I. Suppose a bomb detector at the airport has a 95% chance of detecting a bomb i

Question

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

Question 1:

Here we are given that:

P( detect | bomb ) = 0.95 and P( detect | no bomb ) = 0.01

P( bomb ) = 1/100000 = 0.00001 and therefore P( no bomb ) = 1 - 0.00001 = 0.99999

Now using law of total probability, we get:

P( detect ) = P( detect | bomb )P( bomb ) + P( detect | no bomb ) P( no bomb )

P( detect ) = 0.95*0.00001 + 0.01*0.99999 = 0.0100094

Now using bayes theorem given that there is a detection for the bomb, probability that there is a bomb is computed as:

P( bomb | detect ) = P( detect | bomb )P( bomb ) / P( detect )

P( bomb | detect ) = 0.95*0.00001 / 0.0100094 = 0.000949

Therefore 0.000949 is the required probability here.

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