From this problem, we can know that when K works normally and at least one of A 1 and A2 works normally, the system works normally, so there are three situations: 1 and K is normal, A 1 is normal, A2 is bad, K is normal, A 1 is bad, A2 is normal, 3 K is normal, and A.

If these three situations are added up, P = 0.9 * (0.8 * 0.2+0.2 * 0.8+0.8 * 0.8) = 0.9 * 0.96 = 0.864.

Similarly, from another perspective, K must be working normally. Under normal working conditions, the only thing that K can't work normally is that A 1A2 is broken and can be written.

P = 0.9 * (1-0.2 * 0.2) = 0.9 * 0.96 = 0.864, and the result is the same.