Question

In a meat packaging plant Machine A accounts for 80% of the plant's output, while Machine B accounts for 20% of the plant's output. In total, 5% of the packages are improperly sealed. Also, 2% of the packages are from Machine A and are improperly sealed. a. If a package is selected at random, what is the probability that it will be properly sealed? (2) b. If a package selected at random came from Machine A, what is the probability that it is improperly sealed? (2)

Answer 1

Explanation:

a. The probability that a randomly selected package is properly sealed can be found by subtracting the probability that it is improperly sealed from 1:

P(properly sealed) = 1 - P(improperly sealed)

We know from the problem that 5% of the packages are improperly sealed. Therefore, the probability that a package is properly sealed is:

P(properly sealed) = 1 - 0.05 = 0.95

So the probability that a package selected at random is properly sealed is 0.95.

b. We are given that 80% of the plant's output comes from Machine A and that 2% of the packages from Machine A are improperly sealed. We want to find the probability that a package selected at random from Machine A is improperly sealed. This is a conditional probability, which can be calculated using Bayes' theorem:

P(improperly sealed | Machine A) = P(Machine A and improperly sealed) / P(Machine A)

We already know that P(Machine A) = 0.8 and that P(Machine A and improperly sealed) = 0.02, so we can substitute these values into the formula:

P(improperly sealed | Machine A) = 0.02 / 0.8 = 0.025

So the probability that a package selected at random from Machine A is improperly sealed is 0.025, or 2.5%.