The foreman in an iron foundry knows that the sand used for

The foreman in an iron foundry knows that the sand used for molding iron castings is too dry 5.3% of the time and too wet 1.1% of the time. He also knows that defective castings occur 0.92% of the time when the sand has the correct amount of moisture; 4.1% of the time when the sand is too dry; and 24.6% 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?

Solution

Probability of having perfect amount of moisture = 100 - 5.3 - 1.1 = 93.6%

Proabability that the molding is good = 100% - P(bad molding)

P ( Bad molding) = (93.6%)*(0.92) + (5.3%)*(4.1) + (1.1%)*(24.6) = 1.3490%

Thus, P (Good molding) = 100 - 1.3490 = 98.65098%

b)

P (Good molding) = P ( GM + Perfect moisture) + P (GM + dry sand) + P(GM + Wet sand)

= (99.08 * 93.6) + P(95.90 * 5.3) + (75.4 * 1.1) / 100

= 98.65098%

Conditional probability that sand is too wet

= (75.4 * 1.1) / 98.65098%

= 0.8294 / 98.65098

= 0.840 * 10^-3

c)

Conditional probability of correct moisture content

= (99.08 * 93.6) / 98.65098%

= 0.9400

i.e 94%

Hope this helps. Ask if you have doubts.