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So I am having a problem trying to solve the following question.

Lets say with have $x\times n$ balls, where $x$ is a values in range $0 < x \leq 1$ and we have $n$ bins.

I'm trying to figure out what the expected number of bins containing at least $1$ ball in them, but the $x$ is throwing me off, does anyone know how to tackle this?

$x$ will be a value such that $x\times n$ results in an integer.

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    @Arthur yes, i meant a 0 < x <=1 – 2017-02-27
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    What is the meaning of say $(1/8\sqrt{2})\times 3$ balls? – 2017-02-27
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    @coffeemath I believe that n × x must be another whole number otherwise the values don't work out. So, if you had 3 balls x could be 1/2, 2/3, or 1 – 2017-02-27
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    OK so maybe put that constraint on $x$ into the question... – 2017-02-27

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The probability of one certain bin to be not hit by one certain ball is $$1-\frac{1}{n}.$$

There are $nx$ balls in total, so the probability one one bin to be not hit by any ball is $$\left(1-\frac{1}{n}\right)^{xn}$$

The probability of a certain bin to be non-empty is simply the inverse of this, i.e. $$1-\left(1-\frac{1}{n}\right)^{xn}.$$

The expected number of non-empty bins is then simply the sum of this expression over all bins. I.e. the answer you are looking for is $$n-n\left(1-\frac{1}{n}\right)^{xn}$$

for large $n$ you might want to approximate this with an exponential function.

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    May I ask how the x gets into the exponent place? – 2017-02-27
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    the exponent $xn$ is the number of balls. I edited the answer to make it clearer. Hope this helps. – 2017-02-27