Question 1197428
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Hi
30 Batteries (with 10 defects)  p(defect) = 1/3.  p(no defect) = 2/3
   3 chosen
p( 0 defects) = (2/3)^3 = {{{(8/27)}}}
Using
{{{P (x)= highlight_green(nCx)(p^x)(q)^(n-x) }}} 
P(1 or 2 or 3 defects) =  3(1/3)(2/3)^2 + 3(1/3)^2(2/3) + (1/3)^3
                           = 12/27 + 6/27 + 1/27 = {{{(19/27)}}}
Or  1 - 8/27 = 19/27 (bit easier)
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