Let e mean that only one piece is qualified after the first firing.
Then: P(E)=P(A 1? A2A3? )+P(A 1A2? A3? )+P(a 1a2a 3)= 45× 14× 13+ 15×34× 13+ 15× 14×23 = 320
The probability that only one product is qualified after the first firing is 320.
(2) Note that after A, B and C are fired twice, the qualified events are A, B and C respectively.
Then: P(A)= 1225, P(B)=920, P(C)=25.
Let f mean that all three products are qualified after twice firing, then: P(F)=P(A? b? C)= 1225×920×25=54625
∴ The probability that all three products are qualified after two firings is 54625.