Final answer:
The mean number of defective items in a sample of two items drawn without replacement from an urn with 6 items including 2 defective ones is \(\frac{2}{3}\).
Step-by-step explanation:
To find the mean number of defective items in a sample of two items drawn without replacement from an urn containing 6 items, which includes 2 defective items, we can use the concept of expected value in probability. The events to consider are: drawing two non-defective items (NDND), one defective and one non-defective item in either order (DNDN or NDND), or two defective items (DD).
We calculate the probability for each of these scenarios and multiply each probability by the number of defective items in that scenario (0 for NDND, 1 for DNDN or NDND, and 2 for DD), and then sum these products to get the mean.
The probabilities are:
- P(NDND) = \(\frac{4}{6} \times \frac{3}{5} = \frac{2}{5}\)
- P(DNDN or NDND) = \(\frac{2}{6} \times \frac{4}{5} + \frac{4}{6} \times \frac{2}{5} = \frac{4}{15} + \frac{4}{15} = \frac{8}{15}\)
- P(DD) = \(\frac{2}{6} \times \frac{1}{5} = \frac{1}{15}\)
The mean is:
\(\text{Mean} = (0 \times \frac{2}{5}) + (1 \times \frac{8}{15}) + (2 \times \frac{1}{15}) = 0 + \frac{8}{15} + \frac{2}{15} = \frac{10}{15} = \frac{2}{3}\)
So, the mean number of defective items in the sample is \(\frac{2}{3}\).