Respuesta :
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
A) The number of ways to randomly select 5 of these keyboards for a thorough inspection (without regard to order) is; 53130 ways
B) The number of ways that a sample of 5 keyboards be selected so that exactly two have an electrical defect is; 14535 ways
C) The probability that at least 4 of these will have a mechanical defect if a sample of 5 keyboards is randomly selected is; 0.6566 or 65.66%
Probability Combinations
We are given that;
- Total defective keyboards = 25
- Keyboards with electrical defects = 6
- Keyboards with mechanical defects = 5
A) The number of ways to randomly select 5 of these keyboards for a thorough inspection (without regard to order) is gotten by combination formula which is; nCr
Thus, we have;
25C5 = 25!/(5! × (20 - 5)!)
>> 53130 ways
B) The number of ways that a sample of 5 keyboards can be selected so that exactly two have an electrical defect is;
6C2 × 19C3 = 14535 ways
C) The probability that at least 4 of these will have a mechanical defect if a sample of 5 keyboards is randomly selected is gotten by getting the total number of cases first which is;
N = (19C4 × 6C1) + (19C5 × 6C0)
N = 23256 + 11628
N = 34884
Total possible cases of choosing 5 keyboards is;
25C5 = 53130
Thus;
P(at least 4 of these will have a mechanical defect if a sample of 5 keyboards is randomly selected) = 34884/53130
>> 0.6566
Read more about combinations at; https://brainly.com/question/11871015
