SOA 289 #124

[JavaGeek]

P(X) is the probability that there are X scientists eaten.
Where "B" is the probability that exactly 2 of the "X" scientist eaten exceed 8000 cal. (meaning X-2 were less than 8000 cal.)

The solution is equal to SUM(P(X)*B) or: 0.296492 (should be 0.3, but there are rounding errors as the above table isn't exact...)

[SDY2010]

Thanks! I missed the part of the question that states that exactly 2 are greater than 8,000.

[JackLee]

To me it seems the solution is only calculating....

Calculate "Exactly two of those eaten have at least 8000 calories".


I do not see the solution calculating "the probably that two or more scientists are eaten" --> which is the initial part of the question.

Is it me or.... wording problem?

[daaaave]

If exactly 2 scientists with 8,000 calories are eaten, then at least 2 scientists (in particular, the 2 with 8,000+ calories) are eaten.

[Gandalf]

The wording may be a little odd, but the solution is OK.

Going back to basic probability terms.
Let A = probability 2 or more scientists are eaten
Let B = exactly 2 scientists over 8000 calories are eaten.

P(A and B)=P(B) since B is a subset of A.

You could, instead of the SOA solution, calculate
P(A and B) = P(A)*P(B|A), which is the approach discussed earlier in the thread (in the more detailed case of
P(2 eaten)*P(given 2, both are over 8000)+P(3 eaten)*P(given 3, exactly 2 are over 8000)+...+P(8 eaten)*P(given 8, exactly 2 are over 8000)
then P(2 eaten)+P(3 eaten)+...P(8 eaten)=P(A).
Works, as JavaGeek shows. Not feasible under exam conditions.

ninja'd

[JackLee]

Many thanks for your guys' help! Much appreciated!