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. $$ The derivation of confidence intervals isn't straightforward given the form of the probability mass function. However, they are easy to approximate with Monte Carlo simulations. … group", ylab = "Group size") hist(nb.guests, breaks = 21, probability = TRUE, main = "", xlab = "Guests") par(mfrow = c(1, 1)) ## Theoretical mean and variance c(sum(n * p), sum(n * p * (1-p))) #[1] …

To compute the requested probability, one needs to know the joint distribution of $X$ and $Y$. … In the second approach the requested probability is obtained by integrating the joint density on the domain of $X$ and $Y$ corresponding to situations where the two buddies meet. …

By the independence of $Z_1$ and $Z_2$ the joint probability is equal to the product of the marginal probabilities. … If $t = 0$, then $P(Z_1 < t < Z_2) = 0.25$ as one would expect since events $\{Z_1 < t\}$ and $\{Z_2 > t\}$ are independent and both have probability $0.5$. …

indicating failure, that is $$ I_k = \begin{cases} 1, & \text{if the node $k$ is a failure},\\ 0, & \text{if the node $k$ is a failure}\\ \end{cases} $$ and let $p_k = {\rm Pr}(I_k = 1)$ denote the probability … Compute ${\rm Pr}(I_A = 1 | I_B = 1)$ using the definition of conditional probability, $$ {\rm Pr}(I_A = 1 | I_B = 1) = \frac{{\rm Pr}(I_A = 1, I_B = 1)}{{\rm Pr}(I_B = 1)}, $$ and point 2. To come …

Since the question is flagged as self-study, I'll just provide some hints to (hopefully) help you derive the solution. I'll amend/complete my answer based on your progress. As a starting point, you m …

. $$ If the two random variables are independent, the probability is \begin{align*} P(X_1 \leq X_2) & = \int_{-\infty}^{\infty} \int_{-\infty}^{x_2} f_{X_1}(x_1) f_{X_2}(x_2) dx_1 dx_2\\ & = \int_{ … 2$ follows a shifted exponential distribution with rate $\lambda > 0$ and location parameter $a$, that is $F_{X_2}(x) = 1 - \exp\{-\lambda(x-a)\}$ if $x > a$ and $F_{X_2}(x) = 0 $ otherwise, then the probability …

The Irwin-Hall distribution is the distribution of a random variable defined as the sum of a number of independent random variables, each having a uniform distribution. It won't be useful here since t …

The distribution of the sum of independent random variables/vectors can often be obtained easily using moment generating functions (MGF). In short, the MGF of the sum is the product of the MGFs of th …