问题及解答

The number of customers waiting for gift-wrap service at a department store is an rv $X$ with possible values $0,1,2,3,4$ and corresponding probabilities $.1,.2,.3,.25,.15$. A randomly selected customer will have $1,2$, or $3$ packages for wrapping with probabilities $.6,.3$, and $.1$, respectively. Let $Y$= the total number of packages to be wrapped for the customers waiting in line (assume that the number of packages submitted by one customer is independent of the number submitted by any other customer).

• (a) Determine $P(X=3,\ Y=3)$, i.e., $p(3,3)$.
• (b) Determine $p(4,11)$.
   

(a)

\[
P(X=3,\ Y=3)=p_X(3)\cdot p_Y(3)=0.25\times(0.6\times 0.6\times 0.6)=0.054
\]

(b)

Let $\beta=(i_1,i_2,\ldots,i_X)$ be the vector of the number of packages each customer are waiting for wrapping.

When $X=4$, and $Y=11$. Then one of the element of $\beta$ is 2, and the others are 3. For example, $\beta=(3,3,3,2)$. And other situations are $(3,3,2,3)$, $(3,2,3,3)$, $(2,3,3,3)$. 

\[
\begin{split}
p(4,11)&=p_X(4)\cdot p_Y(11)\\
&=.15\times (.1\times .1\times .1\times .3)\times 4\\
&= 0.000180
\end{split}
\]