Question 1198942
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A person with tuberculosis is given a chest? x-ray. 
Four tuberculosis? x-ray specialists examine each? x-ray independently. 
If each specialist can detect tuberculosis 89?% of the time when it is
present, what is the probability that at least 1 of the specialists 
will detect tuberculosis in this? person?
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The solution is in two steps.


Step 1.  Then the probability that no one of 4 specialists will detect 

         tuberculosis is  {{{(1-0.89)^4}}} = }}}0.11^4}}} = 0.00014641   (rounded).


Step 2.  The probability under the problem's question is the COMPLEMENT to it

         P = 1 - {{{(1-0.89)^4}}} = 1 - 0.00014641 = 0.999853.    <U>ANSWER</U>
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Solved.