Thanks for the explanation. It makes perfect sense. Cancelling common facors saves times which
is exactly what is needed at exam sitting.
Annette
Daniel Ma wrote:
> Form: Reply
> Text: (18 lines follow)
> Hi Annette.
>
> Try the following:
>
> In the denominator, you have Pr(0 item is defective)+Pr(1 item is
>
> defective).
>
> In the numerator, you have Pr(all 100 items are not defective)+Pr(1st item
>
> not defective & one out of item2 thru item100 is defective).
>
> Write down the above probabilities and simplify by dividing each by 0.95
>
> raised to 99. You then should get the answer of 5.90/5.95. Good luck.
>
> Dan Ma
>
>
>
> Daniel Ma/Actuarial
> Original text: (14 lines follow)
> From: ANNETTE.@SMTP.AFLAC (Annette
>
> Thomas){annette.thomas@infinity-insurance.com}, on 6/30/98 4:04 PM:
> Has anyone worked exercise 2-5, Herzog, page 21. My solution is
> .036769/.037082=.9916 which is equivalent to Herzog's solution 5.9/5.95.
>
> I need an explanation of why Herzog states a ratio of probabilities as
> the ratio of two numbers greater than 1.00.
>
> If the solution is truly .059/.0595 or something similar please tell how
> to get there.
>
> Thanks!
>
> Use Proportional Font: true
> Previous From: ANNETTE.@SMTP.AFLAC (Annette Thomas){annette.thomas@infinity-insurance.com}
> Previous To: STUDYGRO@SMTP.AFLAC (studygroup4b){studygroup4b@casact.org}
> Original to: STUDYGRO@SMTP.AFLAC (studygroup4b){studygroup4b@casact.org}
> Attachment Count: 0