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Imagine there are three binary multisets $A$, $B$, and $C$. Lengths are $8, 16,$ and $24$ respectively. Each bit, irrespective of set, has $x$ probability of being $1$.

What is the probability that $A$ will contain no $0s$, while B will contain at least one? what is the probability that $A$ and $B$ will both contain no $0s$, while C contains at least one?

2 Answers 2

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Since each bit is independent, the number of zeros in set of size $n$, follows binomial distribution with $B(n, 1-x)$.

Let $k_A$, $k_B$ and $k_C$ denote number of zeros in corresponding sets.

$ Pr( k_A=0 \text{ and } k_B > 0) = x^{n_A } (1-x^{n_B}) $

$ Pr( k_A=0 \text{ and } k_ B=1 \text{ and } k_C > 0 ) = x^{n_A + n_B} (1-x^{n_C} ) $

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I believe that your distributions are independent since you didn't state any dependence. So,

1.Probability

$P(\{A\text{ contains no zeros}\}\cap \{B \text{ contains at least one zero} \}) = $ $P(\{A\text{ contains no zeros}\})\cdot P(\{B \text{ contains at least one zero} \}).$

The first probability is $x^8\cdot (1-x^{16})$.

2.The same method gives an answer $x^8\cdot x^{16}\cdot (1-x^{24})$.