probability!?

Question

Suppose that an airplane passenger whose itinerary requires a change of airplanes in Ankara,Turkey,has a 4% chance of losing each piece of his or her luggage independently.Suppose that the probability of losing each piece of luggage in this way is 5% at Da Vinci airport in Rome, 5% at Kennedy airport in New York, and 4% at O'Hare airport in Chicago.Dr.May travels from Bombay to San Francisco with one piece of luggage in the baggage compartment. He changes airplanes in Ankara,Rome,NewYOrk, and Chicago
(a)If the luggage does not reach his destination with him, what is the probability that it was lost at Da Vinci airport in Rome?

Answer 1

Dear quz_us,

You have to think of this sequentially. The unconditional probability of Dr. May's luggage being lost in Rome is not 0.05, but rather 0.048, because if it is lost beforehand in Ankara then it can't get lost in Rome. The 0.05 figure for Rome is a conditional probability, meaning the probability of losing a bag GIVEN that it wasn't lost at an earlier stop (or equivalently, that it successfully arrived at Rome).

You can construct the unconditional probabilities for the luggage's last destination as follows (we could also call them joint probabilities of both arriving and being lost at a particular city). For notation, the small letters a, r, n, c, and s are the respective first letters of each destination city. The large letter A indicates that the luggage arrives at a designated city, and similarly L indicates the luggage is lost at a designated city.

P(A(a) & L(a)) = P(A(a)) x P(L(a) | A(a))
= 1 x 0.04
= 0.04 .

P(A(r) & L(r)) = P(A(r)) x P(L(r) | A(r))
= P(A(a) & ~L(a)) x P(L(r) | A(r))
= (1 - P(A(a) & L(a))) x P(L(r) | A(r))
= (1 - 0.04) x 0.05
= 0.048 .

P(A(n) & L(n)) = P(A(n)) x P(L(n) | A(n))
= P(A(a) & ~L(a)) x P(A(r) & ~L(r) | A(a) & ~L(a)) x P(L(n) | A(n))
= (1 - P(L(a))) x (1 - P(L(r))) x P(L(n) | A(n))
= (1 - 0.04) x (1 - 0.05) X 0.05
= 0.0456 .

P(A(c) & L(c)) = P(A(c)) x P(L(c) | A(c))
= P(A(r) & ~L(r)) x P(A(n) & ~L(n) | A(r) & ~L(r)) x P(L(c) | A(c))
= ((1 - 0.04) x (1 - 0.05)) x (1 - 0.05) x 0.04
= 0.034656 .

For the final destination we have
P(A(s)) = P(A(s) & ~L(s))
= 1 - P(A(a) & L(a)) - P(A(r) & L(r)) - P(A(n) & L(n)) - P(A(c) & L(c))
= 1 - 0.04 - 0.048 - 0.0456 - 0.034656
= 0.831744,

or we could compute it as

P(A(s)) = P(A(n) & ~L(n)) x P(A(c) & ~L(c) | A(n) & ~L(n))
= ((1 - 0.04) x (1 - 0.05) x (1 - 0.05)) x (1 - 0.04)
= 0.831744 .

(Since Dr. May can't lose his luggage at San Francisco,
P(A(s) & L(s)) = 0, and P(A(s)) is the same as P(A(s) & ~L(s)).)

Now we can easily calculate the probability that the luggage was lost in Rome, given it did not arrive in San Francisco.

P(L(r) | ~A(s)) = P(A(r) & L(r)) / (1 - P(A(s)))
= 0.048 / (1 - 0.831744)
= 0.048 / 0.168256
= 0.285280 (to six decimal places).

Notice that 1 - P(A(s)) is equal to the sum of the probabilities over the places where the luggage could have arrived and been lost, i.e.,
P(A(a) & P(L(a)) + P(A(r) & L(r)) + P(A(n) & L(n)) + P(A(c) & L(c)), which may make the last calculation more intuitive.

What may also make solving this type of problem more intuitive is to construct a table, rather than explicitly writing out equations. (This is especially true on Yahoo! since mathematical notation is relatively cumbersome here. Even creating tables isn't simple because Yahoo! deletes extra spaces, making it hard to align text in columns, but I'll try.)

Prob. _____ Probability _________ Probability ____
Arrives ____ Arrives, Not Lost ____ Arrives, Lost ___ City __
1.000000 ____ 0.960000 _______ 0.040000 _____ Ankara
0.960000 ____ 0.912000 _______ 0.048000 _____ Rome
0.912000 ____ 0.886400 _______ 0.045600 _____ New York
0.886400 ____ 0.831744 _______ 0.034656 _____ Chicago
0.831744 ____ 0.831744 _______ 0.000000 _____ San Francisco

See that once we know that Dr. May's luggage didn't make it with him to San Francisco, the third column of numbers is all that is relevant to calculating the probabilities of where the luggage was lost. This calculation can be made for each specific city by dividing the number in the row corresponding to a city by the sum of the numbers in the third column. Even without performing the division, we can readily see the relative ordering of the probabilities of where his bag was lost (e.g., Rome is most likely).

Answer 2

Well you stated that from Bombay to SF the doc's luggage was in the baggage compartment. Which means it was on the plane with you. Prob. for Bombay to SF?: 0% ... Unless he's a total idiot. If total idiot, than 20% chance (About 20/100 people are probably total idiots). ... Being high counts as "idiot" too, btw.

Then you say he changed airplanes in all these various places. But you never explicitly stated that he checked the bag in when changing planes ... nor would he likely do that, since most people who can use carry on for one leg of a trip, will keep it up for the rest of the connections (so it won't get lost).

The only way he lost his luggage is if:

A. He was insane enough to check in a bag that fits on the plane during remainder of his flights. (Probability=1/100)

OR

B. The plane went down and the contents was scattered. (Probability=1/20,000)