document.write( "Question 1208645: An integrated circuit manufacturer produces wafers that contain
\n" ); document.write( "20 chips. Each chip has a probability of 0.085 of not placed correctly on the wafer.
\n" ); document.write( "A) Find the probability that a wafer contains at least 3
\n" ); document.write( "incorrectly place chips. (Ans = 0.2390)
\n" ); document.write( "b) What is the probability that a wafer no more than one
\n" ); document.write( "incorrectly place chips on a wafer? (Ans = 0.4836)
\n" ); document.write( "c) What is the expected number of incorrectly placed chips on a
\n" ); document.write( "wafer? (Ans = 1.7)
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Algebra.Com's Answer #847029 by math_tutor2020(3817)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "Please do not type in all upper case.
\n" ); document.write( "I have edited your question to a more readable format.\r
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\n" ); document.write( "\n" ); document.write( "We can use the Binomial probability distribution because:
  • There are two outcomes per trial. Either a chip is placed correctly or it is not.
  • Each trial is independent.
  • The probability of not getting placed correctly is the same for each trial.
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\n" ); document.write( "\n" ); document.write( "Part A\r
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\n" ); document.write( "\n" ); document.write( "n = 20 = sample size = number of chips
\n" ); document.write( "p = 0.085 = probability of a chip being incorrectly placed
\n" ); document.write( "x = number of chips incorrectly placed
\n" ); document.write( "x is a whole number that ranges from x = 0 to x = 20, i.e. it's an integer from the set {0,1,2,...,19,20}\r
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\n" ); document.write( "\n" ); document.write( "B(x) = binomial probability of exactly x chips being incorrectly placed
\n" ); document.write( "B(x) = (nCx)*(p^x)*(1-p)^(n-x)
\n" ); document.write( "The nCx refers to the nCr combination formula. These values are found in Pascal's Triangle. A quick way to calculate the nCr values is to use the Combin function in a spreadsheet. Or you can use a TI Calculator.
\n" ); document.write( "There are many options to calculate nCr.\r
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\n" ); document.write( "\n" ); document.write( "Let's find the probability of exactly x = 0 chips that are incorrectly placed.
\n" ); document.write( "B(x) = (nCx)*(p^x)*(1-p)^(n-x)
\n" ); document.write( "B(x) = (20Cx)*(0.085^x)*(1-0.085)^(20-x)
\n" ); document.write( "B(0) = (20C0)*(0.085^0)*(1-0.085)^(20-0)
\n" ); document.write( "B(0) = (1)*(0.085^0)*(1-0.085)^(20-0)
\n" ); document.write( "B(0) = 0.16920839 approximately\r
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\n" ); document.write( "\n" ); document.write( "Now let's find the probability of exactly x = 1 chip being incorrectly placed.
\n" ); document.write( "B(x) = (20Cx)*(0.085^x)*(1-0.085)^(20-x)
\n" ); document.write( "B(1) = (20C1)*(0.085^1)*(1-0.085)^(20-1)
\n" ); document.write( "B(1) = (20)*(0.085^1)*(1-0.085)^(20-1)
\n" ); document.write( "B(1) = 0.31437624 approximately\r
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\n" ); document.write( "\n" ); document.write( "Repeat similar steps to find that B(2) = 0.27744132 approximately which is the probability of having exactly 2 chips incorrectly placed.\r
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\n" ); document.write( "\n" ); document.write( "Then we could say,
\n" ); document.write( "( B(0)+B(1)+B(2) ) + ( B(3)+B(4)+...+B(19)+B(20) ) = 1
\n" ); document.write( "B(3)+B(4)+...+B(19)+B(20) = 1 - ( B(0)+B(1)+B(2) )
\n" ); document.write( "B(3)+B(4)+...+B(19)+B(20) = 1 - ( 0.16920839+0.31437624+0.27744132 )
\n" ); document.write( "B(3)+B(4)+...+B(19)+B(20) = 1 - 0.76102595
\n" ); document.write( "B(3)+B(4)+...+B(19)+B(20) = 0.23897405
\n" ); document.write( "B(3)+B(4)+...+B(19)+B(20) = 0.2390 which is the answer that your textbook mentioned. \r
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\n" ); document.write( "\n" ); document.write( "If you want to use technology to quickly find the answer, then you can use a TI83 or similar.
\n" ); document.write( "The command you'll use is called BinomCDF
\n" ); document.write( "The order of inputs is: n, p, k
n = 20\r\n" );
document.write( "p = 0.085\r\n" );
document.write( "k = 2
Type in 1-BinomCDF(20,0.085,2) to get the approximate result 0.2390 when rounding to 4 decimal places.
\n" ); document.write( "You can round manually or you can use the Round command in the TI83.
\n" ); document.write( "Note that the BinomCDF(20,0.085,2) portion computes the sum B(0)+B(1)+B(2)\r
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\n" ); document.write( "\n" ); document.write( "Here are some alternative technology options.
  • Search out \"binomial CDF calculator\". This page and this page are two of many results. Feel free to explore your favorite.
  • Use the Probability Calculator in GeoGebra. Select \"binomial\" from the dropdown menu. Type n = 20 and p = 0.085; the goal is to calculate
  • Use the spreadsheet command called BinomDist. The input would be =1-BinomDist(2,20,0.085,true)
  • Use the BinomialDist command in GeoGebra. Note \"binomial\" instead of \"binom\". The input would be 1-BinomialDist(20,0.085,2,true)
Refer to the software's help manual for more information.\r
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\n" ); document.write( "\n" ); document.write( "Part B\r
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\n" ); document.write( "\n" ); document.write( "Refer to the previous part.
\n" ); document.write( "We calculated these approximate values
\n" ); document.write( "B(0) = 0.16920839
\n" ); document.write( "B(1) = 0.31437624
\n" ); document.write( "They add up to 0.48358463 which rounds to 0.4836
\n" ); document.write( "This is the approximate probability of having at most 1 chip being incorrectly placed (i.e. 1 or fewer).\r
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\n" ); document.write( "\n" ); document.write( "If you're using technology, then...
  • BinomCDF(20,0.085,1) is the input on a TI83 or similar.
  • =BinomDist(1,20,0.085,true) is the input for a spreadsheet. Don't forget about the equal sign up front.
  • BinomialDist(20,0.085,1,true) is the input for GeoGebra (either CAS or normal mode). Or you can use the Probability Calculator in GeoGebra.

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\n" ); document.write( "\n" ); document.write( "Part C\r
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\n" ); document.write( "\n" ); document.write( "n = 20
\n" ); document.write( "p = 0.085
\n" ); document.write( "mean = expected value = n*p = 20*0.085 = 1.7
\n" ); document.write( "This result is exact and hasn't been rounded.\r
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\n" ); document.write( "\n" ); document.write( "More practice with Binomial distributions is found here
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