n a particular manufacturing plant, two machines (A and B) produce a particular part. One machine (B) is newer and faster. In one five-minute period, a lot consisting of 32 parts is produced. Twenty-two are produced by machine B and the rest by machine A. Suppose an inspector randomly samples a dozen of the parts from this lot.
a. What is the probability that exactly two parts were produced by machine A?
b. What is the probability that half of the parts were produced by each machine?
c. What is the probability that all of the parts were produced by machine B?
d. What is the probability that seven, eight, or nine parts were produced by machine B?
To solve this problem, we can use the concept of the binomial distribution. The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success.
In this case, each part being produced by machine A or machine B can be considered as a Bernoulli trial with a probability of success (produced by machine A) or failure (produced by machine B).
Before calculating the probabilities, let's define the parameters:
- n: the total number of parts in the sample (12)
- p: the probability of success (produced by machine A)
- q: the probability of failure (produced by machine B), which is equal to 1 - p
Now, let's answer each question:
a. What is the probability that exactly two parts were produced by machine A?
To find the probability that exactly two parts were produced by machine A, we need to calculate the probability of getting exactly two successes (parts produced by machine A) out of twelve trials.
P(X = 2) = C(n, k) * p^k * q^(n-k)
In this case, n = 12 (the total number of trials), k = 2 (the number of successes), p = 10/32 (the probability of success), and q = 1 - p.
Plugging these values into the formula gives us:
P(X = 2) = C(12, 2) * (10/32)^2 * (22/32)^10
Using a calculator or statistical software, you can calculate the final result.
b. What is the probability that half of the parts were produced by each machine?
To find the probability that half of the parts were produced by each machine, we need to calculate the probability of getting exactly six successes (parts produced by machine A) and six failures (parts produced by machine B) out of twelve trials.
P(X = 6) = C(n, k) * p^k * q^(n-k)
In this case, n = 12 (the total number of trials), k = 6 (the number of successes), p = 16/32 (the probability of success, which is half), and q = 1 - p.
Plugging these values into the formula gives us:
P(X = 6) = C(12, 6) * (16/32)^6 * (16/32)^6
Using a calculator or statistical software, you can calculate the final result.
c. What is the probability that all of the parts were produced by machine B?
To find the probability that all of the parts were produced by machine B, we need to calculate the probability of getting twelve failures (parts produced by machine B) out of twelve trials.
P(X = 12) = C(n, k) * p^k * q^(n-k)
In this case, n = 12 (the total number of trials), k = 12 (the number of successes, which is zero), p = 10/32 (the probability of success, which is zero), and q = 1 - p.
Plugging these values into the formula gives us:
P(X = 12) = C(12, 0) * (10/32)^0 * (22/32)^12
Using a calculator or statistical software, you can calculate the final result.
d. What is the probability that seven, eight, or nine parts were produced by machine B?
To find the probability that seven, eight, or nine parts were produced by machine B, we need to calculate the probabilities of getting seven, eight, or nine failures (parts produced by machine B) out of twelve trials and then sum them up.
P(X = 7) + P(X = 8) + P(X = 9) = C(n, k) * p^k * q^(n-k) + C(n, k) * p^k * q^(n-k) + C(n, k) * p^k * q^(n-k)
In this case, n = 12 (the total number of trials), k = 7, 8, and 9 (the number of successes), p = 10/32 (the probability of success), and q = 1 - p.
Plugging these values into the formula and summing the results gives us the final probability.
Note: C(n, k) is the binomial coefficient, which represents the number of ways to choose k objects out of n distinct objects and is calculated as C(n, k) = n! / (k! * (n-k)!), where ! denotes factorial.
You can use a calculator or statistical software to calculate the final results for each probability.