5!/2!=120/2= 60 arrangements
Answer:
The probability that a container will be shipped even though it contains 2 defectives if the sample size is 88, will be 
Step-by-step explanation:
The first step is to count the number of total possible random sets of taking a sample size of 88 engines over 1212 engines of the population, so ![\left[\begin{array}{ccc}1212\\88\end{array}\right] =1212C88=4.7205x10^{135}](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1212%5C%5C88%5Cend%7Barray%7D%5Cright%5D%20%3D1212C88%3D4.7205x10%5E%7B135%7D)
The second step is to count the number of total possible random sets of taking a sample size of 88 engines over 1210 engines (discounting the 2 defective engines) as the possible ways to succeed, so ![\left[\begin{array}{ccc}1210\\88\end{array}\right] =1212C88=4.0596x10^{135}](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1210%5C%5C88%5Cend%7Barray%7D%5Cright%5D%20%3D1212C88%3D4.0596x10%5E%7B135%7D)
Finally we need to compute
, therefore the probability that a container will be shipped is