Let a sample space be $S = \{\omega_1, \omega_2, \ldots, \omega_6\}$. Which of the following assignments of probabilities to each outcome are valid?

| Outcomes | $\omega_1$ | $\omega_2$ | $\omega_3$ | $\omega_4$ | $\omega_5$ | $\omega_6$ |
|---|---|---|---|---|---|---|
| (a) | $

Question

Let a sample space be $S = \{\omega_1, \omega_2, \ldots, \omega_6\}$. Which of the following assignments of probabilities to each outcome are valid?

| Outcomes | $\omega_1$ | $\omega_2$ | $\omega_3$ | $\omega_4$ | $\omega_5$ | $\omega_6$ |
|---|---|---|---|---|---|---|
| (a) | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ |
| (b) | $1$ | $0$ | $0$ | $0$ | $0$ | $0$ |
| (c) | $\frac{1}{8}$ | $\frac{2}{3}$ | $\frac{1}{3}$ | $\frac{1}{3}$ | $-\frac{1}{4}$ | $-\frac{1}{3}$ |
| (d) | $\frac{1}{12}$ | $\frac{1}{12}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{3}{2}$ |
| (e) | $0.1$ | $0.2$ | $0.3$ | $0.4$ | $0.5$ | $0.6$ |

Accepted Answer

> [!TLDR]
> A probability assignment is valid only if every probability is between 0 and 1 (inclusive) and the sum of all probabilities equals exactly 1. Checking these two conditions shows that only assignments (a) and (b) are valid.

The core idea is simple: probability is a measure of how likely an outcome is. The axioms of probability (Kolmogorov’s axioms) give us two non-negotiable rules for any finite sample space:

1. **Non-negativity**: Every outcome’s probability must be $\ge 0$.
2. **Normalization**: The sum of probabilities over all outcomes must equal exactly $1$.

That’s it. No probability can be negative, none can exceed 1 (though that’s actually a consequence of the sum being 1 and non-negativity — if one probability were >1, the sum would exceed 1 unless others were negative, which is forbidden). So we just test each table row against these two rules.

Let’s go through each option.

1. **Option (a)**: Each $\omega_i$ gets $\frac{1}{6}$.  
   - All values are positive, so non-negativity holds.  
   - Sum = $6 	imes \frac{1}{6} = 1$.  
   - Both conditions satisfied. **Valid**.

2. **Option (b)**: $\omega_1 = 1$, all others $0$.  
   - All probabilities are $\ge 0$.  
   - Sum = $1 + 0 + 0 + 0 + 0 + 0 = 1$.  
   - **Valid**. This is a degenerate (but perfectly legal) distribution where only one outcome ever occurs.

3. **Option (c)**: Values are $\frac{1}{8}, \frac{2}{3}, \frac{1}{3}, \frac{1}{3}, -\frac{1}{4}, -\frac{1}{3}$.  
   - Immediately, $\omega_5$ and $\omega_6$ have negative probabilities. That violates non-negativity.  
   - Even if we ignored that, the sum would be $\frac{1}{8} + \frac{2}{3} + \frac{1}{3} + \frac{1}{3} - \frac{1}{4} - \frac{1}{3}$. Let’s compute:  
     $\frac{1}{8} - \frac{1}{4} = -\frac{1}{8}$, and $\frac{2}{3} + \frac{1}{3} + \frac{1}{3} - \frac{1}{3} = \frac{2}{3} + \frac{1}{3} = 1$. So total = $1 - \frac{1}{8} = \frac{7}{8} 
eq 1$.  
   - Fails both conditions. **Invalid**.

4. **Option (d)**: Values: $\frac{1}{12},

Concept understanding — Probability Axioms

Probability Axioms: From Intuition to Precision

The Three Axioms (Kolmogorov's Axioms)

Why These Three Are Enough

Outcomes	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
(a)	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$
(b)	$1$	$0$	$0$	$0$	$0$	$0$
(c)	$\frac{1}{8}$	$\frac{2}{3}$	$\frac{1}{3}$	$\frac{1}{3}$	$-\frac{1}{4}$	$-\frac{1}{3}$
(d)	$\frac{1}{12}$	$\frac{1}{12}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{3}{2}$
(e)	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$

🎓Unlock everything free for 14 days

Unlock everything free for 14 days