The probability that a certain production process will produce a defective part is 0.20. If a box contains 12 parts, the probability that there will be 6 or 7 defective parts is ??
To find the probability of having 6 or 7 defective parts in a box of 12 parts, we can use the binomial probability formula.
The binomial distribution is used when there are two possible outcomes for each trial, with a fixed number of trials and the probability of success remains constant. In this case, the two possible outcomes are producing a defective part (success) or producing a non-defective part (failure), and each part is an independent trial.
The binomial probability formula is:
P(X = k) = (n C k) * p^k * (1-p)^(n-k)
Where:
- P(X = k) is the probability of having exactly k successes
- n is the number of trials
- k is the number of successes
- p is the probability of success for a single trial
- (n C k) is the binomial coefficient, calculated as n! / (k! * (n-k)!)
In this case, we want to calculate the probability of having 6 or 7 defective parts in a box of 12 parts. Let's calculate it step by step:
For k = 6:
P(X = 6) = (12 C 6) * (0.20^6) * (0.80^(12-6))
= (12! / (6! * (12-6)!) * (0.20^6) * (0.80^6)
= (12! / (6! * 6!)) * (0.20^6) * (0.80^6)
For k = 7:
P(X = 7) = (12 C 7) * (0.20^7) * (0.80^(12-7))
= (12! / (7! * (12-7)!) * (0.20^7) * (0.80^5)
= (12! / (7! * 5!)) * (0.20^7) * (0.80^5)
Now, we can calculate the probabilities:
P(X = 6) = (12! / (6! * 6!)) * (0.20^6) * (0.80^6)
P(X = 7) = (12! / (7! * 5!)) * (0.20^7) * (0.80^5)
Finally, we can sum the probabilities to find the probability of having 6 or 7 defective parts:
P(6 or 7 defective parts) = P(X = 6) + P(X = 7)