Question 8.9: A restaurant sells three different types of dessert: chocola......

The random variable X = “choice of dessert” follows a multinomial distribution. More precisely, X_{1} describes whether chocolate brownies were chosen, X_{2} describes whether yoghurt was chosen, X_{3} describes whether lemon tart was chosen and X = {X_{1}, X_{2}, X_{3}}.

(a) Using the PMF of the multinomial distribution, we get

P(X_{1} = n_{1}, X_{2} = n_{2}, . . . , X_{k} = n_{k} ) =\frac{n!}{n_{1}!n_{2}! \ldots n_{k}!} \cdot p^{n_{1}}_{1} \ldots p^{n_{k}}_{k}

P(X_{1} =2 , X_{2} = 1 , X_{3} = 2 ) =\frac{5!}{2!1! 2!} \cdot 0.2^{2} \cdot 0.3^{1} \cdot 0.5^{2}

= 9%.

We would have obtained the same results in R as follows:

dmultinom(c(2,1,2),prob=c(0.2,0.3,0.5))

(b) The probability of choosing lemon tart for the first two guests is 1.We thus need to determine the probability that 3 out of the remaining 3 guests order lemon tart:

P(X_{1} =0 , X_{2} = 0 , X_{3} = 3 ) =\frac{3!}{0!0! 3!} \cdot 0.2^{0} \cdot 0.3^{0} \cdot 0.5^{3}

= 12.5%.

Using dmultinom(c(0,0,3),prob=c(0.2,0.3,0.5)) in R, we get the same result.

(c) The expectation vector is

E(X) = (np_{1}, . . . , np_{k} ) = (20 · 0.2, 20 · 3, 20 · 0.5) = (4, 6, 10).

This means we expect 4 guests to order brownies, 6 to order yoghurt, and 10 to order lemon tart. The covariance matrix can be determined as follows:

Cov\left(X_{i},X_{j}\right) =\left\{\begin{matrix} np_{i}\left(1-p_{i}\right)    if  i = j\\ -np_{i}p_{j}                 if  i \neq j.\end{matrix} \right.

Using n = 20, p_{1} = 0.2, p_{2} = 0.3 and p_{3} = 0.5, we obtain the covariance matrix as:

\left(\begin{matrix} 3.2 & -1.2 & -2 \\ -1.2 & 4.2 & -3 \\ -2 & -3 & 5 \end{matrix} \right)