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Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboards, 8 of which have electrical defects and 17 of which have mechanical defects.
(a) How many ways are there to randomly select 6 of these keyboards for a thorough inspection (without regard to order)?
(b) In how many ways can a sample of 6 keyboards be selected so that exactly three have a mechanical defect?
(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?

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Answer:

a) There are a total of 177100 ways of picking 6 keyboards

b) There are 38080 ways of picking 6 keyboards so that exactly 3 of them will have a mechanical defect

c) The probability that no more than 2 of the keyboards selected will have an electrical defect is 0.7257

Step-by-step explanation:

a) Since we dont care about the order, the total amount of ways to select 6 keyboards from 25 is the combinatoial number of 25 with 6, [tex] {25 \choose 6} = \frac{25!}{6!(25-6)!} = 177100 [/tex] .

b) We have to select 3 keyboards from the set of 8 that have a electrical effect and 3 keyboards from the set of 17 that have a mechanical effect, and then multiply (to count all possible combinations). The result is [tex] {8 \choose 3} * {17 \choose 3} = 38080 [/tex]

c) We have to count the total amount of favourable cases, this is, the total amount of keyboard combinations with at most 2 have electrical defects. After counting all favourable cases, we divide the result by the total amount of cases, 177100. We can compute the total amount of favourable cases by summing the cases in which we have 0, 1 and 2 keyboards with electrical defects (remember that the remaining will have mechanical defects).

For 0, there are [tex] {8 \choose 0} * {17 \choose 6} = 12376 [/tex] cases

For 1, the total amount of cases is [tex] {8 \choose 1} * {17 \choose 5} = 49504

For 2, there are [tex] {8 \choose 2} * {17 \choose 4} = 66640 [/tex] cases

Thus, there are 12376+49504+66640 = 128520 favourable cases from 177100 total. The prbability that no more than 2 of the keyboards in the sample will have an electrical defect is 128520/177100 = 0.7257

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