Hi Mathtalk,
I closed the previous question without a detailed review of your
answer. Then I realized that your answer was not correct!
For the PPX condition, you give a probability of 0.006. However, PPX
cannot exist, because naturally with two placed digits, the third
cannot be a misplaced "X". It would have to be another P or a -.
Therefore, this skews all the rest of the data and therefore, none of
the previous answers are right with the exception of - - - and PPP
probabilities.
The data I have shows that a PP- has a probability of .028, a P-- or
better as .271, and XX- or better as .150. However, I don't know how
they arrived at those answers.
Please clarify your answers and let me know the proper probabilities. Thanks! |
Request for Question Clarification by
mathtalk-ga
on
09 Feb 2004 22:52 PST
Hi, davidj-ga:
Please see the computations provided on your earlier Question.
These are the results with the revised scoring interpretation:
Pr( _ _ _ ) = 0.343
Pr( X _ _ ) = 0.315
Pr( X X _ ) = 0.069
Pr( X X X ) = 0.002
Pr( P _ _ ) = 0.192
Pr( P X _ ) = 0.048
Pr( P X X ) = 0.003
Pr( P P _ ) = 0.027
Pr( P P X ) = 0.000
Pr( P P P ) = 0.001
------
1.000
Assuming the "scores" are ranked in the order shown, then "P P _" or
better would occur with chance 0.027 + 0.001 = 0.028, and "P _ _" or
better with chance 0.271. With regard to the chance of "X X _" or
better, if this were to exclude these cases:
Pr( _ _ _ ) = 0.343
Pr( X _ _ ) = 0.315
Pr( P _ _ ) = 0.192
------
0.850
then in that sense, the chance of "X X _" or better would be 0.150.
However those assessments require that both "X X _" is not better than
"P _ _" and vice versa.
Hope this clears up the confusion!
regards, mathtalk-ga
|