At a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 19% of the time. A random sample of 7 parts produced by this machine is chosen. Find the probability that at most 2 of these parts are defective.Do not round your intermediate computations, and round your answer to three decimal places.

To solve this problem, we can use the binomial probability formula. The binomial probability formula is given by:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

where:

• $n$ is the number of trials (in this case, 7 parts),
• $k$ is the number of successes (defective parts),
• $p$ is the probability of success on a single trial (0.19 for defective parts),
• $\binom{n}{k}$ is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$.

We need to find the probability that at most 2 parts are defective, which means we need to calculate the sum of the probabilities for 0, 1, and 2 defective parts.

1. Calculate $P(X = 0)$: $P(X = 0) = \binom{7}{0} (0.19)^0 (0.81)^7$ $P(X = 0) = 1 \cdot 1 \cdot (0.81)^7$ $P(X = 0) = (0.81)^7$ $P(X = 0) \approx 0.2097152$

2. Calculate $P(X = 1)$: $P(X = 1) = \binom{7}{1} (0.19)^1 (0.81)^6$ $P(X = 1) = 7 \cdot 0.19 \cdot (0.81)^6$ $P(X = 1) = 7 \cdot 0.19 \cdot 0.262144$ $P(X = 1) \approx 0.3486848$

3. Calculate $P(X = 2)$: $P(X = 2) = \binom{7}{2} (0.19)^2 (0.81)^5$ $P(X = 2) = \frac{7!}{2!(7-2)!} \cdot (0.19)^2 \cdot (0.81)^5$ $P(X = 2) = 21 \cdot 0.0361 \cdot 0.3764768$ $P(X = 2) \approx 0.2851792$

Now, sum these probabilities to find the probability that at most 2 parts are defective:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$ $P(X \leq 2) \approx 0.2097152 + 0.3486848 + 0.2851792$ $P(X \leq 2) \approx 0.8435792$

Rounding to three decimal places, the probability that at most 2 parts are defective is:

$P(X \leq 2) \approx 0.844$