Opening a new restaurant

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sarahtee

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You wish to open a new restaurant and are considering locations in Edmonton and St. Albert, but only one location will actually become available. If it is built in Edmonton, the restaurant stands an 80% change of successfully surviving its first year. However, if it is built in St. Albert, its chance of survival falls to 65%. It is estimated that the chance of St. Albert being available is 60% and Edmonton being available is 40%.

Find the probability that the restaurant will:

a) survive its first year.

b) be built in Edmonton and survive its first year.

c) be built in St. Albert, given that it survived its first year.

d) not survive its first year, given that it is built in Edmonton.
 
sarahtee said:
You wish to open a new restaurant and are considering locations in Edmonton and St. Albert, but only one location will actually become available. If it is built in Edmonton, the restaurant stands an 80% change of successfully surviving its first year. However, if it is built in St. Albert, its chance of survival falls to 65%. It is estimated that the chance of St. Albert being available is 60% and Edmonton being available is 40%.

Find the probability that the restaurant will:

a) survive its first year.

b) be built in Edmonton and survive its first year.

c) be built in St. Albert, given that it survived its first year.

d) not survive its first year, given that it is built in Edmonton.
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There are not very many outcomes, but some are oddly defined. Perhaps this is why no one has approached the problem. The term "available" is not clear. Are we to assume that we WILL BUILD somewhere and that the probabilities are mutually exclusive? If not, and based on these probabilities, we have four possibilities for "available", each with its own probability of coming to fruition:

Edmonton -- 0.4^2
St. Albert -- 0.6^2
Both -- 0.6*0.4
Neither -- 0.4*0.6

This entirely changes the question, as 1) We might not be able to build, and 2) We may have to choose if both pop up.

It is for this confusion that I do not believe this is the question. I suspect the last piece of information should be, we WILL build in ONE of the locations, and the probability of this happening in a specific location is Edmonton 40% and St. Albert 60%.

Having Said that, there are only four outcomes.

Edmonton&Survive -- 0.40*0.80 = 0.32
Edmonton&Fail -- 0.40*0.20 = 0.08
St.Albert&Survive -- 0.60*0.65 = 0.39
St.Albert&Fail -- 0.60*0.35 = 0.21

Checking: 0.32 + 0.08 + 0.39 + 0.21 = 1.00 -- Good.

All four questions can be answered very quickly with this display of the entire distribution. What do you get?
 
Edmonton&Survive -- 0.40*0.80 = 0.32
Edmonton&Fail -- 0.40*0.20 = 0.08
St.Albert&Survive -- 0.60*0.65 = 0.39
St.Albert&Fail -- 0.60*0.35 = 0.21

a) survive its first year.

These two:

Edmonton&Survive -- 0.40*0.80 = 0.32
St.Albert&Survive -- 0.60*0.65 = 0.39

0.32 + 0.39 = 0.71

b) be built in Edmonton and survive its first year.

This one:

Edmonton&Survive -- 0.40*0.80 = 0.32

c) be built in St. Albert, given that it survived its first year.

These two:

Edmonton&Survive -- 0.40*0.80 = 0.32
St.Albert&Survive -- 0.60*0.65 = 0.39

0.39 / (0.39+0.32) = 0.54929

d) not survive its first year, given that it is built in Edmonton.

These two:

Edmonton&Survive -- 0.40*0.80 = 0.32
Edmonton&Fail -- 0.40*0.20 = 0.08

0.08 / (0.08+0.32) = 0.20 -- But we knew that already.
 
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