Problem 4 [18 pts (5,44,5)]: At the Airport I. Suppose a bomb detector at the ai

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

1.

Let L be the event that the luggage has bomb and D be the event that the detector claims a bomb in luggage.

Let ~L be the event that the luggage does not has bomb. Then P(~L) = 1 - P(L)

We need to find the conditional probability P(L | D)

P(L) = 1 / 100,000

P(~L) = 1 - (1 / 100,000) =   99999 / 100,000

Given,

P(D | L) = 0.95 and P(D | ~L) = 0.01

By law of total probability,

P(D) = P(L) * P(D | L) + P(~L) P(D | ~L)

= (1 / 100,000 ) * 0.95 + (99999 / 100,000) * 0.01

= 0.01

By Bayes theorem,

P(L | D) = P(D | L) * P(L) / P(D)

= 0.95 * (1 / 100,000) / 0.01 = 0.00095

So, P(L | D) = 0.00095

2.

Probability of false alarm = 0.01

Probability of three bomb detections are all false alarms = 0.013 = 10-6

3.

Assuming this as geometric distribution with probability of success is that there is one real bomb given the detector claims there is bomb, (p = 0.00095)

Expected detected events = 1 / p = 1/0.00095 = 1052.632

4.

Let L be the event that the luggage has liquid and D be the event that the detector claims a liquid in luggage.

Let ~L be the event that the luggage does not has liquid. Then P(~L) = 1 - P(L)

We need to find the conditional probability P(L | D)

P(L) = 1 / 5

P(~L) = 1 - (1 / 5) = 4/5

Given,

P(D | L) = 0.95 and P(D | ~L) = 0.01

By law of total probability,

P(D) = P(L) * P(D | L) + P(~L) P(D | ~L)

= (1 / 5) * 0.95 + (4/5) * 0.01

= 0.198

By Bayes theorem,

P(L | D) = P(D | L) * P(L) / P(D)

= 0.95 * (1 / 5) / 0.198 = 0.9596

So, P(L | D) = 0.9596

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