View Full Version : Statistics and Joint Probability
cowboy
Nov 30, 2005, 12:58 PM
A mortgage company has found that 2% of its mortgage holders default on their mortgage and lose their property. Furthermore, 90% of those who default are late on at least two monthly payments over the life of the mortgage as compared to 45% of those who do not default.
1.) What is the joint probability that a mortgagee has two or more late monthly payments and does not default on the mortgage?
2.) What is the joint probability that a mortgagee has one or less late monthly payments and does not default on the mortgage?
sujan
Mar 28, 2006, 11:38 PM
A mortgage holding company has found that 2% of its mortgage holders default on their mortgage and lose the property. Furthermore, 90% of those who default are late on at least two monthly payments over the life of their mortgage as compared to 45% of those who do not default.
What is the joint probability that a mortgagee has two or more late monthly payments and does not default on the mortgage?
What is the joint probability that a mortgagee has one or less late monthly payments and does not default on the mortgage?
Based on Bayes' Theorem, what is the probability that a mortgagee will not default given one or less payments over the life of the mortgage?
norkong
Nov 29, 2009, 04:44 AM
A mortgage holding company has found that 2% of its mortgage holders default on their mortgage and lose the property. Furthermore, 90% of those who default are late on at least two monthly payments over the life of their mortgage as compared to 45% of those who do not default.
What is the joint probability that a mortgagee has late monthly payments and does not default on the mortgage?
What is the joint probability that a mortgagee has monthly payments and default on the mortgage?
Moriums
Feb 5, 2010, 07:51 PM
What is the joint probability that a mortgagee has one or less late monthly payments and does not default on the mortgage?
janie10
Mar 5, 2010, 09:15 AM
answer is P(D)=.02; therefore P(L>=2given that D)=.90 P(L<=1 given thatD)=.10
P(ND)=.98; therefore P(L>=2(ND)=.45 P(L<=1 given that ND) =.55
janie10
Mar 5, 2010, 09:25 AM
finishing my answer is P(ND P(L>=N conditional on ND)= .98*.45=.441
morgaine300
Mar 5, 2010, 09:15 PM
finishing my answer is P(ND P(L>=N conditional on ND)= .98*.45=.441
While the math is right, you're doing it for the wrong reason. The question is not asking for a conditional. It says "and" - an intersection. But .441 is the intersection not the conditional.