document.write( "Question 1110957: Of the parts produced by a particular machine, 1.2% are defective. If a random sample of 12 parts produced by this machine contains 2 or more defective parts, the machine is shut down for repairs. Find the probability that the machine will be shut down for repairs based on this sampling plan. (Give your answer correct to five decimal places.) \n" ); document.write( "
Algebra.Com's Answer #725960 by rothauserc(4718)\"\" \"About 
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Use the binomial probability formula
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\n" ); document.write( "Probability (P) ( k successes in n trials ) = nCk * p^k * (1-p)^(n-k), where nCk = n! / (k! * (n-k)!
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\n" ); document.write( "for this problem p = 0.012
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\n" ); document.write( "P ( k > or = 2 in 12 trials) = 1 - P(k = 0) - P(k = 1)
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\n" ); document.write( "P(k = 0) = 0.865133
\n" ); document.write( "P(k = 1) = 0.126092
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\n" ); document.write( "P ( k > or = 2 in 12 trials) = 1 -0.865133 -0.126092 = 0.008775
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\n" ); document.write( "P ( k > or = 2 in 12 trials) = 0.00878
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