Question

In a laboratory, 3 technicians are analyzing the active ingredient in the product vials. Amir, Betty, and Carlos analyze 0.55, 0.30, and 0.15 of the vials, respectively. Inaccurate are 0.6 percent of the analyses conducted by Amir, 0.4 percent of the analyses conducted by Betty, and 1.0 percent of the analyses conducted by Carlos in this study. What is the probability that an inaccurate analysis, detected at final verification, was conducted by Carlos

Answer 1

Answer:

The probability is    $P(C | I) = 0.025$    

Step-by-step explanation:

From the question we are told that

      The  proportion analyzed by  Amir is  P(A) =  0.55

      The proportion analyzed by  Betty is  P(B)  =  0.30

     The  proportion analyzed by  Carlos is  P(C) =  0.15

     The inaccuracy by  Amir  is   p = 0.6% =  0.006

     The inaccuracy by  Betty  is  b = 0.4% = 0.004

    The inaccuracy by  Carlos  is c = 0.1% =  0.001

Generally the probability of inaccurate analysis is mathematically represented as

     $P(I) = P(A) * p + P(B) * b + P(C) * c$

=>  $P(I) = 0.55 * 0.006 + 0.30 * 0.004 + 0.15 * 0.001$

=>  $P(I) = 0.006$

Generally the probability that an inaccurate analysis, detected at final verification, was conducted by Carlos is mathematically represented as

       $P(C | I ) = \frac{P(C) * c}{P(I)}$

=>     $P(C | I) = \frac{ 0.15 * 0.001}{ 0.006}$

=>     $P(C | I) = 0.025$