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Bunuel
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P(None of them will successfully hit the target) = (1-P(A))*(1-P(B))*(1-P(C)) = (1-0.3)*(1-0.4)*(1-0.5) = 0.7*0.6*0.5 = 0.21

(C) is the answer
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Since there are 3 probabilities for 3 (different?) targets? Is there any consideration the time 'after a single round of firing', such as we only consider the last two probability instead?

Because of it is not considered (we combined all the 3 Probs), could you explain it why?
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Since there are 3 probabilities for 3 (different?) targets? Is there any consideration the time 'after a single round of firing', such as we only consider the last two probability instead?

Because of it is not considered (we combined all the 3 Probs), could you explain it why?
There is only one target. The 0.3, 0.4, 0.5 are the three cannons’ separate chances to hit that same target in that one round.

“After a single round of firing” means each cannon fires once, and we look at the result of that round. We do not ignore any cannon.

So “none hit” means cannon 1 misses AND cannon 2 misses AND cannon 3 misses, which is why we multiply 0.7 * 0.6 * 0.5.
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