问题及解答

The joint probability distribution of the number $X$ of cars and the number $Y$ of buses per signal cycle at a proposed left turn lane is displayed in the accompanying joint probability table.

%%Table in LaTeX

\begin{table}[htbp]
\centering
\begin{tabular}{cc|p{0.5in}p{0.5in}p{0.5in}}
 & & & $y$ & \\
$p(x,y)$ &  & 0 & 1 & 2 \\\hline
\multirow{3}{*}{$x$}& 0 & .025 & .015 & .010\\
~& 1 & .050 & .030 & .020\\
~& 2 & .125 & .075 & .050\\
~& 3 & .150 & .090 & .060\\
~& 4 & .100 & .060 & .040\\
~& 5 & .050 & .030 & .020\\
\hline
\end{tabular}
\end{table}

• (a) What is the probability that there is exactly one car and exactly one bus during a cycle?
• (b) What is the probability that there is at most one car and at most one bus during a cycle?
• (c) What is the probability that there is exactly one car during a cycle? Exactly one bus?
• (d) Suppose the left turn lane is to have a capacity of five cars and one bus is equivalent to three cars. What is the probability of an overflow during a cycle?
• (e) Are $X$ and $Y$ independent rv's? Explain.

(a) 

The probability that there is exactly one car and exactly one bus during a cycle is

\[
P(X=1\ \text{and}\ Y=1)=p(1,1)=.030
\]

(b)

The probability that there is at most one car and at most one bus during a cycle is

\[
\begin{split}
P(X\leqslant 1\ \text{and}\ Y\leqslant 1)&=\sum_{x\leqslant 1\\ y\leqslant 1}p(x,y)\\
&=p(0,0)+p(0,1)+p(1,0)+p(1,1)\\
&=.025+.015+.050+.030\\
&=.120
\end{split}
\]

(c)

The probability that there is exactly one car during a cycle is

\[
\begin{split}
P(X=1)=p_X(1)=\sum_{y=0}^{2}p(1,y)\\
&=p(1,0)+p(1,1)+p(1,2)\\
&=.050+.030+.020\\
&=.010
\end{split}
\]

The probability that there is exactly one bus during a cycle is

\[
\begin{split}
P(Y=1)=p_Y(1)=\sum_{x=0}^{5}p(x,1)\\
&=p(0,1)+p(1,1)+p(2,1)+p(3,1)+p(4,1)+p(5,1)\\
&=.015+.030+.075+.090+.060+.030\\
&=.0300
\end{split}
\]

(d)

By assumption, for the capacity of the left turn lane, one bus is equivalent to three cars. That is, the total number of cars and buses is $X+3Y$. Then the probability of an overflow during a cycle is $P(X+3Y > 5)$.

Here $X\in\{0,1,2,3,4,5\}$ and $Y\in\{0,1,2\}$. Thus,

\[
\begin{split}
P(X+3Y > 5)&=P(X > 2,\ Y=1)+P(Y=2)\\
&=\sum_{x=3}^{5}p(x,1)+\sum_{x=0}^{5}p(x,2)\\
&=\bigl[p(3,1)+p(4,1)+p(5,1)\bigr]+\bigl[p(0,2)+p(1,2)+p(2,2)+p(3,2)+p(4,2)+p(5,2)\bigr]\\
&=(.090+.060+.030)+(.010+.020+.050+.060+.040+.020)\\
&=.380
\end{split}
\]

(e)

Yes, $X$ and $Y$ are independent random variables.

It is easy to calculate the marginal pmf of $X$ and $Y$.

That is, we get two vector, $p_X=(.05,.1,.25,.3,.2,.1)$ and $p_Y=(.5,.3,.2)$. And

\[
p_X^T\cdot p_Y=\begin{pmatrix}
.05\\
.1\\
.25\\
.3\\
.2\\
.1
\end{pmatrix}\cdot(.5,.3,.2)=\begin{pmatrix}
.025 & .015 & .010\\
.050 & .030 & .020\\
.125 & .075 & .050\\
.150 & .090 & .060\\
.100 & .060 & .040\\
.050 & .030 & .020
\end{pmatrix}
\]

Hence, $X$ and $Y$ are independent random variables.