M01-04

The problem with such an approach, rnn, is that you would come up with a probability in excess of 100 percent, and that makes no sense at all. If Cannon A will hit the mark 30 percent of the time, then it will miss 70 percent of the time. Likewise, if Cannon B will hit 40 percent of the time, it will miss 60 percent of the time, and if Cannon C will hit half the time, it will miss the other half of the time.

.7 + .6 + .5 = 1.8

Are we to understand that the three cannon--apropos of nothing, this is one of those words that can or cannot add an s to the end to pluralize--have a 180 percent probability of not hitting the target if they fire at the same time? Again, that would make no sense. At worst, they could only miss 100 percent of the time, but even for that to be true, they would all have to have an accuracy of 0 percent. That is, if even one cannon could hit the target, then the three of them together would also have a chance of hitting the target.

You might say that you had the right idea but the wrong approach. Also, you should understand that an and probability involves multiplication, while an or probability involves addition. In the problem at hand, the condition is such that any of Cannon A and Cannon B and Cannon C, when fired at once, could hit the target.

I hope that helps a bit. If you have further questions, feel free to ask.

- Andrew

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.