Two random 3000 bit arrays each with exactly two ones(1), what is the probability that they have atleast one bit in common?
Probability question of common 1s
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probability
statistics
probability-theory
2 Answers
1
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 $0.001333$.
0
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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0@Jannat: It isn’t very close to $1$ in $500$; it’s about $2/3$ in $500$. – 2012-11-08