Which of the following can not be a valid ass | vedclass

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Question

(C) For an assignment of probabilities to be valid,it must satisfy two conditions: $1$. Each probability $p(\omega_i)$ must satisfy $0 \le p(\omega_i)

Vedclass · Accepted Answer

$-0.1, 0.2, 0.3, 0.4, 0.2, 0.1, -0.1$

Vedclass · Answer

(C) For an assignment of probabilities to be valid,it must satisfy two conditions: $1$. Each probability $p(\omega_i)$ must satisfy $0 \le p(\omega_i) \le 1$. $2$. The sum of all probabilities must be equal to $1$,i.e.,$\sum_{i=1}^{7} p(\omega_i) = 1$. In option $C$,we have negative values $(-0.1)$,which violates the condition $p(\omega_i) \ge 0$. In option $D$,the sum is $0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8$,which is not equal to $1$. Since the question asks which cannot be a valid assignment,both $C$ and $D$ are invalid. However,usually,in such multiple-choice questions,the most fundamental violation is the negative probability. Thus,$C$ is a primary invalid assignment.

$X = x_i$	$0$	$1$	$2$	$3$	$4$
$P(X = x_i)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

$x$	$0$	$1$	$2$	$3$
$P(x)$	$0.2$	$k$	$k$	$2k$

Outcome	Probability
$\omega_{1}$	$\frac{1}{12}$
$\omega_{2}$	$\frac{1}{12}$
$\omega_{3}$	$\frac{1}{6}$
$\omega_{4}$	$\frac{1}{6}$
$\omega_{5}$	$\frac{1}{6}$
$\omega_{6}$	$\frac{3}{2}$

Which of the following can not be a valid assignment of probabilities for outcomes of sample space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$? (Note: The provided table is an example of a valid assignment).

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