Co 2. Aces INC. owns plants (A, Ag, As, As.As, As) that produce playing asemblin

Question

Co 2. Aces INC. owns plants (A, Ag, As, As.As, As) that produce playing asembling oocur. We know cards. Sometimes errors in printing and that A, As, and As produce each 21 percent of the total output of the playing cards while A, As produce 15 percent each while As produces 7 percent of the playing cards Assume that each of Ai, As, and As have an eror rate of 2.5 percent while Az, As have an error rate of 2 percent while As has an error rate of 1.6 percent. Finally suppose that one playing card is selected randomly from the entire batch and it is found to have errors (whether printing or package assembling doesn't matter.) Determine the probability that this item is produced by each plant Ai, A2, As. As. As,A

Explanation / Answer

consider

Ai= playing card produced by plant 'i' i=1,2,,3,...6

P(A1) = 0.21 P(A2)=0.15, P(A3)=0.21, P(A4)= 0.21, P(A5)=0.15, P(A6) =0.07

B= The playing card having an error

P(B/A1) = 0.025, P( B/A3) = 0.025, P(B/A4) =0.025

P(B/A2) = 0.02, P(B/A5)=0.02, P(B/A6)=0.016

we have compute the probabilities P(Ai/B) , i=1,2,...6

By using Baye's theorem

P(Ai/B) = P(Ai) * P(B/Ai) / sum(P(Ai)*P(B/Ai)) i=1,2,..6

P(A1/B) = (0.21*0.025)/((3*0.21*.025)+(2*0.15*0.02)+0.07*0.016)

= 0.2295

P(A2/B) = (0.15*0.02)/((3*0.21*.025)+(2*0.15*0.02)+0.07*0.016)

= 0.1311

P(A3/B) = (0.21*0.025)/((3*0.21*.025)+(2*0.15*0.02)+0.07*0.016)

= 0.2295

P(A4/B) = (0.21*0.025)/((3*0.21*.025)+(2*0.15*0.02)+0.07*0.016)

= 0.2295

P(A5/B) = (0.15*0.02)/((3*0.21*.025)+(2*0.15*0.02)+0.07*0.016)

= 0.1311

P(A6/B) = (0.07*0.016)/((3*0.21*.025)+(2*0.15*0.02)+0.07*0.016)

= 0.04897

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.