Question

when four given numbers are multipled together, the product is negative.which of the following could be true about the four numbers?
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Answer 1

Answer:

Alon has answered one interpretation of the question. I'll tackle the (less useful) variant where the numbers are chosen without replacement. That is, if we choose four different numbers between 1 and 10 inclusive, what is the probability that their product is divisible by (a) 5, or (b) 10?

To start, there are (104)(104) = 210 ways of choosing the four numbers.

For case (a) we steal Alon's observation: the product is not divisible by 5 precisely when noneof the four numbers is either 5 or 10. There are eight such numbers, and four numbers can be chosen from among them in (84)(84) = 70 ways. Therefore there are 210-70 = 140 ways to choose the numbers such that the product is divisible by 5, and the probability that this happens is 140/210 = 2/3.

For case (b) it's easiest to cheat by asking this question: How is it possible for the product to be divisible by 5 but not by 10? This can happen only if 5 is chosen and none of the other three chosen numbers is even. There are four possibilities for these other three numbers, namely 1, 3, 7, and 9, and there are (43)(43) = 4 ways of choosing three of them. So of the 140 ways to choose four numbers with a product divisible by 5, there are 136 ways that the product is divisible by 10, and the probability is therefore 136/210 or about 0.65.