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Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboards, 13 of which have electrical defects and 12 of which have mechanical defects.

(a) In how many ways can a sample of 7 keyboards be selected so that exactly two have an electrical defect?
(b) If a sample of 7 keyboards is randomly selected, what is the probability that at least 6 of these will have a mechanical defect? (Round your answer to four decimal places.)

User Both FM
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Final answer:

The number of ways to select a sample of 7 keyboards with exactly two having an electrical defect is 61,776. The probability of selecting a sample with at least 6 keyboards having a mechanical defect is 0.00195.

Step-by-step explanation:

(a) To select a sample of 7 keyboards with exactly two having an electrical defect, we need to consider the number of ways we can choose 2 keyboards with electrical defects and 5 keyboards without electrical defects. The number of ways to select 2 out of 13 electrical defects is C(13, 2) = 78. The number of ways to select 5 out of the remaining 12 keyboards without electrical defects is C(12, 5) = 792. Therefore, the total number of ways to select a sample of 7 keyboards with exactly two having an electrical defect is 78 * 792 = 61776.

(b) To find the probability that at least 6 out of 7 randomly selected keyboards have a mechanical defect, we need to consider the number of ways we can choose 6 or 7 keyboards with mechanical defects and 1 or 0 keyboards without mechanical defects. The number of ways to select 6 out of 12 keyboards with mechanical defects is C(12, 6) = 924. The number of ways to select 1 out of the 13 remaining keyboards is C(13, 1) = 13. Therefore, the total number of ways to select a sample of 7 keyboards with at least 6 having a mechanical defect is 924 + 13 = 937. The total number of ways to select a sample of 7 keyboards from the 25 available keyboards is C(25, 7) = 480,700. Therefore, the probability of selecting a sample with at least 6 keyboards having a mechanical defect is 937 / 480700 = 0.00195 (rounded to four decimal places).

User Chamara Abeysekara
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