# Sample Space & Probability

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The manager of a small business receives a shipment of five hard drives. Three hard drives are good and 2 are defective. If a computer repair shop purchases 3 of these stock of hard drives from the small business and the selection was made at random:

1. List the elements of the sample space S using the letters G and D for good and defective respectively.

2. If x represents the number of defective hard drives in a random sample of three hard drives, list in the form of a table, the values of x for each point in the sample space S and the corresponding frequencies.

3. What is the probability that the repair shop purchased at least 1 defective hard drive?

4. What is the probability the repair shop purchased more than 2 defective hard drives?

5. What is the mean number of defective hard drives that the small business receives if all their shipments come in sets of five that contain the same proportion of defective hard drives?

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The expert lists the elements of the sample space using letters for good and defectives.

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Sample Space & Probability

The manager of a small business receives a shipment of five hard drives. Three hard drives are good and 2 are defective. If a computer repair shop purchases 3 of this stock of hard drives from the small business and the selection was made at random,

Q. 1 (5 Points). List the elements of the sample space S using the letters G and D for good and defective respectively.

There are several ways to think about this problem, one of which is a tree diagram to help us map out all possible results AND to help us calculate the various probabilities that are asked for later on. So here's the tree diagram with G meaning that a "good" (non-defective) hard drive was selected and D meaning that a "defective" hard drive was selected, and the fractions (which I call twig probabilities) correspond to the probability of that particular path-segment being the one that is chosen at that point in the tree. By multiplying the three twig probabilities in a particular branch (or path through the tree, from start to finish) we may calculate each of the 7 Branch probabilities.

Outcome

1/3 G Branch #1 (B1) = (G, G, G)

G

2/4 2/3 D Branch #2 (B2) = (G, G, D)

G

2/4 2/3 G Branch #3 (B3) = (G, D, G)

3/5 D

1/3 D Branch #4 (B4) = (G, D, D)

o

2/3 G Branch #5 (B5) = (D, G, G)

2/5 G

3/4 1/3 D Branch #6 (B6) = (D, G, D)

D

1/4 3/3=1 G Branch #7 (B7) = (D, D, G)

D Normally there would be a Branch #8

at the bottom of a tree diagram like this, but since there are assumed to be only 2 defective hard drives and this last branch has already selected both Ds prior to the selection of the 3rd hard drive, you're guaranteed in this last branch that the 3rd hard drive selected is a ...

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