document.write( "Question 1204192: The probability that a certain machine turns out a defective item is 0.05. Find the
\n" ); document.write( "probability that in a run of 75 items, at least two are defective.
\n" ); document.write( "a.)0.8944 b.)0.08425 c.) 0.1056 d.) 0.8359 e.) 0.9158
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Algebra.Com's Answer #840283 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "Answer: 0.8944 (choice A)\r
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\n" ); document.write( "\n" ); document.write( "Explanation\r
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\n" ); document.write( "\n" ); document.write( "We have a binomial probability distribution because of these reasons:
  • There are two options: the item is defective or it isn't.
  • Each item has the same chance of being defective (5% chance)
  • Each item is independent of one another.
Binomial probability distribution formula
\n" ); document.write( "B(x) = (nCx)*(p^x)*(1-p)^(n-x)\r
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\n" ); document.write( "\n" ); document.write( "The nCx refers to the nCr combination formula.
\n" ); document.write( "n = 75 items total = sample size
\n" ); document.write( "p = 0.05 = probability of a defective item
\n" ); document.write( "x = values chosen from the set {0,1,2,3,...,74,75}\r
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\n" ); document.write( "\n" ); document.write( "Use the binomial formula to compute B(0), which is the probability of getting 0 defective items in a batch of 75.
\n" ); document.write( "B(x) = (nCx)*(p^x)*(1-p)^(n-x)
\n" ); document.write( "B(0) = (75C0)*(0.05^0)*(1-0.05)^(75-0)
\n" ); document.write( "B(0) = (1)*(0.05^0)*(1-0.05)^(75-0)
\n" ); document.write( "B(0) = 0.021344 approximately\r
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\n" ); document.write( "\n" ); document.write( "Repeat for x = 1
\n" ); document.write( "B(x) = (nCx)*(p^x)*(1-p)^(n-x)
\n" ); document.write( "B(1) = (75C1)*(0.05^1)*(1-0.05)^(75-1)
\n" ); document.write( "B(1) = (75)*(0.05^1)*(1-0.05)^(75-1)
\n" ); document.write( "B(1) = 0.084252 approximately\r
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\n" ); document.write( "\n" ); document.write( "B(0)+B(1) = 0.021344+0.084252 = 0.105596 represents the probability of getting 0 defective items or 1 defective item.\r
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\n" ); document.write( "\n" ); document.write( "The probability of getting 2 or more defective items is 1 - 0.105596 = 0.894404 which rounds to 0.8944
\n" ); document.write( "I'm using the idea that P(0 or 1 defects) + P(2 or more defects) = 1.\r
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\n" ); document.write( "\n" ); document.write( "There's about an 89.44% chance of getting at least two defective items.\r
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\n" ); document.write( "\n" ); document.write( "Another approach would be to use a calculator like this
\n" ); document.write( "https://www.gigacalculator.com/calculators/binomial-probability-calculator.php\r
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\n" ); document.write( "\n" ); document.write( "Or you could use a TI83/TI84 calculator
\n" ); document.write( "https://www.statology.org/binomial-probabilities-ti-84-calculator/
\n" ); document.write( "The command you'll want is called BinomCDF.
\n" ); document.write( "More specifically you'll input 1-BinomCDF(75,0.05,1)\r
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\n" ); document.write( "\n" ); document.write( "Or you could use the BinomDist spreadsheet function.\r
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\n" ); document.write( "\n" ); document.write( "The spreadsheet command is
\n" ); document.write( "=Round(1-BinomDist(1,75,0.05,1),4)
\n" ); document.write( "The template for BinomDist is:
\n" ); document.write( "BinomDist(X,N,P,C)
\n" ); document.write( "where
\n" ); document.write( "X = number of successes
\n" ); document.write( "N = sample size
\n" ); document.write( "P = probability of success
\n" ); document.write( "C = either 0 or 1 depending if you want a PDF or CDF
\n" ); document.write( "Do not forget about the equal sign up front when typing in =Round(1-BinomDist(1,75,0.05,1),4) or the command won't execute.\r
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\n" ); document.write( "\n" ); document.write( "Computing BinomDist(1,75,0.05,1) will produce the value B(0)+B(1) = 0.105596 that we found earlier.
\n" ); document.write( "Then we subtract it from 1 to get the value of B(2)+B(3)+...+B(74)+B(75) = 0.8944
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