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#1




Stats:Help me in these questions
The questions may be simple but are driving me nuts..
Q1 :A firm accepts either American express or Visa Credit card.A total of 24 percent of its customers carry an American Express card,61 percent carry a Visa card and 11 percent carry both.what percentage of its customers carry a credit card that the establishment will accept? My answer is 82 percent Q2: At a certain stage of criminal inverstigation the inspector is 60% convinced of the guilt of the criminal.Suppose that a new evidence shows that the criminal has certain charatersitics.If 20 percent of the population possesses this chrarteristic,how certain the guilt of the suspect should the inspector now be if it turns out that the suspect has this characteristic. Plz elobrate steps neccessary as well...I am unable to get an answer out of it [] Plz reply soon 
#2




AoA
would you be kind enough to elaborate your calculation of the card statistics as i m unable to construe the stats.
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'Thee woh ik shakhs kay tasawar saay  abb woh ranayee khayal kahaan' 
#3




Quote:
let A be the event where constumer possesses American express P(A) = 0.24 B is the event with costumer having Visa Card P(B) = 0.61 AnB costumer carrying both P(AnB) = 0.11 P (AuB) = P(A) + P(B)  P(AnB) = o.24 + 0.61 0.11 = 0.74 or 74% Quote:
Let C = certainty of possessing that characteristic G = set of events he is of guilty by inspector. P(G/C) = P(GC)/P(C) = P(C/G)*P(G)/{P(C/G)*P(G) + P(C/G')*P(G')} ( Bayes Theorem) P(C/G) = 1 (last line shows ) Required Probability P(G/C) = (1*0.6)/{1*0.6 + 0.2*0.4} = 0.6/(0.6 + 0.08) = 60/68 = 15/17 = 0.8823 or 88% 
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