Actuarial Outpost SOA 289 #124
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#11
01-08-2011, 01:40 AM
 JavaGeek Member Join Date: Feb 2007 Posts: 336

$\begin{tabular}{|c|c|c|}
\hline
X & P(X) & B \\
\hline
0 & 0.06% & 0.00% \\
\hline
1 & 0.78% & 0.00% \\
\hline
2 & 4.1% & 25.0%\\
\hline
3 & 12.4% & 37.5%\\
\hline
4 & 23.2% & 37.5%\\
\hline
5 & 27.8% & 31.3%\\
\hline
6 & 20.9% & 23.5%\\
\hline
7 & 9.0% & 16.4%\\
\hline
8 & 1.7% & 10.9%\\
\hline
\end{tabular}
$

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...)
#12
01-08-2011, 10:37 AM
 SDY2010 Join Date: Nov 2010 Posts: 6

Thanks! I missed the part of the question that states that exactly 2 are greater than 8,000.
#13
08-23-2012, 01:39 AM
 JackLee Member Join Date: Jul 2007 Posts: 139

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?
#14
08-23-2012, 06:52 AM
 daaaave David Revelle Join Date: Feb 2006 Posts: 2,489

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.
__________________

#15
08-23-2012, 06:56 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,471

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
#16
08-23-2012, 10:39 AM
 JackLee Member Join Date: Jul 2007 Posts: 139

Many thanks for your guys' help! Much appreciated!

 Tags allosaur, dinoz arr wierd, exam c, probability, soa 289

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