Statistics Q #2 Watch
I have another stats Q i need help with.
This question is about calculating conditional probabilities such as of A\B and the factthat conditional probabilities referring to A\B and B \ A are not equal, as oftensupposed by people in 'real life', It is also about appreciating the numerical values ofprobabilities in practical situations.
A fire insurance company has high-risk, medium-risk and low-risk policy holders who have, respectively, probabilities 0.02, 0.01 and 0.0025 of making claims within a given year. The proportions of the numbers of policy holders in the three categories are 0.10, 0.20 and 0.70, respectively.
(i) By considering the law of total probability, obtain the probability of a randomly selected policy holder making a claim; comment on whether your answer is plausible. By saying 'randomly selected', the above proportions can be used as probabilities.
Ok well the law of total probability is
P(B) = P(B|A)P(A) + P(B|A`)P(A`)
but i don't know what is B and what is A? I'm not sure what the difference between P(A) and P(A`) is either.
p(A') is the probability that it doesnt
so P(A') = 1 - p(A)
in this quesion event B is the policy holder making a claim, but the "law" cant be applied like this
p(making a claim) = p(claims|high risk)*p(high risk) + p(claims|med risk)*p(medrisk) + p(claims|lowrisk)*p(low risk)
= 0.02*0.1 + .01*.2+.0025*.7