Q10. part (i) a) Determine the probability function of S. How did they arrive at the conclusion that S takes the values 0,1,2,3,4? Also, how the probability of P(S=3) is calculated? I could not understand this part... http://imgur.com/SwnQq1W
as per Probabilities Max no. of claims=2, max cars involves=2, so max S totals 4 min is 0. P(S=3)=3 cars involved in accidents=P(2 accidents)[P(3 cars in 2 accidents)] P(2 accidents)=0.2,P(3 cars in 2 accidents)=0.7*0.3+0.3*0.7
Can someone please explain this question to me? I am really having difficulty understanding the wording of this question. I will try to breakdown my understanding here: 1) The first 'table' shows that a policyholder can make 0 claims, 1 claim or 2 claims in a year. It shows probability distribution for those claims. 2) The second 'table' tells us how many cars can be involved in each accident that results in a claim. So AM I right in saying that, if there is 1 claim, there can be 2 cars involved in the accident or 1 car. And if there are 2 claims/accidents there can be one car involved in the accident or 2 cars involved in the accident So S can take values of: 0 - so that means there are no claims, hence no cars involved in each accident. 1 - so that means there was 1 claim and 1 car was involved in the accident or 2 separate claims by a policy holder and in each claim there was 1 car involved. However when the solution gives P(S=1), it seems to me that they have taken P(n=1) * P(x=1). Why did they not take the other scenario of having 2 separate claims and in each claim there was 1 car involved? The wording is really confusing me on this one.
if No. of claims=2, There will be at least 2 cars involve in the total claims. Cars involved in first claim-second claim :1-1, 1-2, 2-1, 2-2. (Note: each accident results in a claim) Answer to bold fonts: For x=2 and n=1, S will be 2. P (S=2)=P (n=1) × P (x=2) + P (n=2)×P (x=1)×P (x=1) or P(one claim with 2 cars) + P(two claims with one car involve in each)
