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I feel like this is fairly basic question, but I just can't wrap my head around it.

Two red, four green blocks are placed randomly in a row. Find out probability density function and cumulative distribution function for the random variable X, when X is the number of green blocks between the red blocks.

I started with thinking about the different ways the blocks could be laid out.

4 green blocks
RGGGGR - 1! * 4! * 1! = 24

3 greens blocks
RGGGRG - 1! * 3! * 1! * 1! = 6

RGGRGG - 1! * 2! * 1! * 2! = 4

RGRGGG - 1! * 1! * 1! * 3! = 6

RRGGGG - 2! * 4! = 96

But at this point I'm completely stuck. I don't think this is even heading the right way.

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    This won't have$a$pdf in the sense of something integrated with respect to Lebesgue measure, i.e. where the probability of being between $a$ and $b$ is $\int_a^b f(x)\,dx$. It will have a pdf with respect to "counting measure", i.e. a "probability mass function". The probability distribution is discrete.2012-10-03

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There are $\binom62=15$ equally likely possible positions for the pair of red blocks. Only one of those $15$ puts them on the ends, with $4$ green blocks between them. Two of the fifteen have $3$ green blocks between them: the leftmost red block must be the first or second block in the string. Similarly, there are three positions that have $2$ green blocks between the red ones, four that have only $1$ green block between the red ones, and five that have the red blocks adjacent.

From here it’s straightforward to get the pdf and cdf.

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    @Zavior: You’re welcome!2012-10-03