• We are pleased to announce that the winner of our Feedback Prize Draw for the Winter 2024-25 session and winning £150 of gift vouchers is Zhao Liang Tay. Congratulations to Zhao Liang. If you fancy winning £150 worth of gift vouchers (from a major UK store) for the Summer 2025 exam sitting for just a few minutes of your time throughout the session, please see our website at https://www.acted.co.uk/further-info.html?pat=feedback#feedback-prize for more information on how you can make sure your name is included in the draw at the end of the session.
  • Please be advised that the SP1, SP5 and SP7 X1 deadline is the 14th July and not the 17th June as first stated. Please accept out apologies for any confusion caused.

Chapter 7 Question 7.3

S

Sarahlouise_23

Member
Hi All,

I'm having some real difficulty understanding this question. I understand how to calculate P(S < 0), P(S < 1) and P(S < 2) but I don't understand how you know that 3 and 4 are also possible options.

Any help would be greatly appreciated. Thanks!
 
please edit '<' to '<='

Let A:Amount
Now, you know P(S<=2)=0.925
P(S<=3)= P(S<=2)+P(S=3)=0.975
P(S=3)=P(N=2)*P(A=1)*P(A=2)+P(N=2)*P(A=2)*P(A=1)......................as you claim amounts to 3 only if 2 claim occurs with amount 1 and 2(first 1 then 2 or first 2 then 1)

and,P(S=4)=P(N=2)*P(A=2)*P(A=2)
then,P(S<=4)=1
 
Hi All,

I'm having some real difficulty understanding this question. I understand how to calculate P(S < 0), P(S < 1) and P(S < 2) but I don't understand how you know that 3 and 4 are also possible options.

Any help would be greatly appreciated. Thanks!

Please consider the possibilities given below:
1. For N=2, One claim of 1 unit and another of 2 units will give S=3 units
2. For N=2, Both the claims of 2 units will give S=4 units.

Hope it helps.
 
reverting back to this questions not clear as to how did we get P(S<=2)...

Kindly assist
 
reverting back to this questions not clear as to how did we get P(S<=2)...

Kindly assist
P(S<=2)=P(S=0)+P(S=1)+P(S=2) ... All the mutually exclusive.
Now,
P(S=0)=P(N=0) ... As there is no nil claim, zero can occur only when no claim occurs.
P(S=1)=P(N=1)*P(X=1) ... 1 claim of 1 amount
P(S=2)=P(N=1)*P(X=2)+P(N=2)*P(X=1)*P(X=1) ... 1 of amount 2, 2 of amount 1 each.
 
hi Hemant, precisely my point for P(S = 2).. what we are trying to get is the probability that there are 2 units in S
1. This is possible when we have one claim of 2 units i.e. P(N=1)*P(X=2) AND
2. that there are 2 no of claims of 1 unit each i.e. P(N=2)* P(X=1)... this is where my logic goes...

However the solution that you have provided for the 2nd part is P(N=2)*P(X=1)*P(X=1).... the second P(X=1) is what is not clear.

I have tried to get the solution using a Binomial tree diagram..

Kindly guide.

Thanks.
 
Hi Pratik,
As there are 2 claims, you need to consider 2 outcomes.
You considered P(X=1) 1 time, and that inherently assumed that P(any X) in 2nd claim. Which is wrong. We want 2nd claim as 1 too(two 1s make 2). So, P(X=1) needed for 2nd claim.
 
hi Hemant, precisely my point for P(S = 2).. what we are trying to get is the probability that there are 2 units in S
1. This is possible when we have one claim of 2 units i.e. P(N=1)*P(X=2) AND
2. that there are 2 no of claims of 1 unit each i.e. P(N=2)* P(X=1)... this is where my logic goes...

However the solution that you have provided for the 2nd part is P(N=2)*P(X=1)*P(X=1).... the second P(X=1) is what is not clear.

I have tried to get the solution using a Binomial tree diagram..

Kindly guide.

Thanks.
For one claim of size 2 : P(S=2|N=1)*P(N=1) = P(X1=2)*P(N=1)
For two claims each of size 1 : (P(S=2|N=2)*P(N=2) = P(X1=1)*P(X2=1)*P(N=2)
Now add both to get P(S=2)
 
Back
Top