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Probability ques
a nw tst fr detect pros canc is neg in 95% of ptx who do not have the disease. If the tst is usde wth 8 consecutive bld sample taken from ptx without the dsesas, wht is the prob of getting at lest one post tst rslt?
appologize for the misspels... |
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The answer is
1-(0.95^8) You have to multiply because the events are independent. I thought also to go the 0.5^8 route however it gives a very small probability when I actually calculated it, but I still don't know why this is wrong. Anyone can enlighten us? |
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Hi,
Oh, yes you are right 1-(.95^8) Because you want at least one test positive ,it means 1 positive or 2 positive or 3 positive,... . (.95^8) means all cases are negative and 1-(.95^8) means you calculate all situations except all negative. You can't use 0.5^8 because this means all tests positive not at least one positive. |
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Okay now it makes sense, thnx.
At least one test positive so 0.95^8 = prob All test are negative. 0.05^8 = prob All test are positive. 1-(0.95^8) = prob of 1 or more positives. 1-(0.05^8) = prob of 1 or more negatives. |
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