# Probability question of common 1s

Two random 3000 bit arrays each with exactly two ones(1), what is the probability that they have atleast one bit in common?

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It’s easiest here to calculate the probability that they have no $1$’s in common and subtract from $1$. There are $\binom{3000}2$ such arrays. No matter what the first array is, there are $\binom{2998}2$ that have no $1$’s in common with it. Thus, the probability of having no $1$’s in common is

$$\frac{\binom{2998}2}{\binom{3000}2}=\frac{2998\cdot2997}{3000\cdot2999}\approx0.998667\;,$$

and the probability of having at least one $1$ in common is about $0.001333$.

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There are $3000\choose 2$ possibilities for the second array, among these are $2998\choose 2$ that do not have a bit in common with the first. Thus the probabilitiy is $$1- \frac{2998\choose 2}{3000\choose 2}=1-\frac{2998\cdot2997}{3000\cdot2999}.$$

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the approximate answer that should come is 1/500 but i am not getting..how to get that –  Jannat Arora Nov 8 '12 at 17:25
@Jannat: It isn’t very close to $1$ in $500$; it’s about $2/3$ in $500$. –  Brian M. Scott Nov 8 '12 at 17:27