Question

A factory received a shipment of 26 compressors, and the vendor who sold the items knows there are 4 compressors in the shipment that are defective. Before the receiving foreman accepts the delivery, he samples the shipment, and if too many of the compressors in the sample are defective, he will refuse the shipment.

For each of the following, give your responses as reduced fractions.

If a sample of 4 compressors is selected, find the probability that all in the sample are defective.

If a sample of 4 compressors is selected, find the probability that none in the sample are defective.

Answer 1

Let's calculate the probabilities step by step:

Total number of compressors = 26
Number of defective compressors = 4

(a) Probability that all 4 compressors in the sample are defective:

To find this probability, we need to consider that we have 4 defective compressors and 22 non-defective compressors in the shipment. When drawing the first compressor, there are 4 defective out of 26, so the probability is 4/26. For the second compressor, there are now 3 defective out of 25, so the probability is 3/25. Similarly, for the third compressor, it's 2/24, and for the fourth, it's 1/23.

Now, we multiply these probabilities together since each draw is independent:

(4/26) * (3/25) * (2/24) * (1/23)

Now, calculate this product to find the probability:

(a) Probability = (4/26) * (3/25) * (2/24) * (1/23) ≈ 0.0001393 (rounded)

So, the probability that all 4 compressors in the sample are defective is approximately 0.0001393.

(b) Probability that none of the 4 compressors in the sample are defective:

To find this probability, we consider that when drawing the first compressor, there are 22 non-defective out of 26, so the probability is 22/26. For the second compressor, it's 21/25, for the third, it's 20/24, and for the fourth, it's 19/23.

Now, multiply these probabilities together:

(22/26) * (21/25) * (20/24) * (19/23)

Calculate this product to find the probability:

(b) Probability = (22/26) * (21/25) * (20/24) * (19/23) ≈ 0.5471 (rounded)

So, the probability that none of the 4 compressors in the sample are defective is approximately 0.5471.