A factory makes toys The probability that a toy is defective is 008 A random sample of 50 toys istestedGiven that there is at least one defective toy in the sample find the probability that there are

We are given that the probability of a toy being defective is 0.08, so the probability of a toy not being defective is 0.92.

Let X be the number of defective toys in the sample. We want to find P(X ≤ 3 | X ≥ 1).

Using Bayes' theorem:

P(X ≤ 3 | X ≥ 1) = P(X ≥ 1 | X ≤ 3) * P(X ≤ 3) / P(X ≥ 1)

We can find each of these probabilities as follows:

P(X ≥ 1 | X ≤ 3) = 1 (if there are 3 or fewer defective toys, then there must be at least one defective toy) P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = (0.92)^50 + 50*(0.08)(0.92)^49 + (5049/2)(0.08)^2(0.92)^48 + (504948/6)(0.08)^3(0.92)^47 ≈ 0.9999 (using a calculator) P(X ≥ 1) = 1 - P(X = 0) = 1 - (0.92)^50 ≈ 0.9999

Substituting into the formula:

P(X ≤ 3 | X ≥ 1) = 1 * 0.9999 / 0.9999 ≈ 1

Therefore, given that there is at least one defective toy in the sample, the probability that there are at most three defective toys is approximately 1