answer:L'equazione logica Y = A'BC' + ABC + AB'C + ABC' dove A, B, C sono le variabili di ingresso, corrisponde alla tabella della verità data.

answer:Assuming X and Y are independent, we can use the formula for cov(X,Y) to obtain: cov(X,Y) = E[XY] - E[X]E[Y] First, we need to find E[XY]. Using the definition of covariance, we have: E[XY] = ∑x∑y xy P(X=x, Y=y) Since X follows a Poisson distribution with parameter λ and Y follows a binomial distribution with parameters n and p, we have: P(X=x, Y=y) = P(X=x)P(Y=y) = (e^(-λ) λ^x / x!) (n choose y) p^y (1-p)^(n-y) Substituting this into the expression for E[XY], we get: E[XY] = ∑x∑y xy (e^(-λ) λ^x / x!) (n choose y) p^y (1-p)^(n-y) = (e^(-λ) / x!) * λ * p * (n ∑y y(y-1) (n-1 choose y-1) p^(y-1) (1-p)^(n-y) = λnp (1-p) + λn(n-1)p^2 Therefore, cov(X,Y) = λnp(1-p) + λn(n-1)p^2 - λ*n*p = λnp(1-p) - λnp(1-p) = 0. Thus, under the assumption that X and Y are independent, the covariance between X and Y is zero.

answer:If X and Y are dependent, we cannot make any general statement about the covariance between X and Y without further assumptions about their relationship. One common way to model dependent variables is through a bivariate Poisson distribution. In this case, the joint probability mass function of X and Y is given by: P(X=x, Y=y) = (λ1λ2)^{x+y} / (x+y)! * exp(-λ1-λ2) * ρ^{xy} where λ1 and λ2 are the means of X and Y, respectively, and ρ is the correlation coefficient between X and Y. Using this joint pmf, we can compute the covariance between X and Y as: cov(X,Y) = E[XY] - E[X]E[Y] = λ1λ2ρ - λ1λ2 = λ1λ2(ρ - 1) So, if we assume that X and Y follow a bivariate Poisson distribution with correlation coefficient ρ, then the covariance between X and Y is proportional to their means and the correlation coefficient. If ρ = 1, then X and Y are perfectly positively correlated and the covariance is equal to the product of their means. If ρ = -1, then X and Y are perfectly negatively correlated and the covariance is equal to the negative product of their means. If ρ = 0, then X and Y are uncorrelated and the covariance is equal to zero.

answer:One way to model the dependence between X and Y is to assume a Poisson-binomial distribution, where Y is the sum of n independent Bernoulli trials with success probability p and X depends on Y through the conditional distribution P(X=x|Y=y) = Poisson(λy). In this case, the joint distribution of X and Y is given by: P(X=x,Y=y) = P(X=x|Y=y) P(Y=y) = Poisson(λy) (n choose y) p^y (1-p)^(n-y) The marginal distributions are: P(X=x) = ∑_{y=0}^n Poisson(λy) (n choose y) p^y (1-p)^(n-y) and P(Y=y) = (n choose y) p^y (1-p)^(n-y) Using these distributions, we can compute the covariance between X and Y: cov(X,Y) = E[XY] - E[X] E[Y] where E[XY] = ∑_{x=0}^∞ ∑_{y=0}^n x y Poisson(λy) (n choose y) p^y (1-p)^(n-y) = ∑_{y=0}^n y Poisson(λy) (n choose y) p^y (1-p)^(n-y) ∑_{x=0}^∞ x P(X=x|Y=y) = ∑_{y=0}^n λy y (n choose y) p^y (1-p)^(n-y) = λ ∑_{y=0}^n y^2 (n choose y) p^y (1-p)^(n-y) Using the identity: ∑_{y=0}^n y^2 (n choose y) p^y (1-p)^(n-y) = n(n-1)p^2 + np(1-p), we have: E[XY] = λ (n(n-1)p^2 + np(1-p)) Also, E[X] = ∑_{x=0}^∞ x P(X=x) = λ and E[Y] = ∑_{y=0}^n y (n choose y) p^y (1-p)^(n-y) = np Therefore, cov(X,Y) = λ (n(n-1)p^2 + np(1-p)) - λ np = λ (n-1) (np^2 - p) So, if we assume that X and Y follow a Poisson-binomial distribution, then the covariance between X and Y is proportional to the mean of X (λ), the sample size of Y (n), and the difference between the mean and variance of Y (np(1-p)). Note that this Poisson-binomial assumption may not be appropriate or realistic depending on the context and nature of the dependence between X and Y.