Question 203046: If two numbers are randomly chosen without replacement from {1, 2, 3, 4, 5}, what is the probability their sum is greater than their product? Express your answer as a common fraction.
Found 2 solutions by jsmallt9, albertemc:
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! If you check out the various combinations of sums and products of the numbers from 1 to 6, I think you will find that the only way for the sum of two numbers to be greater than the product of those two is if one of the numbers is 1. In other words, when selecting from {2, 3, 4, 5, 6}, the sum of any two numbers selected is never greater than the product.
So the problem is now: If two numbers are randomly chosen without replacement from {1, 2, 3, 4, 5, 6}, what is the probability that at least one of the numbers is 1? And the easiest way to find this is to find the probability that no 1's are selected and then subtracting that from 1:
P(at least 1 one) = 1 - P(no ones)
P(no ones) = (5/6)*(5/6) = 25/36
P(at least 1 one) = 1 - 25/36 = 11/36
P(sum greater than product) = P(at least 1 one) = 11/36
Answer by albertemc(1) (Show Source):
You can put this solution on YOUR website! 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) :)
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