Question 1183756
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The quality assurance engineer of a receiving-sets manufacturer inspects receiving-sets in lots of 50. 
He selects 5 of the 50 receiving-sets at random and inspects them thoroughly. 
Assuming that 6 of the 50 receiving-sets in the current lot are defective, 
find the probability that exactly 2 of the 5 receiving-sets selected by the engineer are defective.
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I solved a TWIN problem several days ago under this link

https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1183648.html



Now, I copy-paste that solution and change/recalculate the numbers in it accordingly.



<pre>
P = {{{(C[6]^2*C[50-6]^3)/C[50]^5}}} = {{{(C[6]^2*C[44]^3)/C[50]^5}}} = {{{(15*13244)/2118760}}} = 0.0093762.      <U>ANSWER</U>



The numerator of P,  {{{C[6]^2*C[44]^3}}},  is the number of favorable combinations of 5 of the 50 receiving-sets, 
that contain exactly 2 defective receiving-sets and 3 good receiving-set.



It is the favorable set.



The denominator is the total number of 5 receiving-sets selected by the engineer from 50 receiving-sets.


The probability, as always in such problem, is the ratio  {{{favorable/total}}}.
</pre>

Solved.


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&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Elementary-Probability-problems-related-to-Combinations.lesson>Elementary Probability problems related to combinations</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Elementary-Probability-problems-related-to-combinations-REVISITED.lesson>Elementary Probability problems related to combinations REVISITED</A> 

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The solution by @Theo is incorrect.


He mistakenly assumed that the problem is on Binomial distribution.


In reality, it is not so.