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Interesting thing about archers hitting the target 
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#1
Jun1309, 10:23 AM

P: 365

Hello, again!
I got one very interesting question. We got three archers, and the probability of the ones to hit the target is: A_{1}, A_{2}, A_{3}. What if the task is to find the probability that the target will be hit at least from one archer. So at least one archer to hit the target. Is it P(A_{1} U A_{2} U A_{3}) = A_{1} + A_{2} +A_{3} ? Or [tex](1 P(\bar{A_{1}} \cap \bar{A_{2}} \cap \bar{A_{3}})) = 1  \bar{A_{1}}* \bar{A_{2}} * \bar{A_{3}}[/tex], where [tex]\bar{A}[/tex] is opposite of A? Or maybe, both are valid? Thanks in advance. 


#2
Jun1309, 10:47 AM

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#3
Jun1309, 10:53 AM

P: 365

Thanks for the post, HallsofIvy.
In this case, the shootings of the archers are independent cases. So that's why we do not need [itex]P(A_1\cup A_2\cup A_3)= P(A_1)+ P(A_2)+ P(A_3) P(A_1\cap A_2) P(A_1\cap A_3) P(A_2\cap A_3)+ P(A_1\cap A_2\cap A_3)[/itex] since [itex]P(A_1\cap A_2) P(A_1\cap A_3) P(A_2\cap A_3)+ P(A_1\cap A_2\cap A_3)=0000=0[/itex]. That's why I said independent cases. But what if P(A_{1})=0.8, P(A_{2})=0.9, P(A_{3})=0.75 In that case the sum [itex]P(A_1\cup A_2\cup A_3)= P(A_1)+ P(A_2)+ P(A_3)=0.8+0.9+0.75=2.45[/itex] This is strange. 


#4
Jun1309, 11:55 AM

P: 330

Interesting thing about archers hitting the target
The fact that two cases A and B are independent doesn't mean P(A n B) = 0. It means that P(A n B) = P(A)P(B).



#5
Jun1309, 01:00 PM

P: 365

If two cases are independent, that means that they do not have something in common, right? So A n B = 0, right?



#6
Jun1309, 02:52 PM

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P: 1,361

[tex] A \cap B = \emptyset [/tex] if the two sets are disjoint , which is not the same as independent. 


#7
Jun1309, 04:36 PM

P: 38

With regards to the original question, it would be easiest to take the probability that every archer misses and subtract it from one. 


#8
Jun1409, 03:36 AM

P: 365

Thanks for the replies.
@Tibarn, in this case the occurrence of one does not influence the occurrence of the other. [tex]P(A/B)=\frac{m_{A\cap B} }{ m_{B}}[/tex] out of there [tex]P(A/B)=\frac{\frac{m_{A\cap B}}{n}}{\frac{m_{B}}{n}}[/tex] and [tex]P(A/B)=\frac{P(A\cap B)}{ P(B)}[/tex] The cases are independent if the occurrence of one does not influence the occurrence of the other. So, if two cases are independent, then P(A/B)=P(A). Out of there P(A n B)=P(A)*P(B) Now, let's get back to the task. I think I mean, disjoint. So cases A,B are disjoint. P(A U B)=P(A) + P(B)  P(A n B) In this case [itex]m_{A\cap B}=0[/itex] because the cases: I  the 1st archer will hit the target II  the 2nd archer will hit the target DO NOT have something in common. [tex] P(A U B) = \frac{m_A+m_Bm_{A\cap B}}{n}[/tex] [tex] P(A U B) = \frac{m_A+m_B0}{n}=P(A)+P(B)[/tex] Is this true? Are those cases disjoint?? 


#9
Jun1409, 11:30 AM

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#10
Jun1409, 02:37 PM

P: 365

Are those cases disjoint?



#11
Jun1409, 08:37 PM

P: 38

If both archers take one shot, then we have four possible events: 1. Both miss 2. Archer A hits, B misses 3. Archer A misses, B hits 4. Both hit. In this case, [tex]A \cap B[/tex] is case 4, where both archers hit. If you're going to do probability by cases, it's important that you get all of them. 


#12
Jun1509, 08:53 AM

P: 365

Now, I understood. Thank you very much for the help.
Regards. 


#13
Jun1509, 10:29 AM

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You may be confusing "mutually exclusive" with "independent". "Independent" means what happens in one case does not affect what happens in the other [itex]P(A\cap B)= P(A)P(B)[/itex] or, equivalently, P(AB)= P(A). "Mutually exclusive" means [itex]P(A\cap B)= 0[/itex] so that P(AB)= 0. Not at all the same thing!



#14
Jun1509, 10:56 AM

P: 365

Yes, you're right. I did a little research, and find out that the events in this case aren't exclusive, but they are independent. Because if they are exclusive P(A) or P(B) will be equal 0 so that P(A n B)=0.



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