Question

A certain grocery store in Tuscaloosa receives 20%, 40%, and 40% of its packaged salad from...

  1. A certain grocery store in Tuscaloosa receives 20%, 40%, and 40% of its packaged salad from Suppliers A, B and C, respectively. It is known that excess moisture (which causes a slimy texture) occurs in 12% of packaged salad from Supplier A, in 8% of packaged salad from Supplier B, and in 10% of packaged salad from Supplier C.

  1. What is the probability of randomly selecting a salad that is both from Supplier C and has excess moisture?
  2. Find the probability that a randomly selected packaged salad has excess moisture.
  3. If it is known that a packaged salad came from Bursitis Hospital, what is the probability that it has excess moisture?
  4. If it is known that a packaged salad has excess moisture, what is the probability that it came from Supplier A?

e.   If it is known that a packaged salad has excess moisture, what is the probability that it came from Supplier B?

f.    If it is known that a packaged salad has excess moisture, what is the probability that it came from Supplier C?

Homework Answers

Answer #1

a) P(supplier C and excess moisture) = P(excess moisture | supplier C) * P(Supplier C) = 0.1 * 0.4 = 0.04

B) P(excess moisture) = 0.12 * 0.2 + 0.08 * 0.4 + 0.1 * 0.4 = 0.096

C) P(excess moisture | B) = 0.08

D) P(A | excess moisture) = P(Excess moisture | A) * P(A)/P(excess moisture) = (0.12 * 0.2)/0.096 = 0.25

E) P(B | excess moisture) = P(Excess moisture | B) * P(B)/P(excess moisture) = (0.08 * 0.4)/0.096 = 0.3333

F) P(C | excess moisture) = P(excess moisture | C) * P(C)/P(excess moisture) = (0.1 * 0.4)/0.096 = 0.4167

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