Covariance in multinomial distribution:

• positive covariance is impossible because trials are constrained by $\sum X_i = n$; an increase in one outcome necessitates a decrease in others.
• In a multinomial distribution with $n$ trials and probabilities $p_1, \dots, p_k$, the marginal distribution of each $X_i$ is $\operatorname{Binomial}(n, p_i)$ with $\operatorname{Var}(X_i) = n p_i (1 - p_i)$. Because the sum $\sum_{i=1}^k X_i = n$ is fixed, the variables are negatively correlated, and the exact covariance between any two distinct categories $i \ne j$ is $\operatorname{Cov}(X_i, X_j) = -n p_i p_j$.
• Omits both the negative sign and sample size $n$.
• Lacks the sample size parameter $n$.

Mode of sum of independent binomial variates:

• $3, 2$ would imply bimodal distribution, which occurs only if $(n+1)p$ is an integer.
• By the additive property of the binomial distribution, since $p$ is identical, $Z = X + Y \sim \operatorname{Binomial}(n_1 + n_2, p) = \operatorname{Binomial}(4 + 5, 1/3) = \operatorname{Binomial}(9, 1/3)$. The mode is given by $\lfloor (n + 1)p \rfloor$. Here $(9 + 1) \times (1/3) = 10/3 = 3.333$. Since $10/3$ is non-integral, there is a unique mode equal to the integer part: $\lfloor 3.333 \rfloor = 3$.
• $3, 4$ is incorrect.
• $4$ is the ceiling, not the mode.

Conditional distribution of independent Poisson variables:

• classical theorem in probability states that if $X \sim \operatorname{Poisson}(\lambda_1)$ and $Y \sim \operatorname{Poisson}(\lambda_2)$ independently, the conditional distribution of $X$ given $X + Y = n$ is a Binomial distribution with parameters $n$ and success probability $p = \frac{\lambda_1}{\lambda_1 + \lambda_2}$.
• The unconditional distribution of $X + Y$ is Poisson, but the conditional distribution of $X \mid (X+Y)$ is Binomial.
• Negative binomial arises from a Poisson mixture with gamma, not this conditional structure.
• Multinomial arises if conditioning more than two variables.

Calculation of coefficient of variation for Poisson distribution:

• $100 \times 3$ multiplies SD by $100$ without dividing by the mean.
• $100 \times 9$ multiplies mean by $100$.
• For a Poisson distribution with mean $\lambda = 9$, the variance is also $\sigma^2 = \lambda = 9$, so standard deviation is $\sigma = \sqrt{9} = 3$. The coefficient of variation is $\text{CV} = \frac{\sigma}{\mu} \times 100 = \frac{3}{9} \times 100 = 100 \times \left(\frac{1}{3}\right) = 33.33\%$.
• $100 \times (1/9)$ erroneously divides variance by mean.