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Ail Bane
Dec 4, 2007, 05:35 AM
Two airlines offer conveniently scheduled flights to the airport nearest your corporate headquarters. Historically, flights have been scheduled as reflected in this transition matrix.

http://img339.imageshack.us/img339/4914/95624961zx5.png

If your last flight was on B, what is the probability your second next flight (that is, the flight after the next flight) will be on A?

For the flight after next, I arrived at .667 for B and .333 for A. I tried to make a tree diagram to get further but I came up with smaller numbers. Any help or pointers welcomed.

terryg752
Dec 4, 2007, 06:13 AM
Possibilities are

B - A - A: Probability = .2 times .6 = .12

B - B - A : Probability = .8 times .2 = .16


Total = .12 + .16 = .28

Ail Bane
Dec 4, 2007, 07:15 AM
Oh, I get it. You did P[(AA)*(AB)] to get .12 and P[(AB)*(BB)] to get .16.

Wow, it was that simple? I must have over complicated it with a tree diagram. :( I was try to use some of the Markov formulations on this one.

I had a question, why would the two possibilities have to be added?

terryg752
Dec 4, 2007, 11:36 AM
I will give you an informal answer because it is easier to understand:

Probabilty of your going to London immediately= p1

Probabilty of your going to Paris immediately = p2

Probability of your going to either London or to Paris = p1 + p2

Don't you think it makes sense?

Note! The two events are mutually exclusive. You either go to London or go to Paris. This is an essential condition for adding probabilties.

Similarly, you throw a dice. Probabilty of 1 = 1/6, probability of 2 = 1/6
What is probability of 1 or 2? 1/6 + 1/6.

Or: there are 10 balls in a basket. 4 are red, 2 are green. You close your eyes and pick up 1 ball.

Probability of drawing red or green = 6/10. (4/10 + 2/10)

On the other hande, if you throw dice twice, probability of first 1 and then 2 is
1/6 times 1/6 = 1/36.

In this case both the events can occur and are independent of each other (occurence of one does not affect the probabilty of the other).

Ail Bane
Dec 4, 2007, 07:09 PM
I seem to get it now. You were using the Union principle, while I was using the exclusion one.

So if I went to London first then got a choice of London again or Paris:

London (current flight) - London (next)- Paris (flight after next)
London (current flight) - Paris (next) - Paris (flight after next)

Just add (L*P) and (P*P) to get the flight after next, right?

terryg752
Dec 4, 2007, 07:31 PM
Correct

Ail Bane
Dec 5, 2007, 01:32 AM
Correct

Thanks again!! The help is much appreciated!!