What is the probability that in a KENO draw, the digit 8 does not appear in any of the 20 numbers drawn from the integers ranging from 1 to 80?
Among the integers ranging from 1 to 80, we can choose 20 numbers in $\binom{80}{20}$ ways. The digit 8 appears in 9 numbers, which are 8, 18, 28, 38, 48, 58, 68, 78, 88. Therefore, the favorable cases occur when we choose 20 numbers from the remaining 71 numbers.

The probability $p$ we are looking for is:

$$
p=\frac{\binom{71}{20}}{\binom{80}{20}}=\frac{\frac{71!}{51! \cdot 20!}}{\frac{80!}{60! \cdot 20!}}=\frac{71!}{51!} \cdot \frac{60!}{80!}
$$

Simplifying by 71! and 51!, we are left with the product of 9 integers in both the numerator and the denominator:

$$
p=\frac{52 \cdot 53 \cdot 54 \cdot 55 \cdot 56 \cdot 57 \cdot 58 \cdot 59 \cdot 60}{72 \cdot 73 \cdot 74 \cdot 75 \cdot 76 \cdot 77 \cdot 78 \cdot 79 \cdot 80}
$$

After performing the possible simplifications:

$$
p=\frac{3 \cdot 29 \cdot 53 \cdot 59}{4 \cdot 5 \cdot 37 \cdot 73 \cdot 79}=\frac{27249}{4267580} \approx 0.063748
$$

The sought probability is approximately $0.06$.