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Ace Machine Works estimates that the probability its lathe tool is properly adjusted is 0.8. When the lathe is properly adjusted, there is a 0.9 probability that the parts produced pass inspection. If the lathe is out of adjustment, however, the probability of a good part being produced is only 0.2. A part randomly chosen is inspected and found to be acceptable.

What is the posterior probability that the lathe tool is properly adjusted?

Answer :

Using Bayes' Theorem, the posterior probability that the lathe tool is properly adjusted given a part passes inspection is approximately 0.947 (94.7%). We determined this by combining the prior probabilities and the conditional probabilities to find the total probability of a part passing inspection.

To find the posterior probability, we can use Bayes' Theorem. Let's define the events:

  • A : Lathe tool is properly adjusted
  • A' : Lathe tool is out of adjustment
  • B : Part passes inspection

Prior Probabilities

The probability that the lathe tool is properly adjusted is 0.8. Therefore, P(A) = 0.8 and P(A') = 0.2.

Conditional Probabilities

  • When the lathe is properly adjusted, the probability that a part passes inspection is P(B|A) = 0.9.
  • When the lathe is out of adjustment, the probability that a part passes inspection is P(B|A') = 0.2.

Applying Bayes' Theorem:

Bayes' Theorem states:

  • [tex]\[P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\][/tex]

First, we need to find P(B), the total probability that a part passes inspection:

  • [tex]\[P(B) = P(B|A) \cdot P(A) + P(B|A') \cdot P(A')\][/tex]

Substituting values:

  • [tex]\[P(B) = 0.9 \cdot 0.8 + 0.2 \cdot 0.2 = 0.72 + 0.04 = 0.76\][/tex]

Now, using Bayes' Theorem, we can calculate P(A|B):

  • [tex]\[P(A|B) = \frac{0.9 \cdot 0.8}{0.76} = \frac{0.72}{0.76} \approx 0.947\][/tex]

Thus, the posterior probability that the lathe tool is properly adjusted given that a part is found to be acceptable is approximately 0.947 (94.7%).

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Rewritten by : Barada

Answer:

The posterior probability that the lathe tool is properly adjusted is 94.7%

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In your problem we have that:

-A is the probability that the part chosen is found to be acceptable.

The problem states that the probability its lathe tool is properly adjusted is 0.8. When it happens, there is a 0.9 probability that the parts produced pass inspection. There is also a 0.2 probability of the lathe is out of adjustment, when it happens the probability of a good part being produced is only 0.2.

So, P(A) = P1 + P2 = 0.8*0.9 + 0.2*0.2 = 0.72 + 0.04 = 0.76

Where P1 is the probability of a good part being produced when lathe tool is properly adjusted and P2 is the probability of a good part being produced when lathe tool is not properly adjusted.

- P(B) is the the probability its lathe tool is properly adjusted. The problem states that P(B) = 0.8

P(A/B) is the probability of A happening given that B has happened. We have that A is the probability that the part chosen is found to be acceptable and B is the probability its lathe tool is properly adjusted. The problem states that when the lathe is properly adjusted, there is a 0.9 probability that the parts produced pass inspection. So P(A/B) = 0.9

So, probability of B happening, knowing that A has happened, where B is the lathe tool is properly adjusted and A is that the part randomly chosen is inspected and found to be acceptable is:

[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.8*0.9}{0.76} = \frac{0.72}{0.76} = 0.947 = 94.7%[/tex]

The posterior probability that the lathe tool is properly adjusted is 94.7%