Problem:you have two guns, probability that first gun hits the target is 0.9, and probability that second gun hits the target is 0.8.you shot first shot by first gun and than you shoot by second gun.what is the probability that target will be damaged(these shots are independent).my logic:P($B \cup A$)=P(A)+P(B)-P(A$\cap$B).but then I thought differently:I calculated the probability that first hits second not+first hits second hits, first doesn't hit second does.but when I calculated results I got different results.please help me to deduce which is the real solution.

  • $\begingroup$ .72+.08+.18+.02=1. Also .9+.8-.72+.02=1 $\endgroup$ – Peter May 14 '20 at 5:17
  • $\begingroup$ so what do you mean?in every experiment target is damaged? $\endgroup$ – dato nefaridze May 14 '20 at 5:20

The target is undamaged when both shots miss. That each shot misses is independent of the other. The probability that both miss ($P(\neg A \cap \neg B)$ is $(1-0.9)(1-0.8) = 0.02$. There is a 2% chance the target is undamaged. Therefore, there is a ($P(A \cup B) = 1 - P(\neg (A \cup B)) = 1 - P(\neg A \cap \neg B)$), $1 - 0.02 = 0.98$, so 98% chance the target is damaged.

(There are many notations for the complement of a state. I have used "$\neg$" above for the complement. Notice that we are using De Morgan's Laws to go from $\neg (A \cup B) = \neg A \cap \neg B$.)


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