Question 1208645: An integrated circuit manufacturer produces wafers that contain
20 chips. Each chip has a probability of 0.085 of not placed correctly on the wafer.
A) Find the probability that a wafer contains at least 3
incorrectly place chips. (Ans = 0.2390)
b) What is the probability that a wafer no more than one
incorrectly place chips on a wafer? (Ans = 0.4836)
c) What is the expected number of incorrectly placed chips on a
wafer? (Ans = 1.7)
Answer by math_tutor2020(3817)   (Show Source):
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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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Part A
n = 20 = sample size = number of chips
p = 0.085 = probability of a chip being incorrectly placed
x = number of chips incorrectly placed
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}
B(x) = binomial probability of exactly x chips being incorrectly placed
B(x) = (nCx)*(p^x)*(1-p)^(n-x)
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.
There are many options to calculate nCr.
Let's find the probability of exactly x = 0 chips that are incorrectly placed.
B(x) = (nCx)*(p^x)*(1-p)^(n-x)
B(x) = (20Cx)*(0.085^x)*(1-0.085)^(20-x)
B(0) = (20C0)*(0.085^0)*(1-0.085)^(20-0)
B(0) = (1)*(0.085^0)*(1-0.085)^(20-0)
B(0) = 0.16920839 approximately
Now let's find the probability of exactly x = 1 chip being incorrectly placed.
B(x) = (20Cx)*(0.085^x)*(1-0.085)^(20-x)
B(1) = (20C1)*(0.085^1)*(1-0.085)^(20-1)
B(1) = (20)*(0.085^1)*(1-0.085)^(20-1)
B(1) = 0.31437624 approximately
Repeat similar steps to find that B(2) = 0.27744132 approximately which is the probability of having exactly 2 chips incorrectly placed.
Then we could say,
( B(0)+B(1)+B(2) ) + ( B(3)+B(4)+...+B(19)+B(20) ) = 1
B(3)+B(4)+...+B(19)+B(20) = 1 - ( B(0)+B(1)+B(2) )
B(3)+B(4)+...+B(19)+B(20) = 1 - ( 0.16920839+0.31437624+0.27744132 )
B(3)+B(4)+...+B(19)+B(20) = 1 - 0.76102595
B(3)+B(4)+...+B(19)+B(20) = 0.23897405
B(3)+B(4)+...+B(19)+B(20) = 0.2390 which is the answer that your textbook mentioned.
If you want to use technology to quickly find the answer, then you can use a TI83 or similar.
The command you'll use is called BinomCDF
The order of inputs is: n, p, kn = 20
p = 0.085
k = 2 Type in 1-BinomCDF(20,0.085,2) to get the approximate result 0.2390 when rounding to 4 decimal places.
You can round manually or you can use the Round command in the TI83.
Note that the BinomCDF(20,0.085,2) portion computes the sum B(0)+B(1)+B(2)
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.
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Part B
Refer to the previous part.
We calculated these approximate values
B(0) = 0.16920839
B(1) = 0.31437624
They add up to 0.48358463 which rounds to 0.4836
This is the approximate probability of having at most 1 chip being incorrectly placed (i.e. 1 or fewer).
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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Part C
n = 20
p = 0.085
mean = expected value = n*p = 20*0.085 = 1.7
This result is exact and hasn't been rounded.
More practice with Binomial distributions is found here
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