Question 1208645
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We can use the Binomial probability distribution because:<ul><li>There are two outcomes per trial. Either a chip is placed correctly or it is not.</li><li>Each trial is independent.</li><li>The probability of not getting placed correctly is the same for each trial.</li></ul>------------------------------------------------------------------------------------


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 <a href="https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php">nCr combination formula</a>. These values are found in <a href="https://www.mathsisfun.com/pascals-triangle.html">Pascal's Triangle</a>. A quick way to calculate the nCr values is to use the <a href="https://support.microsoft.com/en-us/office/combin-function-12a3f276-0a21-423a-8de6-06990aaf638a">Combin</a> function in a spreadsheet. Or you can use a <a href="https://education.ti.com/en/customer-support/knowledge-base/ti-83-84-plus-family/product-usage/34718">TI Calculator</a>. 
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) = <font color=blue>0.16920839 approximately</font>


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) = <font color=blue>0.31437624 approximately</font>


Repeat similar steps to find that B(2) = <font color=blue>0.27744132 approximately</font> which is the probability of having exactly 2 chips incorrectly placed.


Then we could say,
<font size=5>(</font> <font color=blue>B(0)+B(1)+B(2)</font> <font size=5>)</font> + <font size=5>(</font> <font color=red>B(3)+B(4)+...+B(19)+B(20)</font> <font size=5>)</font> = 1
<font color=red>B(3)+B(4)+...+B(19)+B(20)</font> = 1 - <font size=5>(</font> <font color=blue>B(0)+B(1)+B(2)</font> <font size=5>)</font>
B(3)+B(4)+...+B(19)+B(20) = 1 - ( <font color=blue>0.16920839+0.31437624+0.27744132</font> )
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) = <font color=red>0.2390</font> 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 <a href="https://www.statology.org/binomial-probabilities-ti-84-calculator/">BinomCDF</a>
The order of inputs is: n, p, k<pre>n = 20
p = 0.085
k = 2</pre>Type in <font color=red>1-BinomCDF(20,0.085,2)</font> to get the approximate result <font color=red>0.2390</font> 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.<ul><li>Search out "binomial CDF calculator". <a href="https://stattrek.com/online-calculator/binomial">This page</a> and <a href="https://www.gigacalculator.com/calculators/binomial-probability-calculator.php">this page</a> are two of many results. Feel free to explore your favorite.</li><li>Use the <a href="https://geogebra.github.io/docs/manual/en/Probability_Calculator/">Probability Calculator in GeoGebra</a>. Select "binomial" from the dropdown menu. Type n = 20 and p = 0.085; the goal is to calculate *[Tex \large P( 3 \le \text{x} \le 20 )]</li><li>Use the spreadsheet command called <a href="https://support.google.com/docs/answer/3093987?hl=en">BinomDist</a>. The input would be <font color=red>=1-BinomDist(2,20,0.085,true)</font></li><li>Use the <a href="https://geogebra.github.io/docs/manual/en/commands/BinomialDist/">BinomialDist</a> command in GeoGebra. Note "binomial" instead of "binom". The input would be <font color=red>1-BinomialDist(20,0.085,2,true)</font></li></ul>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 <font color=red>0.4836</font>
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...<ul><li><font color=red>BinomCDF(20,0.085,1)</font> is the input on a TI83 or similar.</li><li><font color=red>=BinomDist(1,20,0.085,true)</font> is the input for a spreadsheet. Don't forget about the equal sign up front.</li><li><font color=red>BinomialDist(20,0.085,1,true)</font> is the input for GeoGebra (either CAS or normal mode). Or you can use the <a href="https://geogebra.github.io/docs/manual/en/Probability_Calculator/">Probability Calculator in GeoGebra</a>. </li></ul>
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Part C


n = 20
p = 0.085
mean = expected value = n*p = 20*0.085 = <font color=red>1.7</font>
This result is exact and hasn't been rounded.


More practice with Binomial distributions is found <a href="https://www.algebra.com/statistics/Binomial-probability/Binomial-probability.faq.question.1208472.html">here</a>
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