Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboards, 6 of which have electrical defects and 19 of which have mechanical defects. a. How many ways are there to randomly select 5 of these keyboards for a thorough inspection (without regard to order)?
b. In how many ways can a sample of 5 keyboards be selected so that exactly two have an electrical defect?
c. If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect?

A. 53130

B. 1 4535

C.  0.656578

Explanation:

a.

Failed keyboards = 25

The number of electrical defects = 6

The number of mechanical defects = 19

using Combination Formulas since the question stated that order isn't required.

Combination formula is nCr = n! / r! * (n - r)!,

In the above formula,  n = The number of the population

r= the number of items chosen randomly whenever there is selection

In the question, n= 25, r= 5

so putting this in the formula we have:

nCr =  25! / 5! (25-5)!

nCr =  25! / 5! (20)!

so, 25! is 25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 divided by 5! (20)!

where 5! = 5 x 4 x 3 x 2 x 1

and (20)! = 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

25C5 = 53130

b.

To do this, we will find the combination of the value of keyboards that have have electrical defects then multiply it be the value of the combination of keyboards that have mechanical defects.

for electrical defects, n= 6

6C2 =  6! / 2! (6-2)!

Note we use 2 because of the value given from the question (In how many ways can a sample of 5 keyboards be selected so that exactly two have an electrical defect?)

6C2 =  6! / 2! (4)!

6C2 = 6 x 5 x 4 x 3 x 2 x 1 / 2 x 1 (4 x 3 x 2 x 1 )

6C2 = 720 / 2(24‬)

6C2 = 15‬

for mechanical defects, n = 19, selecting the other 3, so r = 3

19C3 =  19! / 2! (19-3)!

19C3 = 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 / 3 x 2 x 1 ( x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1)

19C3 = 121,645,100,408,832,000‬ / 2 (355,687,428,096,000‬)

19C3 =969‬

19C3  x 6C2 =  969 x 15‬ = 1 4535

C.

For mechanical defect = we add to events together

19C4 =  19! / 4! (19-4)! + (19C5 = 19! / 5! (19-5)!)

19C4=  23256

19C5  = 11628

so adding these events 23256  +  11628  = 34884

Since the total of 25 combinating 5 selections, that is 25C5 =  53130

Then the probability that at least 4 of these will have a mechanical defect

= 34884 / 53130

= 0.656578

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