Question

The foreman in an iron foundry knows that the sand used for molding iron castings is too dry 5.5% of the time and too wet 3.3% of the time. He also knows that defective castings occur 0.89% of the time when the sand has the correct amount of moisture; 7.1% of the time when the sand is too dry; and 39% of the time when the sand is too wet. Suppose that a casting is selected at random from the batch just produced. Use four decimals. (a) What is the probability that the molding is good? (b) If the molding is good, what is the probability that the sand is too wet? (c) If the molding is good, what is the probability that the moisture content of the sand is correct?

Answer 1

Please don't hesitate to give a "thumbs up" for the answer in case you're satisfied with the answer.

Here' every stat given in the question. lets solve the problem using this:

p(dry) = .055

p(wet) = .033

p(normal) = 1-.055-.033 = 1-.088 = .912

p(defect cast | normal) = .0.0089

p(defect cast | dry) = .0.071

p(defect cast | wet) = .0.39

a. P(molding is good) = p(dry)*(1-p(defect cast | dry)) + p(wet)*(1-p(defect cast | wet)) + p(normal)*(1-p(defect cast | normal))

= .055*(1-.071)+.033*(1-.39) + .912*(1-.0089)

= 0.9751

b. P(given molding is good , sand is too wet) = p( molding normal (good) but sand wet) / p(normal)

= .033*(1-.39) / .9751 ( calculated in part a)

= 0.0206

c. P(given molding is good, moisture content in sand is correct) = p(normal)

= p( molding is good and sand is normal) / p(molding is good)

= .912*(1-.0089) /.9751 ( as calculated in 1st step of the solution)

= 0.9270