Questions: A quality control inspector has drawn a sample of 19 light bulbs from a recent production lot. If the number of defective bulbs is 1 or more, the lot fails inspection. Suppose 30% of the bulbs in the lot are defective. What is the probability that the lot will fall inspection? Round your answer to four decimal places.

Solution Steps

To solve this problem, we need to use the binomial distribution. The binomial distribution describes the number of successes in a fixed number of independent Bernoulli trials. Here, a "success" is drawing a defective bulb. We need to calculate the probability that at least one bulb is defective in a sample of 19 bulbs, given that 30% of the bulbs are defective.

1. Calculate the probability of drawing 0 defective bulbs.
2. Subtract this probability from 1 to get the probability of drawing 1 or more defective bulbs.

We need to find the probability that a lot of light bulbs will fail inspection. The lot fails inspection if at least one of the 19 sampled bulbs is defective. Given that 30% of the bulbs are defective, we need to calculate the probability of drawing 1 or more defective bulbs from the sample.

Using the binomial distribution, the probability of drawing exactly 0 defective bulbs (\(P(X = 0)\)) is given by: \[ P(X = 0) = \binom{n}{0} p^0 (1-p)^{n-0} \] where \(n = 19\) and \(p = 0.30\). This simplifies to: \[ P(X = 0) = (1 - 0.30)^{19} = 0.0011398895185373125 \approx 0.0011 \]

The probability of drawing 1 or more defective bulbs is the complement of drawing 0 defective bulbs: \[ P(X \geq 1) = 1 - P(X = 0) \] Substituting the value from Step 2: \[ P(X \geq 1) = 1 - 0.0011 = 0.9989 \]

Final Answer