When a machine is working properly only 4% of the units it produces are defective.When it slips out of adjustment, the rate jumps to 25% and the machine has to be shut down and adjusted.On 90% of the mornings when production restarts, the machine is in proper adjustment.You turn on the machine this morning and produce the first unit.If the first unit is not defective, what is the probability that the machine is out of adjustment?
A).10
B).360
C).080
D).536
E).618

Question

Quizplus · Accepted Answer

Given that the first unit produced is not defective, we need to find the probability that the machine is out of adjustment. Let's denote:- $A$ as the event that the machine is out of adjustment,- $A'$ as the event that the machine is in proper adjustment,- $D'$ as the event that a unit is not defective.We are looking for $P(A|D')$, the probability that the machine is out of adjustment given that the first unit is not defective.Using Bayes' theorem:$$P(A|D') = \frac{P(D'|A)P(A)}{P(D')}$$Given:- $P(A) = 0.10$ (10% chance the machine is out of adjustment),- $P(A') = 0.90$ (90% chance the machine is in proper adjustment),- $P(D'|A) = 0.75$ (25% defective means 75% not defective if out of adjustment),- $P(D'|A') = 0.96$ (4% defective means 96% not defective if in proper adjustment).We also need $P(D')$, the overall probability of producing a non-defective unit, which can be found using the law of total probability:$$P(D') = P(D'|A)P(A) + P(D'|A')P(A')$$$$P(D') = 0.75 \times 0.10 + 0.96 \times 0.90$$$$P(D') = 0.075 + 0.864$$$$P(D') = 0.939$$Now, we can calculate $P(A|D')$:$$P(A|D') = \frac{0.75 \times 0.10}{0.939}$$$$P(A|D') = \frac{0.075}{0.939}$$$$P(A|D') \approx 0.080$$Therefore, the correct answer is C, which corresponds to a probability of approximately 0.080, or 8%.

When a Machine Is Working Properly Only 4% of the Units