8 There are six cards, each with a number $1, 2, 3, 4, 5, 6$ written on it. Each time, one card is drawn, the number on it is noted, and then it is put back. This is done 4 times. The probability that the difference between the maximum and minimum numbers drawn is equal to 5 is $\qquad$.
To make the difference between the maximum and minimum numbers drawn equal to 5, both 6 and 1 must be drawn in 4 draws. There are $6^{4}$ possible outcomes in 4 draws. Among these $6^{4}$ outcomes, there are $5^{4}$ outcomes where 1 is not drawn, $5^{4}$ outcomes where 6 is not drawn, and $4^{4}$ outcomes where neither 1 nor 6 is drawn. Therefore, the number of outcomes where both 1 and 6 are drawn is $6^{4}-2 \times 5^{4}+4^{4}=302$ (outcomes).
Thus, the required probability is $\frac{302}{6^{4}}=\frac{151}{648}$.