Question 203046
Sorry jsmall I'ma afraid your answer is wrong.


The question is asking : if 2 numbers are picked randomly WITHOUT REPLACEMENT. 

hence, while you may use the multiplicative probability properties, the second probability of drawing a number that is not 1 is not 5/6  (further more , the student is asking about the numbers (1,2,3,4,5)  , 6 is not included


But lets assume we are dealing with (1,2,3,4,5,6) when you pick your first choice, you have 5/6 chance of picking numbers not 1, 

namely, 2,3,4,5,6.

in your second choice, since you are drawing without replacement, you have only  5 options left, assumed you have picked 4 for your first choice, you now have 1,2,3,5,6 to choose from, however, you don't want to choose one, so you have 4/5 chance of picking numbers not 1.


hence, the probability of not drawing 2 numbers whose sum is greater than product is (5/6)*(4/5) = (4/6) = 2/3.

probability of drawing 2 numbers whose sum is greater than product is then 1/3



Alternatively, you may see that there are (6 choose 2) = 15 ways to pick 2 random numbers out of 6.  And there are 5 ways to choose 2 numbers such that 1 is included (1,2) (1,3) ..... (1,5). since the probability of drawing any two numbers are the same, therefore, 5/15 = 1/3 is the probability of drawing 2 numbers whose sum is greater than their product.



So I think you have all the tools you need to solve the problem with just (1,2,3,4,5) :)