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Found this in a SAT II Math book:

If two integers are selected at random without replacement from the integers 1 through 6 inclusive, what is the probability that the sum of the two integers is even and greater than seven?

I am unsure about how to tackle and "even" and "greater than seven" clauses sine I only know how to use combinations to solve for things like drawing two odd numbers out of the set at the same time or more simple problems of the like. Can someone show me how to solve this problem?

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Listing the sample space. In this case it is possible to list the sample space of possible outcomes:

 --    12    13+   14    15+   16
 21    --    23    24+   25    26*+
 31+   32    --    34    35*+  36*
 41    42+   43    --    45*   46*+
 51+   52    53*+  54*   --    56*
 61    62*+  63*   64*+  65*   --

Dashes -- indicate outcomes that are impossible because sampling is done without replacement. Results greater than seven are marked with a $*$, and those with even sums are marked with $+$. There are 6 with both marks so the answer is 6/30 = 1/5.

This is a good opportunity to illustrate the general addition rule: $P(G \cup E) = P(G) + P(E) - P(G \cap E) = 12/30 + 12/30 - 6/30 = 18/30 = 3/5.$ There are 18 outcomes with one mark or the other.

Simulation. Because there is disagreement among the answers [at the time of original posting, since resolved], I did a simulation with a million performances of this experiment. Notice that the only 'legal' sums are 8 and 10. The result is 0.20 to two place accuracy. The last line of output shows the simulated distribution of the sum (exact values are 1/15, 2/15 and 3/15--some repeated).

 > m = 10^6;   s = numeric(m)
 > for (i in 1:m) {s[i] = sum(sample(1:6, 2)) }
 > mean(s==8 | s==10)
 [1] 0.199251
 > round(table(s)/m, 3)
 s
     3     4     5     6     7     8     9    10    11 
 0.067 0.067 0.133 0.134 0.200 0.133 0.134 0.066 0.067 
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If an odd number is picked, then the second number should also be odd. If an even number is picked, then a second even number should also be picked. Then just do it with brute force.

When 1 is picked, $p=0$.

When 2 is picked, the second number can be 6, $p = \frac{1}{6} \times \frac{1}{5} = \frac{1}{30}$.

When 3 is picked, the second number can be 5, $p = \frac{1}{6} \times \frac{1}{5} = \frac{1}{30}$.

When 4 is picked, the second number can be 6, $p = \frac{1}{6} \times \frac{1}{5} = \frac{1}{30}$.

When 5 is picked, the second number can be 3, $p = \frac{1}{6} \times \frac{1}{5} = \frac{1}{30}$.

When 6 is picked, the second number can be 2,4, $p = \frac{1}{6} \times \frac{2}{5} = \frac{1}{15}$.

Finally, add those possibilities up, $p=\frac{1}{5}$.

Bah, I am really sorry, I made mistake on counting before the edit. Bruce Trumbo is correct.

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