Which of the following cannot 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 | $\omega_1$ | $\omega_2$ | $\omega_3$ | $\omega_4$ | $\omega_5$ | $\omega_6$ | $\omega_7$

Question

Which of the following cannot 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) | $\frac{1}{14}$ | $\frac{2}{14}$ | $\frac{3}{14}$ | $\frac{4}{14}$ | $\frac{5}{14}$ | $\frac{6}{14}$ | $\frac{15}{14}$ |

Accepted Answer

> [!TLDR]
> A valid probability assignment must satisfy two axioms: each probability is between 0 and 1, and the sum of all probabilities equals 1. Assignments (c), (d), and (e) violate one or both of these rules.

The question asks which of the given assignments **cannot** be a valid probability distribution over the seven outcomes. This is a direct application of the **Kolmogorov probability axioms** — the two that matter here are:

1. **Non-negativity**: For every outcome $\omega_i$, $P(\omega_i) \ge 0$.
2. **Normalization**: The sum of probabilities over the entire sample space equals 1: $\sum_{i=1}^7 P(\omega_i) = 1$.

A third axiom (countable additivity) is automatically satisfied for a finite sample space if the sum condition holds. So we simply check each assignment against these two rules.

Let’s go through them one by one.

1. **Assignment (a)**: $0.1, 0.01, 0.05, 0.03, 0.01, 0.2, 0.6$  
   All values are non-negative. Sum:  
   $0.1 + 0.01 = 0.11$  
   $+ 0.05 = 0.16$  
   $+ 0.03 = 0.19$  
   $+ 0.01 = 0.20$  
   $+ 0.2 = 0.40$  
   $+ 0.6 = 1.00$  
   Sum is exactly 1. So (a) is valid.

2. **Assignment (b)**: $\frac{1}{7}$ each  
   All are positive. Sum = $7 	imes \frac{1}{7} = 1$. Valid.

3. **Assignment (c)**: $0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7$  
   All non-negative. Sum: $0.1+0.2=0.3$, $+0.3=0.6$, $+0.4=1.0$, $+0.5=1.5$, $+0.6=2.1$, $+0.7=2.8$.  
   The sum is $2.8$, far greater than 1. This violates normalization. So (c) is invalid.

4. **Assignment (d)**: $-0.1, 0.2, 0.3, 0.4, -0.2, 0.1, 0.3$  
   Here $\omega_1$ has probability $-0.1$ and $\omega_5$ has $-0.2$ — both negative. This violates non-negativity. Even if we ignore that, the sum:  
   $(-0.1)+0.2=0.1$, $+0.3=0.4$, $+0.4=0.8$, $+(-0.2)=0.6$, $+0.1=0.7$, $+0.3=1.0$ — the sum happens to be 1, but the negative entries alone make it invalid. So (d) is invalid.

5. **Assignment (e)**: $\frac{1}{14}, \frac{2}{14}, \frac{3}{14}, \frac{4}{14}, \frac{5}{14}, \frac{6}{14}, \frac{15}{14}$  
   All

Concept understanding — Probability Axioms

Probability Axioms: From Intuition to Precision

The Three Axioms (Kolmogorov's Axioms)

Why These Three Are Enough

A Quick Check

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)	$\frac{1}{14}$	$\frac{2}{14}$	$\frac{3}{14}$	$\frac{4}{14}$	$\frac{5}{14}$	$\frac{6}{14}$	$\frac{15}{14}$

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