An inspection procedure at a manufacturing plant involves picking three items at random and then accepting the whole lot if at least two of the three items are in perfect condition. If in reality 90% of a certain very large lot are perfect, what is the probability that the lot will be accepted? Round your answer to the nearest thousandth (3 decimal places).
To find the probability that the lot will be accepted, we need to calculate the probability of picking at least two perfect items out of three.
The probability of picking a perfect item at random is 90% or 0.9 since 90% of the lot is perfect. The probability of picking a defective item is then 1 - 0.9 = 0.1.
We can use the concept of binomial probability to calculate the probability of picking a specific number of perfect items in a given number of trials. In this case, we want to find the probability of picking at least two perfect items out of three trials.
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
P(X = k) is the probability of getting k success (in our case perfect items)
C(n, k) is the combination function (n choose k) which calculates the number of ways to choose k items out of n
p is the probability of success (picking a perfect item)
n is the number of trials (number of items we are picking)
Let's calculate the probability for each possible outcome:
P(2 out of 3 perfect items) = C(3, 2) * (0.9^2) * (0.1^1) = 3 * 0.81 * 0.1 = 0.243
P(3 out of 3 perfect items) = C(3, 3) * (0.9^3) * (0.1^0) = 1 * 0.729 * 1 = 0.729
The probability of accepting the lot is the sum of these probabilities:
P(accepting the lot) = P(2 out of 3 perfect items) + P(3 out of 3 perfect items)
= 0.243 + 0.729
= 0.972
So, the probability that the lot will be accepted is 0.972 or approximately 0.972 (rounded to three decimal places).