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You are essentially being asked to compute the conditonal probability that person was not a drunk driver given that the accident was not fatal.

Actually, I think Kenneth is right. Here is another way to look at it. 1000 people were involved in accidents 10 were fatalities, 6 of them attributed
to to nondrunk drivers. A total 0f 800 -6 = 794...

Actually, it's ex:n + .5(1-nPx).

You're very welcome. Good Luck with The Exam.

Since 6 is even it must occur on or after the fifth toss. In the former case we have E[X given Y= 5]Pr[X=5 and Y=5]/Pr[Y=5] =5(1/3) =(16/3)
In the latter case, we have E[X given X>5 and Y=5]Pr[X>5...

I wrote up two different solutions for this similar problem myself. Just keep in mind that 6 is even. You have to take that into consideration when you compute probabilities. If you are still having...

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You are missing the fact that once you are told that the second toss resulted
in a 6, the expected number of tosses required to obtain your first 5 changes.

What makes you think X and Y are...

Sounds like a plan.
Good Luck

Dr. Broverman is very good.

I'm sorry to hear that. I hope your decision wasn't based strictly on your experience on this forum.Best Regards & Good Luck

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I am pleased to hear that you are interested in making the switch.

Thank You. I appreciate your efforts. Are you planning a study guide for CAS Exam 6? Loss-Reserve Estimation for P&C actuaries is of particular interest to me

I guess we both do.

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More like inability to read between the lines in your case.

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Isn't being a prospective actuary all about making predictions?
Studying stuff like stochastic calculus and martingales and working
in a wall street type of environment 70-80 hours a week doesn't...

Most high school (and some college) students can't even handle calc 1.
Moreover, most high school teachers aren't even qualified to teach real analysis! What country are you living in?

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