A sensor that detects dangerous levels of carbon monoxide (CO) in industrial app

Question

A sensor that detects dangerous levels of carbon monoxide (CO) in industrial applications is being tested. The manufacturer claims that it gives an alarm in 99% of such dangerous situations. However, it also responds to hydrogen, alarming 40% of the time when significant levels are present, and sometimes it gives an erroneous alarm in 0.3% of situations when neither carbon monoxide nor hydrogen are present. Dangerous levels of CO are present in 6% of situations in a particular industry. Significant levels of hydrogen are present in 3%. You may assume that CO and hydrogen never occur together (although in practice this is not true).

A. What is the probability that the sensor's alarm will trigger? Enter your answer to four decimal places (0.XXXX)

B. What is the probability that dangerous levels of carbon monoxide are present if the alarm goes off? Enter your answer to four decimal places (0.XXXX)

C. What is the probability that there are dangerous levels of carbon monoxide if the alarm does not go off? Enter your answer to FOUR decimal places (X.XXX)

Please show work, not just answers. I want to be able to understand the concepts that were used in order to solve the problem. Will rate/review.

Explanation / Answer

Solution

Preparatory Work

Let C represent the event that dangerous levels of carbon monoxide are present and H represent the event that dangerous levels of hydrogen are present.

Naturally, CC and HC represent the respective complementary events, i.e., CC and HC respectively represent that levels of carbon monoxide and hydrogen are not dangerous.

Let A represent the event that the alarm goes off and hence AC represents the event that the alarm does not go off.

Back-up Theory

If probability of an event B is influenced by occurence of another event A, then

P(B) = P(B/A).P(A) + P(B/AC).P(AC) ………. (1)

P(B/A) = P(B&A)/P(A) ………………….....…(2)

P(A/B) = {P(B/A).P(A)}/P(B)

Now, to convert the given data into the above terminology,

P(A/C) = 0.99 and so, P(AC/C) = 0.01

P(A/H) = 0.40 and so, P(AC/H) = 0.60

P(A/HCCC) = 0.30

P(C) = 0.06   P(H) = 0.03 P(C&H) = 0.

Now, to answer the questions,

A) P(A) = P(A/C)xP(C) + P(A/CC)xP(CC) + P(A/H)xP(H) + P(A/HC)xP(HC)

              = (0.99x0.06) + (0.3x0.94) + (0.4x0.03) + (0.3x0.97) = 0.6444 ANSWER

B) P(C/A) = {P(A/C).P(C)}/P(A) = (0.99x0.06)/0.6444 [from (A) above] = 0.0922 ANSWER

C) P(C/AC) = {P(AC/C). P(C)}/P(AC) = (0.3x0.06)/0.3556 [1 - (A) above] = 0.0640 ANSWER

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