Question

please solve all problems included

8. (Bonus for undergrad, required for graduate students) A chemical engineer is interested in determining whether a certain trace inquiry is present in a product. An experiment has a probability of .80 of detecting the impurity if it is present. The probability of not detecting the impurity if it is absent is 90. The prior probabilities of the impurity being present and absent are .40 and 60, respectively. Three separate experiments result in only two detections. a. (5 points) If the impurity is present, what is the probability that three separate experiments result in only two detections? b. (5 points) If the impurity is absent, what is the probability that three separate experiments result in only two detections? c. (5 points) What is the probability that three separate experiments result in only two detections? d. (5 points) What is the probability that the impurity is present if three separate experiments result in only two detections?

Answer 1

a)

As a part of determining the presence of impurity in a product, the following information is given. The Probability that impurity is present in the product, P(1)=0.4 The Probability that impurity is absent in the product, P(l') = 0.6 The probability of detecting the impurity (D) when it is present is 0.8. Similarly, the probability of not detecting the impurity (D') when it is present is 0.9.

P(D2,3|I) = 0.384

It is also given that 2 out of 3 experiments resulted in two detections. Now, find the probability that an impurity is detected in two of the three tests when the impurity is present. That is, find P(D231) Therefore, P(0.311)=()(0.8)*(1 – 0,8) = 0.384

b)

P(D2,3|I') = 0.027

Also, the probability of not detecting an impurity if it is not present is 0.9. That is, find P(D231') Therefore, P(D;317)= ((0.9) (1–0,9) = 0.027

c)

P(D2,3) = 0.384 * 0.4 + 0.027*0.6 = 0.1698

d)

Now, find the posterior probability that impurity is present. P(D231)P(1) P(1\D23) - P(D,, 1) P(1)+P(D2,3 0.384 x 0.4 0.384 x 0.4+0.027 x 0.6 = 0.905 The probability that an impurity is present given that an impurity is detected in two of three tests is 0.905

P(I| D2,3) = 0.384*0.4/(0.384 * 0.4 + 0.027*0.6)

= 0.905