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Old 02-10-2014
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Default calculate probability

Assume that any single single laboratory test in a 3 test battery obtained from healthy volunaers having 5% chance of falling outside the normal range.Which of the following is closest probability that healthy volunteer has same result on all test????

5
40
85
95
100
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Old 02-11-2014
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Originally Posted by medicalbiology View Post
Assume that any single single laboratory test in a 3 test battery obtained from healthy volunaers having 5% chance of falling outside the normal range.Which of the following is closest probability that healthy volunteer has same result on all test????

5
40
85
95
100
I think you mean: the probability that healthy volunteer will have "normal" result for all 3 tests..
which is 95%x95%x95% = 85%
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Old 02-12-2014
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Originally Posted by whitecoat View Post
I think you mean: the probability that healthy volunteer will have "normal" result for all 3 tests..
which is 95%x95%x95% = 85%
How did you do that?can you explain please
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Old 02-12-2014
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Originally Posted by arvine_14 View Post
How did you do that?can you explain please
In this question, the test results are independent events
(i.e. the result of one test tells you nothing about the other 2 results) and the rule is we combine probabilities for independent events by multiplication.

The chance of "having the first test normal" equals 95%, which also equals the chance of having the second test normal as well as the third.

So the chance for having them all normal = the chance of having the first one normal x the chance of having the second one normal x the chance of having the third one normal.
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Old 02-12-2014
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Originally Posted by whitecoat View Post
In this question, the test results are independent events
(i.e. the result of one test tells you nothing about the other 2 results) and the rule is we combine probabilities for independent events by multiplication.

The chance of "having the first test normal" equals 95%, which also equals the chance of having the second test normal as well as the third.

So the chance for having them all normal = the chance of having the first one normal x the chance of having the second one normal x the chance of having the third one normal.
Thanks a lot whitecoat. It was a crystal clear explanation and quite simple too. I forgot how to solve this type of questions actually. But thanks anyways. :-)
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