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Why can the typical example 1 on the left be directly multiplied, but the class problem on the right should be solved by 1-the probability of the opposite event? What's the title on the right?
Because the two events are different in nature, they are represented by C.

The premise of the event on the left is that A holds and B holds, so C=A to B..

As long as one person on the right hits, even if the target is hit, including two people hitting or one person hitting and one person missing, there are always three situations: A crosses B, A crosses B ~, A crosses B ~, A crosses B ~, and the event holds C=(A crosses B)+(A crosses B ~)+(A crosses B ~).

So the exclusion method is adopted.

Suspected event C~ The place where the target was hit was that the target was not hit, that is, both A and B missed C~=A~ across B~

P(C)= 1-P(C~)