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Statistics / Mathematical Statistics with Applications 7 / Chapter 5 / Problem 1E

Mathematical Statistics with Applications | 7th Edition | ISBN: 9780495110811 | Authors: Dennis Wackerly; William Mendenhall; Richard L. Scheaffer

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Contracts for two construction jobs are randomly assigned

Chapter 5, Problem 1E

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QUESTION:

Problem 1E

Contracts for two construction jobs are randomly assigned to one or more of three firms, A, B, and C. Let Y1 denote the number of contracts assigned to firm A and Y2 the number of contracts assigned to firm B. Recall that each firm can receive 0, 1, or 2 contracts.

a Find the joint probability function for Y1 and Y2.

b Find F(1, 0).

Questions & Answers

QUESTION:

Problem 1E

a Find the joint probability function for Y1 and Y2.

b Find F(1, 0).

ANSWER:

Answer:

Step 1 of 2:

(a)

The contract for two construction jobs are randomly assigned to one or more of three firms, $A,\ B,\ and\ C.$ Let ${Y}_{1}$ denote the number of contracts assigned to firm $A$ and ${Y}_{2}$ the number of contracts assigned to firm $B.$

Recall that each firm can receive $0,\ 1,\ or\ 2$ contracts.

We need to find the joint probability function for ${Y}_{1}$ and ${Y}_{2}.$

Let ${Y}_{1}$ and ${Y}_{2}$ be discrete random variables. The joint (or bivariate) probability function for ${Y}_{1}$ and ${Y}_{2}$ is given by,

$p({y}_{1},\ {y}_{2})\ =P({Y}_{1}\ =\ {y}_{1},\ {Y}_{2}\ ={y}_{2})\ \ \ \ \ \ \ \ \ \ \ \ -\infty{}\ {<y}_{1}<\infty{},\ -\infty{}\ {<y}_{2}<\infty{}.$

…………(1)

Let $(i,\ j)$ denote the pair of event that $i$ contracts are assigned to firm $A$ and $j$ contracts are assigned to firm $B.$

Here $i\ and\ j$ can take values $0,1,\ and\ 2$ since there are only two construction jobs.

To find the total number of events in the sample space, we need to find out the number of pairs.

So by the product rule, a total number of ways is $3\times{}3\ =9.$

Therefore the sample space $S,$

$\{(0,0),\ (0,1),\ (0,2),\ (1,0),\ (1,1),\ (1,2),\ (2,0),\ (2,1),\ (2,2)\}\ and\ n(s)\ =9$

Hence the probability that $0$ contracts is assigned to firm $A,$ and 0 contracts assigned to firm $B.$

Using equation (1),

$p(0,\ 0)\ =P({Y}_{1}\ =\ 0,\ {Y}_{2}\ =0)$

$=\frac{{(}_{\ 0}^{\ 2}\ )\times{}\ {(}_{\ 0}^{\ 2}\ )}{n(s)}$ [ ${(}_{\ 0}^{\ 2}\ )\ ={C}_{r}^{n}$ ]

$p(0,\ 0)=\frac{1\times{}\ 1}{9}=\frac{\ 1}{9}$

Similarly, we can write,

$p(0,\ 1)\ =P({Y}_{1}\ =\ 0,\ {Y}_{2}\ =1)$

$=\frac{{(}_{\ 0}^{\ 2}\ )\times{}\ {(}_{\ 1}^{\ 2}\ )}{9}$

$p(0,\ 0)=\frac{1\times{}\ 2}{9}=\frac{\ 2}{9}$

$p(0,\ 2)\ =P({Y}_{1}\ =\ 0,\ {Y}_{2}\ =2)$

$=\frac{{(}_{\ 0}^{\ 2}\ )\times{}\ {(}_{\ 2}^{\ 2}\ )}{9}$

$p(0,\ 2)=\frac{1\times{}\ 1}{9}=\frac{\ 1}{9}$

Similarly, we can write the probability,

$p(1,\ 0)\ =P({Y}_{1}\ =\ 1,\ {Y}_{2}\ =0)\ =\frac{{(}_{\ 1}^{\ 2}\ )\times{}\ {(}_{\ 0}^{\ 2}\ )}{9}=\frac{2}{9}$

$p(1,\ 1)\ =P({Y}_{1}\ =\ 1,\ {Y}_{2}\ =1)\ =\frac{{(}_{\ 1}^{\ 2}\ )\times{}\ {(}_{\ 1}^{\ 1}\ )}{9}\ =\ \frac{2}{9}$

$p(1,\ 2)\ =P({Y}_{1}\ =\ 1,\ {Y}_{2}\ =2)\ =0=not\ possible\ only\ 2\ contracts\ can\ be\ assigned$

$p(2,\ 0)\ =P({Y}_{1}\ =\ 2,\ {Y}_{2}\ =0)\ =\frac{{(}_{\ 2}^{\ 2}\ )\times{}\ {(}_{\ 0}^{\ 2}\ )}{9}=\frac{1}{9}$

$p(2,\ 1)\ =P({Y}_{1}\ =\ 2,\ {Y}_{2}\ =1)\ =0==not\ possible\ only\ 2\ contracts\ can\ be\ assigned$

$p(2,\ 2)\ =P({Y}_{1}\ =\ 2,\ {Y}_{2}\ =2)\ =0=not\ possible\ only\ 2\ contracts\ can\ be\ assigned$

Lets update the above values in the joint table.

Hence the joint distribution of ${Y}_{1}\ and\ {Y}_{2}\ is,$

${y}_{2}$ \ ${y}_{1}$	0	1	2	Total
0	$\frac{\ 1}{9}$	$\frac{\ 2}{9}$	$\frac{\ 1}{9}$	$\frac{\ 4}{9}$
1	$\frac{\ 2}{9}$	$\frac{\ 2}{9}$	0	$\frac{\ 4}{9}$
2	$\frac{\ 1}{9}$	0	0	$\frac{\ 1}{9}$
Total	$\frac{\ 4}{9}$	$\frac{\ 4}{9}$	$\frac{\ 1}{9}$	1