A manufacturing process has 100 customer orders to fill. Each order requires one component part that is purchased from a supplier. However, typically, 2% of the components are identified as defective, and the components can be assumed to be independent.
a. If the manufacturer stocks 100 components, what is the probability that the 100 orders can be filled without reordering components?
b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?
a. Well, let's break it down. The probability of a component being defective is 2%, which means the probability of a component NOT being defective is 98%. Since the components are assumed to be independent, the probability of all 100 components being non-defective is (0.98)^100. So, the probability that the 100 orders can be filled without reordering components is approximately:
P(100 orders filled) = (0.98)^100
b. Now, if the manufacturer stocks 102 components, they have a little bit more breathing room. The probability of all 100 components being non-defective is still the same, (0.98)^100. However, this time we need to consider the possibility of using more than 100 components. The probability of using 101 components (1 extra component) is 100C1 * (0.98)^100 * (0.02)^1. Similarly, the probability of using 102 components (2 extra components) is 100C2 * (0.98)^100 * (0.02)^2. So, the probability that the 100 orders can be filled without reordering components is approximately:
P(100 orders filled) = (0.98)^100 + 100C1 * (0.98)^100 * (0.02)^1 + 100C2 * (0.98)^100 * (0.02)^2
Remember, this is just an approximation, not an exact value. But hey, at least you'll be entertained while thinking about probabilities!
To solve this problem, we need to use the binomial distribution, as each component can be considered a "success" (i.e., a non-defective component) or "failure" (i.e., a defective component). The probability of success (getting a non-defective component) is 1 - 0.02 = 0.98, and the probability of failure (getting a defective component) is 0.02.
a. If the manufacturer stocks 100 components, we want to calculate the probability that all 100 orders can be filled without reordering components. In other words, we want to find the probability of getting 100 successes (non-defective components) in 100 trials (orders).
Using the binomial probability formula, the probability of getting exactly k successes in n trials is given by:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
where (n choose k) represents the number of possible combinations of k successes in n trials, and p is the probability of success in one trial.
In this case, since we want to fill all 100 orders without reordering components, we need to find the probability of getting 100 successes (non-defective components) in 100 trials:
P(X = 100) = (100 choose 100) * 0.98^100 * (1 - 0.98)^(100 - 100)
Using a combination calculator or formula, (100 choose 100) = 1.
P(X = 100) = 1 * 0.98^100 * (1 - 0.98)^0
Simplifying this equation, we get:
P(X = 100) = 0.98^100
Evaluating this expression, we find that the probability that the 100 orders can be filled without reordering components, given that the manufacturer stocks 100 components, is approximately 0.1338 or 13.38%.
b. If the manufacturer stocks 102 components, the calculation is similar, but now we want to find the probability of getting 100 or more successes in 100 trials:
P(X ≥ 100) = P(X = 100) + P(X = 101) + P(X = 102)
P(X = 101) = (100 choose 101) * 0.98^101 * (1 - 0.98)^(100 - 101)
P(X = 101) = 0.98^101 * (1 - 0.98)^(-1)
P(X = 102) = (100 choose 102) * 0.98^102 * (1 - 0.98)^(100 - 102)
P(X = 102) = 0.98^102 * (1 - 0.98)^(-2)
Using a combination calculator, (100 choose 101) = 100 and (100 choose 102) = 0, as it is not possible to choose more successes than there are trials.
P(X ≥ 100) = 0.98^100 + 0.98^101 * (1 - 0.98)^(-1)
Evaluating this expression, we find that the probability that the 100 orders can be filled without reordering components, given that the manufacturer stocks 102 components, is approximately 0.1869 or 18.69%.