Question

Suppose that a batch of 100 items contains 6 that are defective and 94 that are not defective. Let X be the number of defective items in a randomly selected sample of 10 items from the batch. (a) Determine P(X=0). (b) Determine P(X>2).

Answer 1

The probability of getting 0 defective items out of 10 is 0.7374. The probability of getting more than 2 defective items out of 10 is 0.0349.

Step-by-step explanation:

To find the probability, we need to use the concept of binomial distribution. The probability of getting 0 defective items out of 10 can be calculated using the formula:

P(X = 0) = (n C x) * p^x * (1 - p)^(n - x)

where n is the sample size (10), x is the number of defective items (0), and p is the probability of a defective item (6/100). Plugging in these values, we get:

P(X = 0) = (10 C 0) * (0.06)^0 * (0.94)^(10 - 0)

P(X = 0) = 0.7374

To find the probability of getting more than 2 defective items out of 10, we need to calculate the probability of getting 3, 4, 5, 6, 7, 8, 9, or 10 defective items and sum them up:

P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

Using the same formula as before, we can calculate each probability and sum them up to find P(X > 2) = 0.0349.

The probability of getting 0 defective items out of 10 is 0.7374. The probability of getting more than 2 defective items out of 10 is 0.0349.