20. Among $1, 2, 3, \cdots, 50$, if 10 consecutive numbers are randomly selected, then the probability that there are exactly 3 prime numbers is
【Solution】Solution: (1) The prime numbers between 1 and 50 are: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,$
$37, 41, 43, 47$;
There are exactly 3 prime numbers, which are clearly consecutive, otherwise the 10 natural numbers would not be consecutive;
(2) The difference between the largest and smallest of 3 consecutive prime numbers is at least 6, otherwise there would not be exactly 3 prime numbers;
(3) Starting from 7, the list is as follows:
\begin{tabular}{|c|c|c|c|}
\hline Consecutive primes & \begin{tabular}{c}
10 consecutive \\
groups
\end{tabular} & Consecutive primes & \begin{tabular}{c}
10 consecutive \\
groups
\end{tabular} \\
\hline $7, 11, 13$ & 2 & $17, 19, 23$ & 4 \\
\hline $11, 13, 17$ & 2 & $37, 41, 43$ & 4 \\
\hline $13, 17, 19$ & 2 & $41, 43, 47$ & 4 \\
\hline
\end{tabular}

As shown in the table above, the number of groups is: $2 \times 3 + 4 \times 3 = 18$ (groups)
(4) The difference between the largest and smallest of 3 consecutive prime numbers is 8, clearly: the number of 10 consecutive natural number groups $=$ the smallest prime number + 10
- the largest prime number, the 10 consecutive natural numbers with exactly 3 prime numbers being $23, 29, 31$ are: $23 + 10 - 31 = 2$ (groups), the 10 consecutive natural numbers with exactly 3 prime numbers being $29, 31, 37$ are: $29 + 10 - 37 = 2$ (groups), totaling 4 groups;
(5) Taking any 10 consecutive numbers results in 41 groups, of which the probability of having exactly 3 prime numbers is: $\left(18 + 4 \right) \div 41 = \frac{22}{41}$; Answer: The probability of having exactly 3 prime numbers is $\frac{22}{41}$. Therefore, the answer is: $\frac{22}{41}$.