: Three machines A, B and C produce respectively 60%, 30% and 10 % of the total number of items of a factory. The percentage of defectives output of these machines are 2%, 3% and 4%, respectively. If an item is selected at random and is found defective. Find the probability that the selected item was produced by machine C.

• (4/100* 10/100)/(120+40+90/10000)=4/25
• 10 years agoHelpfull: Yes(1) No(0)
• This problem is based on Bayes theorem
  Let P(A)=60%=0.6
  P(B)=0.3
  P(C)=0.1
  M=selected is defective
  Then P(M/A)=0.020)*0.6=0.012
  Similarly P(M/B)=0.009 & P(M/C)=0.004
  Required is P(C/M)=0.16
• 10 years agoHelpfull: Yes(0) No(0)
• probability of defective of c is to find so by navie bayes theorem: let it be p(x)
  p(x/c)=? i.e probability of c is known and x is unknown.
  so, p(x/c)= p(c/x)*p(c) / [p(c/x)*p(c)+p(a/x)*p(a)+p(a/x)*p(a)]
  =10/100*4/100 / [10/100*4/100+60/100*2/100+30/100*3/100]
  =4/25
• 1 year agoHelpfull: Yes(0) No(0)