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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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    `ok so I'm trying to expand this out to four numbers... A = 10% B = 20% C = 30% D = 40% 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) .10 + .20 + .30 + .40 = 1 P(A) + P(B) + P(C) + P(D) - .02 - .03 - .04 - .06 - .08 - .12 P(A * B) - P(A * C) - P(A * D) - P(B * C) - P(B * D) - P(C * D) + .0024 = .06724 67.24%`2012-04-10
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    Your formula is incorrect; you are missing terms such as $P(A\cap B\cap C)$ which will occur with a positive sign, and $P(A\cap B\cap C \cap D)$ will be subtracted, not added the way you have it. More easily, $$\begin{align*}P(A\cup B \cup C \cup D)&=1-P(A^cB^cC^cD^c)\\&=1-(1-P(A))(1-P(B))(1-P(C))(1-P(D)).\end{align*}$$2012-04-11
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    @DilipSarwate so basically all I'm doing with 10%/20%/30%/40% is finding 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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