When a new machine is functioning properly, only 2% of the items produced are defective. If 5 parts are randomly selected from this production process, what is the probability that at most 1 defective item is in the sample?
To find the probability that at most 1 defective item is in the sample, we can use the binomial probability formula. The formula for the binomial probability is:
P(x) = (n C x) * p^x * (1-p)^(n-x)
Where:
- P(x) is the probability of having x successes in n trials.
- (n C x) is the binomial coefficient, calculated as n! / (x! * (n-x)!)
- p is the probability of success in one trial.
- q is the probability of failure in one trial, calculated as (1-p).
- n is the number of trials.
In this case, we have:
- n = 5 (since we are selecting 5 parts).
- x can take on values 0 and 1 (since we want at most 1 defective item).
- p = 0.02 (since 2% of the items produced are defective).
- q = 1 - p = 1 - 0.02 = 0.98.
Now we can calculate the probability of having 0 defective items and the probability of having 1 defective item, and then sum them up to find the total probability.
First, let's calculate the probability of having 0 defective items:
P(x=0) = (5 C 0) * (0.02)^0 * (0.98)^(5-0)
Calculating this binomial probability, we get: P(x=0) = (1) * (1) * (0.98)^5 = 0.9039 (rounded to 4 decimal places).
Next, let's calculate the probability of having 1 defective item:
P(x=1) = (5 C 1) * (0.02)^1 * (0.98)^(5-1)
Calculating this binomial probability, we get: P(x=1) = (5) * (0.02) * (0.98)^4 = 0.07959 (rounded to 5 decimal places).
Now, we can sum up these probabilities to find the total probability:
P(at most 1 defective item) = P(x=0) + P(x=1) = 0.9039 + 0.07959 = 0.98349 (rounded to 5 decimal places).
Therefore, the probability that at most 1 defective item is in the sample is approximately 0.98349.