Chapter 7 Question 7.3

[Sarahlouise_23]

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!

 

[Hemant Rupani]

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

 

[satyam]

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.

 

[Pratik]

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

Kindly assist

 

[Hemant Rupani]

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.

 

[Pratik]

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.

 

[Hemant Rupani]

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.

 

[Chandrima]

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)