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Ok this is probably an easy one,

Person A hits a target 20% of the time Person B hits a target 40% of the time

What are the odds, and formula, that either one of them hits the target?

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    @DilipSarwate so basically all I'm doi$n$g with 10%/20%/30%/40% is fi$n$ding the probability of it NOT happening... .90 * .80 * .70 * .60 = .3024 and then 1 - .3024 = .6976 = 69.76% chance of any one of the 4 hapening?2012-04-12

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For brevity, lets name the event that $A$ hits the target "$A$", and likewise with $B$. By the inclusion-exclusion principle, we know that $P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B).$ (This is also called the "addition law".)

Also, the implicit assumption is that $A$'s success in hitting the target is independent of $B$'s, i.e. the probability $A$ hits the target doesn't change depending on whether $B$ does, or vice versa. Therefore, $P(A\text{ and }B)=P(A)\cdot P(B).$ Now you can compute the value of $P(A\text{ or }B)$.

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    @DilipSarwate so... this is what i've got after reading, is this correct? P(A or B or C or D) = P(A) + P(B) + P(C) + P(D) - P(A * B) - P(A * C) - P(A * D) - P(B * C) - P(B * D) - P(C * D) + P(A * B * C * D) ugh this comments box screws up the formatting but yea...2012-04-10
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"either one of them" is not clear.

  • if "either" means "one, or both" then the probability is $0.2$ that A hits and in case he doesn't ($0.8\cdot$) the other one might with a chance of $0.4$. So $P_\text{or}=0.2+0.8\cdot 0.4=0.52$

  • if "either" means "one, but not both", then the probability is that of A hitting, but B missing ($0.2\cdot0.6$) plut that of A missing but B hitting ($0.8\cdot0.4$). therefore $P_\text{xor}= 0.2\cdot 0.6 + 0.8\cdot 0.4 = 0.44 $

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    There is an implicit assumption of independence here that should be mentioned explicitly.2012-04-10