Answer:
The number of ways to select 6 keyboards is 177100.
The number of ways so that exactly three have a mechanical defect is 28560.
The probability that no more than 2 electrical defect is 0.806.
Step-by-step explanation:
Consider the provided information.
A repair facility currently has 25 failed keyboards, 7 of which have electrical defects and 18 of which have mechanical defects.
Part(A)
It is given that we need to find How many ways are there to randomly select 6 of these keyboards for a thorough inspection.
We need to select 6 keyboards out of 25.
Thus, the number of ways are:
![_{25}C_6=\frac{25!}{(25-6)!6!}](https://tex.z-dn.net/?f=_%7B25%7DC_6%3D%5Cfrac%7B25%21%7D%7B%2825-6%29%216%21%7D)
![_{25}C_6=\frac{25!}{19!6!}](https://tex.z-dn.net/?f=_%7B25%7DC_6%3D%5Cfrac%7B25%21%7D%7B19%216%21%7D)
![_{25}C_6=177100](https://tex.z-dn.net/?f=_%7B25%7DC_6%3D177100)
Hence, the number of ways to select 6 keyboards is 177100.
Part(B)
In how many ways can a sample of 6 keyboards be selected so that exactly three have a mechanical defect?
We need to select 6 keyboards if exactly three have a mechanical defect that means the remain 3 have a electrical defect.
3 keyboard with electrical defects: ![_{7}C_3](https://tex.z-dn.net/?f=_%7B7%7DC_3)
3 keyboard with mechanical defects: ![_{18}C_3](https://tex.z-dn.net/?f=_%7B18%7DC_3)
This can be written as:
![_{7}C_3\times_{18}C_3=35\times816](https://tex.z-dn.net/?f=_%7B7%7DC_3%5Ctimes_%7B18%7DC_3%3D35%5Ctimes816%20)
![_{7}C_3\times _{18}C_3=28560](https://tex.z-dn.net/?f=_%7B7%7DC_3%5Ctimes%20_%7B18%7DC_3%3D28560)
Thus, the number of ways so that exactly three have a mechanical defect is 28560.
Part(C)
If a sample of 6 keyboards is randomly selected, what is the probability that no more than 2 of these will have an electrical defect?
No more than 2 of these will have an electrical defect that means the number of electrical defect keyboards can be 0, 1 or 2.
P(No more than 2 electrical defect)={P(0 electrical defect)+P(1 electrical defect)+P(2 electrical defect)}
This can be written as:
![\frac{_{7}C_0\times _{18}C_6}{_{25}C_6}+\frac{_{7}C_1\times _{18}C_5}{_{25}C_6}+\frac{_{7}C_2\times _{18}C_4}{_{25}C_6}](https://tex.z-dn.net/?f=%5Cfrac%7B_%7B7%7DC_0%5Ctimes%20_%7B18%7DC_6%7D%7B_%7B25%7DC_6%7D%2B%5Cfrac%7B_%7B7%7DC_1%5Ctimes%20_%7B18%7DC_5%7D%7B_%7B25%7DC_6%7D%2B%5Cfrac%7B_%7B7%7DC_2%5Ctimes%20_%7B18%7DC_4%7D%7B_%7B25%7DC_6%7D%20)
![\frac{\left(18564+7\cdot8568+21\cdot3060\right)}{177100}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cleft%2818564%2B7%5Ccdot8568%2B21%5Ccdot3060%5Cright%29%7D%7B177100%7D)
![0.806324110672\approx0.806](https://tex.z-dn.net/?f=0.806324110672%5Capprox0.806)
Hence, the probability that no more than 2 electrical defect is 0.806.