Answer:
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
