Which of the following can not be valid assignment of probabilities for outcomes of sample space S = { ${ω_{1}, ω_{2}, ω_{3}, ω_{4}, ω_{5}, ω_{6}, ω_{7}}$

Assignment $ω_{1}$ $ω_{2}$ $ω_{3}$ $ω_{4}$ $ω_{5}$ $ω_{6}$ $ω_{7}$
(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) $\frac{1}{1 4}$ $\frac{2}{1 4}$ $\frac{3}{1 4}$ $\frac{4}{1 4}$ $\frac{5}{1 4}$ $\frac{6}{1 4}$ $\frac{1 5}{1 4}$

Question

Verified by Toppr

Answer 1

(a)
$ω_{1}$ $ω_{2}$ $ω_{3}$ $ω_{4}$ $ω_{5}$ $ω_{6}$ $ω_{7}$
0.1 0.01 0.05 0.03 0.01 0.2 0.6
Here each of the numbers p( $ω_{i}$ ) is positive and less than $1$
Sum of probabilities
$= p (ω_{1}) + p (ω_{2}) + p (ω_{3}) + p (ω_{4}) + p (ω_{5}) + p (ω_{6}) + p (ω_{7})$
$= 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6$ = 1$$
Thus the assignment is valid
(b)

$ω_{1}$ $ω_{2}$ $ω_{3}$ $ω_{4}$ $ω_{5}$ $ω_{6}$ $ω_{7}$
$\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$ $\frac{1}{7}$
Here each of the number p( $ω_{i}$ ) is positive and less than $1$
Sum of probabilities
= $p (ω_{1}) + p (ω_{2}) + p (ω_{3}) + p (ω_{4}) + p (ω_{5}) + p (ω_{6}) + p (ω_{7})$
= $\frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} = 7 \times \frac{1}{7} = 1$
Thus the assignment is valid
(c)

$ω_{1}$ $ω_{2}$ $ω_{3}$ $ω_{4}$ $ω_{5}$ $ω_{6}$ $ω_{7}$
0.1 0.2 0.3 0.4 0.5 0.6 0.7
Here each of the numbers p( $ω_{i}$ ) is positive and less than $1$
Sum of probabilities
=

$p (ω_{1}) + p (ω_{2}) + p (ω_{3}) + p (ω_{4}) + p (ω_{5}) + p (ω_{6}) + p (ω_{7})$
$= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7$
$= 2.8$ $\neq = 1$
Thus the assignment is not valid
(d)

$ω_{1}$ $ω_{2}$ $ω_{3}$ $ω_{4}$ $ω_{5}$ $ω_{6}$ $ω_{7}$
-0.1 0.2 0.3 0.4 -0.2 0.1 0.3
Here p $(ω_{1})$ and p $(ω_{5})$ are negative
Hence the assignment is not valid
(e)

$ω_{1}$ $ω_{2}$ $ω_{3}$ $ω_{4}$ $ω_{5}$ $ω_{6}$ $ω_{7}$
$\frac{1}{1 4}$ $\frac{2}{1 4}$ $\frac{3}{1 4}$ $\frac{4}{1 4}$ $\frac{5}{1 4}$ $\frac{6}{1 4}$ $\frac{1 5}{1 4}$
$p (ω_{7}) = \frac{1 5}{1 4} > 1$
Hence the assignment is not valid

Answer 2

(a)

$ω_{1}$	$ω_{2}$	$ω_{3}$	$ω_{4}$	$ω_{5}$	$ω_{6}$	$ω_{7}$
0.1	0.01	0.05	0.03	0.01	0.2	0.6

Here each of the numbers p(

ω_{i}

) is positive and less than

1

Sum of probabilities

= p (ω_{1}) + p (ω_{2}) + p (ω_{3}) + p (ω_{4}) + p (ω_{5}) + p (ω_{6}) + p (ω_{7})

= 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6

= 1$$
Thus the assignment is valid
(b)

$ω_{1}$	$ω_{2}$	$ω_{3}$	$ω_{4}$	$ω_{5}$	$ω_{6}$	$ω_{7}$
$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$	$\frac{1}{7}$

Here each of the number p(

ω_{i}

) is positive and less than

1

Sum of probabilities
=

p (ω_{1}) + p (ω_{2}) + p (ω_{3}) + p (ω_{4}) + p (ω_{5}) + p (ω_{6}) + p (ω_{7})

=

\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)

$ω_{1}$	$ω_{2}$	$ω_{3}$	$ω_{4}$	$ω_{5}$	$ω_{6}$	$ω_{7}$
0.1	0.2	0.3	0.4	0.5	0.6	0.7

Here each of the numbers p(

ω_{i}

) is positive and less than

1

Sum of probabilities
=

p (ω_{1}) + p (ω_{2}) + p (ω_{3}) + p (ω_{4}) + p (ω_{5}) + p (ω_{6}) + p (ω_{7})

= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7

= 2.8

\neq = 1

Thus the assignment is not valid
(d)

$ω_{1}$	$ω_{2}$	$ω_{3}$	$ω_{4}$	$ω_{5}$	$ω_{6}$	$ω_{7}$
-0.1	0.2	0.3	0.4	-0.2	0.1	0.3

Here p

(ω_{1})

and p

(ω_{5})

are negative
Hence the assignment is not valid
(e)

$ω_{1}$	$ω_{2}$	$ω_{3}$	$ω_{4}$	$ω_{5}$	$ω_{6}$	$ω_{7}$
$\frac{1}{1 4}$	$\frac{2}{1 4}$	$\frac{3}{1 4}$	$\frac{4}{1 4}$	$\frac{5}{1 4}$	$\frac{6}{1 4}$	$\frac{1 5}{1 4}$

p (ω_{7}) = \frac{1 5}{1 4} > 1

Hence the assignment is not valid