Chapter 10 Counting Methods and Probability - 10.3 Define and Use Probability - 10.3 Exercises - Skill Practice - Page 702: 17

$\frac{1}{12271512}$

We know that if we want to select $r$ objects out of $n$ disregarding the order, we can do this in $_nC_r=\frac{n!}{r!(n-r)!}$ ways. Here we have $n=48$ and $r=6$. Hence we have $_{48}C_6=12271512$ total possible selections. Thus the probability: $\frac{1}{12271512}$