# Probability problem

• Sep 27th 2012, 04:11 PM
dumbledore
Probability problem
Air Canada and United Continental sell seats on each others cross-border flights, coordinating fare structures and discounts, while competing for passengers. In 2011 they proposed a closer collaboration involving sharing revenue and coordinating schedules on 19 Canada-US routes.In Canada, deals of this type are evaluated by the Competition Bureau, and if there are objections to its rulings, a final decision is made by the Competition Tribunal.In mid 2011 the Canadian Competition Bureau ruled against the deal on the grounds that it would monopolize 10 key Canada-U.S. routes and significantly reduce competition on nine others, possibly resulting in increased prices.In August 2011 Air Canada objected to the Competition Bureau ruling saying that its judgement was "fundamentally misconceived", and that the proposed joint venture would result in "substantial gains in efficiency".A few days later, WestJet intervened in the case on the side of the Competition Bureau, arguing that the deal between United Continental and Air Canada would prevent it from competing on equal terms on trans-border routes. WestJet has a deal with American Airlines but it is more restrictive than the one proposed between Air Canada and United Continental.The case then went to the Competition Tribunal for a decision as to whether or not to allow the collaboration agreement between Air Canada and United Continental.
Put yourself in the position of an investment banker early in 2011, prior to the above events taking place. You know that Air Canada might propose a closer collaboration agreement with United Continental and believe this will be profitable to Air Canada. You need to calculate the chance that such a deal will eventually go through. You assess that they will propose the collaboration agreement with a probability of 0.6. You also know that, if they do, the Competition Bureau might oppose the deal with a probability of 0.8. If that happens, Air Canada might object with a probability of 0.9. If they do, WestJet could intervene.

You assess the probability of WestJet intervening in the case at 0.75, which you believe will affect the decision of the Competition Tribunal. If WestJet intervenes, you believe the chance of the Competition Tribunal blocking the deal is 0.85, and without WestJet it is 0.7.

What is the probability that there will be a closer collaboration agreement between Air Canada and United Continental that is approved by the Competition Tribunal?

I constructed my own probability tree (can be found here: View image: IMG 0229 ). Is it correct? And if so how do I further it to find the answer to the problem?
• Sep 27th 2012, 09:51 PM
chiro
Re: Probability problem
Hey dumbledore.

This looks pretty good what you have calculated in terms of the tree.

Also when you say the probability that the Competition Tribunal accepts it, are you finding all probabilities of it not being rejected in any branch?

If this is the case, then you need to look at all possibilities where it is approved and add up these possibilities (you can do this because all events are disjoint from each other). If the above is the case you have five ways of getting accepted which means you have to calculate five probabilities.

Have you covered how to calculate probabilities given a probability tree?
• Sep 28th 2012, 06:26 AM
dumbledore
Re: Probability problem
Thanks for the response. In regards to your first question I'm not sure. I can only go on what I've been given. I was taught to multiply the probabilities on the branches but when I do that they dont add to 1. I get 0.2754, 0.0486, 0.0756, and 0.0324 when I do that.
• Sep 28th 2012, 06:46 AM
chiro
Re: Probability problem
If you add all of them, they will add to 1 but adding just the accepted ones won't add to one (not by a long shot given your probability tree). You are correct in multiplying the probabilities, but make sure you only multiply ones with a leaf (i.e. that terminate).

You have four probabilities of acceptance throughout the various stages and using R I get the following results:

> 0.6*0.2
[1] 0.12
> 0.6*0.8*0.1
[1] 0.048
> 0.6*0.8*0.9*0.25*0.3
[1] 0.0324
> 0.6*0.8*0.9*0.75*0.15
[1] 0.0486

This is just looking at all leaves where you get acceptance but if this is not what you want, then please inform us what shouldn't be included.
• Sep 28th 2012, 06:57 AM
dumbledore
Re: Probability problem
solved