Suppose a shipment of 500 machine parts contains 62 defective and 438 non-defective machine parts. From the shipment you take a random sample of 25. You are interested in the number of defective machine parts out of 25 trials and corresponding probabilities.
I believe the correct probability is .08. Is this correct? Thanks!!
.08 is the probability for defective parts.
To determine if your belief is correct, we need to calculate the probability of obtaining a specific number of defective machine parts out of 25 trials.
We can approach this problem using the binomial probability formula, which is given by:
P(X = k) = (n C k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) represents the probability of getting exactly k successes (defective machine parts) out of n trials.
- (n C k) represents the number of combinations of n items taken k at a time.
- p represents the probability of success on a single trial (defective machine parts).
- (1 - p) represents the probability of failure on a single trial (non-defective machine parts).
- n represents the number of trials (in this case, 25).
Let's calculate the probability of getting 0 defective machine parts (k = 0) out of 25 trials.
P(X = 0) = (25 C 0) * (0.62^0) * (0.38^25)
= 1 * 1 * 0.000292
The probability of getting 0 defective machine parts out of 25 trials is approximately 0.000292.
Since the probability of getting 0 defective machine parts is very small, it is unlikely that the correct probability is 0.08. However, let's calculate it to confirm.
P(X = 1) = (25 C 1) * (0.62^1) * (0.38^24)
= 25 * 0.62 * 0.000758
The probability of getting 1 defective machine part out of 25 trials is approximately 0.011825.
As you can see, the calculated probability is not 0.08. The actual probability depends on the desired number of defective parts, but it is unlikely to reach 0.08 based on the provided information.