Question

Professor X travels from Houston to Istanbul with stop overs in New York and Londa. At each stop he luggage is transferred from one plane to another. In each airport, including Houston, chances are that with probability p her luggage is not placed in the right plane. Professor X finds that his suitcase has not reached Istanbul. What are the chances that the mishap took place in Houston, New York, and London, respectively?

Answer 1

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Chance that mishap took place in Houston is p, as Houston is the first airport for professor X

Chance that mishap took place in New York (2nd Airport)

= P( mishap didn't happen at Houston )*p(mishap happened at NY)

= (1-p)*p

P( mishap at london) = p( no mishap at Houston)*p(no mishap at NY)*p(mishap at london)

= (1-p)*(1-p)*p

The above aren't the answers though, as we have given a condition , we need to calculate chances of mishap at each of the cities given Professor X doesn't get his suitcase at istanbul. So, please read below to find the answer.

Now, total probability that mishap happened at any of the airports is additional of all these probabilities

= p+(1-p)p+(1-p)*(1-p)*p

Hence, given Professor X didn't get his suitcase, chances that mishap happened in cities are:

P(mishap at Houston, given prof X didn't get suitcase at Istanbul) = p/( p+(1-p)*p + (1-p)^2*p)

P(mishap at NY, given prof X didn't get suitcase at Istanbul)

= (1-p)*p/(p+(1-p)*p+(1-p)^2*p)

P(mishap at london, given prof didn't get suitcase at Istanbul)

= (1-p)^2*p/(p+(1-p)*p+(1-p)^2*p)