen}).$ Hence, here (with $6.4\%=0.06$) we have $P(\text{never spent any money in gambling})=1-P(\text{spent some money in gambling})\\ P(\text{never spent any money in gambling})=1-0.064\\ P(\text{never spent any money in gambling})=0.936$