November 1996
cascoinc ( poduska@cascoconsulting.com )
Wed, 28 Apr 1999 09:41:05 -0700
Q. 7
Think of this one inductively -- If the A is used on the first roll, the
posterior probability of A is 1/2*[(1/2*1/3*1/3)+(1/2*1/3*1)]/
{1/2*[(1/2*1/3*1/3)+(1/2*2/3*1)]+1/2* [(1/2*2/3*1/3)+(1/2*2/3*1)]}
= 1/3 Posterior probability of B is 2/3.
If the A is used on the first two rolls, the posterior probability of A
is 1/3*{[(1/2*1/3*1/3)+(1/2*1/3*1)]^2}/
(1/3*{[(1/2*1/3*1/3)+(1/2*2/3*1)]^2}+2/3*
{[(1/2*2/3*1/3)+(1/2*2/3*1)]}^2)
= 1/9 = (1/3)^2 = (1/3)^2 Posterior probability of B is (2/3)^2.
So, on the nth trial, the probability of A = (1/3)^(n-1) (Take it out a
couple more steps to be sure of nth trial.)
After infinite trials (using the same die=no longer random), the
posterior probability of B goes to 1, probability of A goes to 0.
The expected frequency for B is 2/3. The expected severity is 4
=(1/2*1/3*2)+(1/2*2/3*8)+(1/2*1*2)
Answer: 2/3*4=8/3=2.667
Q. 17 I did the same as you when I took this exam. The point I was
missing is the the probability is uniform over the entire circle around
each target. If you draw a picture, you'll see that S is within the
circle for all three targets, meaning that it's equally likely that the
shooter was aiming at any of the targets. So the posterior estimate is
the central point of the triangle.