问题及解答

A certain market has both an express checkout line and a superexpress checkout line. Let $X_1$ denote the number of customers in line at the express checkout at a particular time of day and let $X_2$ denote the number of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of $X_1$ and $X_2$ is as given in the accompanying table.

%%Table in LaTeX

\begin{table}[htbp]
\centering
\begin{tabular}{cc|p{0.5in}p{0.5in}p{0.5in}p{0.5in}}
 & & & $x_2$ & & \\
 & & 0 & 1 & 2 & 3\\\hline
\multirow{5}{*}{$x_1$}& 0 & .08 & .07 & .04 & .00\\
~& 1 & .06 & .15 & .05 & .04\\
~& 2 & .05 & .04 & .10 & .06\\
~& 3 & .00 & .03 & .04 & .07\\
~& 4 & .00 & .01 & .05 & .06\\
\hline
\end{tabular}
\end{table}

• (a) What is $P(X_1=1,\ X_2=1)$, that is, the probability that there is exactly one customer in each line?
• (b) What is $P(X_1=X_2)$, that is, the probability that the numbers of customers in the two lines are identical?
• (c) Let $A$ denote the event that there are at least two more customers in one line than in the other line. Express $A$ in terms of $X_1$ and $X_2$, and calculate the probability of this event.
• (d) What is the probability that the total number of customers in the two lines is exactly four? At least four?
   

(a)

\[
P(X_1=1,\ X_2=1)=p(1,1)=.15
\]

(b)

\[
\begin{split}
P(X_1=X_2)&=\sum_{x=0}^{3}p(x,x)\\
&=p(0,0)+p(1,1)+p(2,2)+p(3,3)\\
&=.08+.15+.10+.07\\
&=.40
\end{split}
\]

(c) 

The event $A$ is

\[A=\{X_1\geqslant X_2+2\ \text{or}\ X_2\geqslant X_1+2\}.\]

Hence, the probability of event $A$ is

\[
\begin{split}
P(A)&=P(|X_1-X_2|\geqslant 2)\\
&=\sum_{x-y\geqslant 2}p(x,y)+\sum_{y-x\geqslant 2}p(x,y)\\
&=\sum_{|x-y|\geqslant 2}p(x,y)\\
&=1-\sum_{|x-y| < 2}p(x,y)\\
&=1-\sum_{|x-y|\leqslant 1}p(x,y)\\
&=1-\sum_{x=0}^{3}p(x,x)-\sum_{|x-y|=1}p(x,y)\\
&=1-.40-\Bigl[p(1,0)+p(0,1)+p(2,1)+p(1,2)+p(3,2)+p(2,3)+p(4,3)\Bigr]\\
&=.60-\Bigl[.06+.07+.04+.05+.04+.06+.06\Bigr]\\
&=.60-.38\\
&=.22
\end{split}
\]

(d)

The probability that the total number of customers in the two lines is exactly four is

\[
\begin{split}
P(X_1+X_2=4)&=\sum_{x+y=4}p(x,y)\\
&=p(1,3)+p(2,2)+p(3,1)+p(4,0)\\
&=.04+.10+.03+.00\\
&=.17
\end{split}
\]

The probability that the total number of customers in the two lines is at least four is

\[
\begin{split}
P(X_1+X_2\geqslant 4)&=\sum_{x+y\geqslant 4}p(x,y)\\
&=p(1,3)+p(2,2)+p(3,1)+p(4,0)\\
&\quad +p(2,3)+p(3,2)+p(3,3)+p(4,1)+p(4,2)+p(4,3)\\
&=.04+.10+.03+.00\\
&\quad +.06+.04+.07+.01+.05+.06\\
&=.46
\end{split}
\]