15. Given the set $A=\{1,2,3,4\}$, every second, Andel randomly picks a number from $A$ (repetition is allowed), and stops picking when the sum of the last two numbers picked is a prime number. What is the probability that the last number picked is “1”?
15. Let's arrange the numbers in a sequence, and let the probability of a sequence starting with “ $i$ ” and ending with “ 1 ” be $P_{i}(i=1,2,3,4)$.

If a sequence starts with 1, then in the first second, only “ 1 ” can be chosen. Since the selection is arbitrary, there is a $\frac{1}{4}$ probability of selecting “1”. But after selecting “1”, if “ 1 ” is chosen again in the second second, the sequence ends, resulting in a sequence that starts with “1” and ends with “1”. If “ 2 ” is chosen in the second second, 1+2=3 is a prime number, and the sequence also ends, but it does not end with “1”. If “ 3 ” is chosen in the second second, the sequence does not stop. At this point, the probability of ending with “1” is $P_{3}$. If “ 4 ” is chosen in the second second, the situation is the same as with “ 2 ”.

From the above analysis, the event of a sequence starting with “1” and ending with “1” includes two mutually exclusive events: the first number is “1” and the second number is “1”; the first number is “1” and the second number is “3” and ends with “1”. Therefore, $\quad P_{1}=\frac{1}{4}\left(\frac{1}{4}+P_{3}\right)$.
Similarly, we get: $P_{2}=\frac{1}{4} \times\left(\frac{1}{4}+P_{2}+P_{4}\right)$,
(2) $P_{3}=\frac{1}{4}\left(P_{1}+P_{3}\right)$,
(3) $P_{4}=\frac{1}{4} \times\left(\frac{1}{4}+P_{2}+P_{4}\right)$
From (1) (2)(3)(4), we get $P_{1}=\frac{3}{44}, P_{3}=\frac{1}{44}, P_{2}=P_{4}=\frac{1}{8}$
Thus, the probability of ending with “1” is $P_{1}+P_{2}+P_{3}+P_{4}=\frac{3}{44}+\frac{1}{8}+\frac{1}{44}+\frac{1}{8}=\frac{15}{44}$.