a women checks a suitcase to be carried successively by thre

a women checks a suitcase to be carried successively by three connceting airlines ,Airy Airlines gets the bag first, Airy mishandles 1% of the baggage it receives, Blur airlines is supposed to get the suitcase next,if blur gets it on time ,the probability that the bag will reach the next airline properly is 97% Clearsky the third airline , has probability of 2% of not doing its part, The suit case does not arrive on time ,find the probability that of the airline is responsible

Solution

let E1,E2,E3 denote three connecting airlines,that is Airy, Blur and Clearskt respectively

since the baggage(suit case) has to be received properly on time by all the three airlines we have

p(E1) = P(E2) = P(E3) = 1/3

let A be the event of mishandling the baggage by the airlines,

therefore P(A/E1) = 1% = .01

P(A/E2) = 2% =.02 (since given that 2% of work not done by the third airlines)

P(A/E3) = 97% = .97 (since if the total probability is taken as one)

hence, P(A) = P(E1) P(A/E1) + P(E2)P(A/E2) + P(E3) P(A/E3)

= 1/3 X [.01 +.02 + .97 ] = 0.333

But we have to find which airline is responsible for not receiving baggage on time

P(E1/A) = [ P(E1) P(A/E1) ] / [ P(A) ] = [ 1/3X.01] / [0.333] = 0.0101

Similarly,

P(E2/A) = 0.2020

P(E3/A) = 0.9797

of three airlines we find the third airline Clearsky has higher probability value compared to others, hence this airline is responsible for suit case not arriving on time.