Respuesta :
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:
[tex]_{25}C_6=\frac{25!}{(25-6)!6!}[/tex]
[tex]_{25}C_6=\frac{25!}{19!6!}[/tex]
[tex]_{25}C_6=177100[/tex]
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: [tex]_{7}C_3[/tex]
3 keyboard with mechanical defects: [tex]_{18}C_3[/tex]
This can be written as:
[tex]_{7}C_3\times_{18}C_3=35\times816 [/tex]
[tex]_{7}C_3\times _{18}C_3=28560[/tex]
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:
[tex]\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} [/tex]
[tex]\frac{\left(18564+7\cdot8568+21\cdot3060\right)}{177100}[/tex]
[tex]0.806324110672\approx0.806[/tex]
Hence, the probability that no more than 2 electrical defect is 0.806.
