This is my second time posting this question, Problem 4 [18 pts (5,44,5)]: At th

Question

This is my second time posting this question,

Problem 4 [18 pts (5,44,5)]: At the Airport I. Suppose a bomb detector at the airport has a 95% chance of detecting a bomb if there is one in a piece of luggage, but it has a 1% chance of falsely detecting a bomb within an arbitrary innocent piece of luggage. Suppose only 1 in 100,000 pieces of luggage actually contains a bomb. Calculate the conditional probability of a piece of luggage containing a bomb, given that the detector is claiming there is such a bomb inside 2. What is the probability that three bomb detections are all false alarms? 3. How many detection events must occur until there is one real bomb among them in expecta- tion? 4. Now suppose there is a liquid detector with a 95% chance of detecting liquid if luggage contains some, and a 1% chance of detecting liquid if there is none. 1 in 5 pieces of luggage actually contains a liquid. What is the probability that a piece of luggage contains a liquid, given that the detector claims there is liquid?

Explanation / Answer

Here Pr(bomb) = 1/100000

Here, The Pr(true Positive) = 0.95

Pr(False positive) = 0.01

(1) Now, we have to calculate that

Pr(luggage contain Bomb l detector claiming There is bomb inside)

= Pr(luggage contain bomb) * Pr(There is bomb detected by detector l There is a bomb) / [ Pr(luggage contain bomb) * Pr(There is bomb detected by detector l there is a bomb) + Pr(luggage not contain Bomb) * Pr(there is bomb detected by detector l there is no bomb)]

= (1/100000 * 0.95)/ (1/100000 * 0.95 + 99999/100000 * 0.01)

= 9.4911 x 10-4  

Q.2 Here false alarm means that if there is no bomb then also detector will falsely detect the bomb.

so Pr(false alarm) = Pr(bomb detected l there was no bomb) = 1 - 9.4911 x 10-4

Pr(3 bomb detections are all false alarms) = (1 - 9.4911 x 10-4 )3 = 0.9972

Q3 NUmber of detection events required to to expect one real bomb = 1/ (9.4911 x 10-4) = 1053.62 = 1054 times

Q.4 Now here Pr(Liquid) = 1/5 = 0.2

Here Pr(Liquid l detector claim is liquid) = Pr(detector claim is liquid l liquid) * Pr(Liquid)/ [ Pr(detector claim is liquid l liquid) * Pr(Liquid) + Pr(Not liquid) * Pr(detector claim is liquid l not liquid) ]

= 0.20 * 0.95 / (0.20 * 0.95 + 0.80 * 0.01) = 0.9596

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