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问题及解答

[Exer14-1] Exercise 1 of Book {Devore2017B} P.211

Posted by haifeng on 2020-05-27 18:31:28 last update 2020-05-27 18:42:01 | Edit | Answers (1)

A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let $X$ denote the number of hoses being used on the self-service island at a particular time, and let $Y$ denote the number of hoses on the full-service island in use at that time. The joint pmf of $X$ and $Y$ appears in the accompanying tabulation.

 

  $y$
$p(x,y)$ 0 1 2
  0 .10 .04 .02
$x$ 1 .08 .20 .06
  2 .06 .14 .30

 

%%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 & .10 & .04 & .02\\
~& 1 & .08 & .20 & .06\\
~& 2 & .06 & .14 & .30\\
\hline
\end{tabular}
\end{table}


 

  • (a) What is $P(X=1\ \text{and}\ Y=1)$?
  • (b) Compute $P(X\leqslant 1\ \text{and}\ Y\leqslant 1)$.
  • (c) Give a word description of the event $\{X\neq 0\ \text{and}\ Y\neq 0\}$ and compute the probability of this event.
  • (d) Compute the marginal pmf of $X$ and of $Y$. Using $p_{X}(x)$, what is $P(X\leqslant 1)$?
  • (e) Are $X$ and $Y$ independent rv's? Explain.
     

 

1

Posted by haifeng on 2020-05-27 21:30:42

(a)

\[
P(X=1\ \text{and}\ Y=1)=p(1,1)=.20
\]


(b)

\[
\begin{split}
P(X\leqslant 1\ \text{and}\ Y\leqslant 1)&=\sum_{x\leqslant 1\\ y\leqslant 1}p(x,y)\\
&=p(0,0)+p(0,1)+p(1,0)+p(1,1)\\
&=.10+.04+.08+.20\\
&=.42
\end{split}
\]


(c)

At least one hose is in use at each pump.

\[
\begin{split}
P(X\neq 0\ \text{and}\ Y\neq 0)&=\sum_{x\neq 0\\ y\neq 0}p(x,y)\\
&=\sum_{x\geq 1\\ y\geq 1}p(x,y)\\
&=p(1,1)+p(1,2)+p(2,1)+p(2,2)\\
&=.20+.06+.14+.30\\
&=.70
\end{split}
\]


(d)

The marginal pmf of $X$ is equal to $p_{X}(x)=\sum_{y}p(x,y)$. Then

\[
\begin{eqnarray}
p_X(0)&=&\sum_{y=0}^{2}p(0,y)=p(0,0)+p(0,1)+p(0,2)=.10+.04+.02=.16\\
p_X(1)&=&\sum_{y=0}^{2}p(1,y)=p(1,0)+p(1,1)+p(1,2)=.08+.20+.06=.34\\
p_X(2)&=&\sum_{y=0}^{2}p(2,y)=p(2,0)+p(2,1)+p(2,2)=.06+.14+.30=.50\\
\end{eqnarray}
\]

 

The marginal pmf of $Y$ is equal to $p_{Y}(y)=\sum_{x}p(x,y)$. Then

\[
\begin{eqnarray}
p_Y(0)&=&\sum_{x=0}^{2}p(x,0)=p(0,0)+p(1,0)+p(2,0)=.10+.08+.06=.24\\
p_Y(1)&=&\sum_{x=0}^{2}p(x,1)=p(0,1)+p(1,1)+p(2,1)=.04+.20+.14=.38\\
p_Y(2)&=&\sum_{x=0}^{2}p(x,2)=p(0,2)+p(1,2)+p(2,2)=.02+.06+.30=.38\\
\end{eqnarray}
\]

 

\[
\begin{split}
P(X\leqslant 1)&=P(X=0)+P(X=1)\\
&=p_X(0)+p_X(1)\\
&=.16+.34\\
&=.50
\end{split}
\]


(e)

No, $X$ and $Y$ are not independent random variables. Since

\[
p(0,0)=.10\neq .0384=.16\times .24=p_X(0)\cdot p_Y(0)
\]