A machine has a 95% chance of being properly adjusted each morning. When properly adjusted, the product qualification rate is 98%. When malfunctioning, the qualification rate is 55%. The problem asks for the probability that the machine was properly adjusted given that the first product of the day is qualified. This requires using Bayes' Theorem.

1. Problem Description

A machine has a 95% chance of being properly adjusted each morning. When properly adjusted, the product qualification rate is 98%. When malfunctioning, the qualification rate is 55%. The problem asks for the probability that the machine was properly adjusted given that the first product of the day is qualified. This requires using Bayes' Theorem.

2. Solution Steps

Let A be the event that the machine is properly adjusted.

Let Q be the event that the first product of the day is qualified.

We are given the following probabilities:

$P(A) = 0.95$ (probability the machine is properly adjusted)

$P(A') = 1 - P(A) = 1 - 0.95 = 0.05$ (probability the machine is not properly adjusted, i.e., malfunctioning)

$P(Q|A) = 0.98$ (probability the product is qualified given the machine is properly adjusted)

$P(Q|A') = 0.55$ (probability the product is qualified given the machine is not properly adjusted)

We want to find $P(A|Q)$, the probability that the machine is properly adjusted given that the product is qualified.

We can use Bayes' Theorem:

$P(A|Q) = \frac{P(Q|A)P(A)}{P(Q)}$

First, we need to find $P(Q)$, the probability that the product is qualified. We can use the law of total probability:

$P(Q) = P(Q|A)P(A) + P(Q|A')P(A')$

$P(Q) = (0.98)(0.95) + (0.55)(0.05)$

$P(Q) = 0.931 + 0.0275$

$P(Q) = 0.9585$

Now we can plug the values into Bayes' Theorem:

$P(A|Q) = \frac{P(Q|A)P(A)}{P(Q)}$

$P(A|Q) = \frac{(0.98)(0.95)}{0.9585}$

$P(A|Q) = \frac{0.931}{0.9585}$

$P(A|Q) \approx 0.97130$

3. Final Answer

The probability that the machine was properly adjusted, given that the first product of the day is qualified, is approximately 0.

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