Apr 2008 q7

[Viki2010]

can anyone tell me the logic here for setting:

P(n=2) = 1* 0.2
P(n=3) = 1 * 0.8 * 0.2 etc?

In my view the prob. should be P(n=2)= 1, P(n=3) = 1 * 0.8 etc.

what is 0.2 representing? it is not obvious to me....

 

[Calum]

What solution are you looking at here? I agree with you, and so does the examiner's report...the probability of retention in year n is simply 0.8^(n-2), for n>=2

 

[Viki2010]

What solution are you looking at here? I agree with you, and so does the examiner's report...the probability of retention in year n is simply 0.8^(n-2), for n>=2

I was looking at ASET

 

[John Potter]

We need the expected time before they change auditors again. They have just changed to A and must stay for 2 years. There is then a 20% chance that they will leave.

So 1 year + 20% chance of just one more year

Or an 80%*20% chance of just 2 more years
Or an 80%*80%*20% chance of just 3 more years
etc.

So, expected time = 1+ 1*0.2 +2*0.8*0.2 +3*0.8^2*0.2 +..... = 6

John

 

[Viki2010]

We need the expected time before they change auditors again. They have just changed to A and must stay for 2 years. There is then a 20% chance that they will leave.

So 1 year + 20% chance of just one more year

Or an 80%*20% chance of just 2 more years
Or an 80%*80%*20% chance of just 3 more years
etc.

So, expected time = 1+ 1*0.2 +2*0.8*0.2 +3*0.8^2*0.2 +..... = 6

John

but if they must stay for 2 years, than prob of leaving is 0 and not 20%.

 

[John Potter]

Yes, that's why we definitely have the first 1 with no probabilty attached. Remember that it's expected TIME.

How about another way of thinking of it...

What are the possibilities? 2 years, 3 years, 4 years, etc.

What are the associated probabilities?
2 years... you'd have to leave straight away prob = 0.2
3 years... you'd have to leave after one more year prob = 0.8*0.2
4 years... you'd have to leave after 2 more years prob = 0.8^2*0.2
etc.

So, expected time = 2*0.2 +3*0.8*0.2 +4*0.8^2*0.2 +..... = 6

John