Question

Which of the following can not be valid assignment of probabilities for outcomes of sample space S = {$\{{\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,{\omega }_5,{\omega }_6,{\omega }_7\}$

Assignment	${\omega }_1$	${\omega }_2$	${\omega }_3$	${\omega }_4$	${\omega }_5$	${\omega }_6$	${\omega }_7$
(a)	0.1	0.01	0.05	0.03	0.01	0.2	0.6
(b)	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$
(c)	0.1	0.2	0.3	0.4	0.5	0.6	0.7
(d)	-0.1	0.2	0.3	0.4	-0.2	0.1	0.3
(e)	$\frac{1}{14}$	$\frac{2}{14}$	$\frac{3}{14}$	$\frac{4}{14}$	$\frac{5}{14}$	$\frac{6}{14}$	$\frac{15}{14}$

Answer 1

(a)

${\omega }_1$	${\omega }_2$	${\omega }_3$	${\omega }_4$	${\omega }_5$	${\omega }_6$	${\omega }_7$
0.1	0.01	0.05	0.03	0.01	0.2	0.6

Here each of the numbers p(${\omega }_i$) is positive and less than $1$
Sum of probabilities
$=p\left({\omega }_1\right)+p\left({\omega }_2\right)+p\left({\omega }_3\right)+p\left({\omega }_4\right)+p\left({\omega }_5\right)+p\left({\omega }_6\right)+p\left({\omega }_7\right)$
$=0.1+0.01+0.05+0.03+0.01+0.2+0.6$= 1$
Thus the assignment is valid
(b)

${\omega }_1$	${\omega }_2$	${\omega }_3$	${\omega }_4$	${\omega }_5$	${\omega }_6$	${\omega }_7$
$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$

Here each of the number p(${\omega }_i$) is positive and less than $1$
Sum of probabilities
= $p\left({\omega }_1\right)+p\left({\omega }_2\right)+p\left({\omega }_3\right)+p\left({\omega }_4\right)+p\left({\omega }_5\right)+p\left({\omega }_6\right)+p\left({\omega }_7\right)$
= $\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=7\times \frac{1}{7}=1$
Thus the assignment is valid
(c)

${\omega }_1$	${\omega }_2$	${\omega }_3$	${\omega }_4$	${\omega }_5$	${\omega }_6$	${\omega }_7$
0.1	0.2	0.3	0.4	0.5	0.6	0.7

Here each of the numbers p(${\omega }_i$) is positive and less than $1$
Sum of probabilities
=
$p\left({\omega }_1\right)+p\left({\omega }_2\right)+p\left({\omega }_3\right)+p\left({\omega }_4\right)+p\left({\omega }_5\right)+p\left({\omega }_6\right)+p\left({\omega }_7\right)$
$=0.1+0.2+0.3+0.4+0.5+0.6+0.7$
$=2.8$ $\ne 1$
Thus the assignment is not valid
(d)

${\omega }_1$	${\omega }_2$	${\omega }_3$	${\omega }_4$	${\omega }_5$	${\omega }_6$	${\omega }_7$
-0.1	0.2	0.3	0.4	-0.2	0.1	0.3

Here p$\left({\omega }_1\right)$ and p$\left({\omega }_5\right)$ are negative
Hence the assignment is not valid
(e)

${\omega }_1$	${\omega }_2$	${\omega }_3$	${\omega }_4$	${\omega }_5$	${\omega }_6$	${\omega }_7$
$\frac{1}{14}$	$\frac{2}{14}$	$\frac{3}{14}$	$\frac{4}{14}$	$\frac{5}{14}$	$\frac{6}{14}$	$\frac{15}{14}$

$p\left({\omega }_7\right)=\frac{15}{14}>1$
Hence the assignment is not valid