Let a sample space be $$S=\left\{\omega_{1}, \omega_{2}, \ldots, \omega_{6}\right\}$$ . Which of the following assignments of probabilities to each outcome are valid? $$\begin{array}{ccccccc} \text { Outcomes } & \omega_{1} & \omega_{2} & \omega_{3} & \omega_{4} & \omega_{5} & \omega_{6} \text { (a) } & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \text { (b) } & 1 & 0 & 0 & 0 & 0 & 0 \text { (c) } & \frac{1}{8} & \frac{2}{3} & \frac{1}{3} & \frac{1}{3} & -\frac{1}{4} & -\frac{1}{3} \text { (d) } & \frac{1}{12} & \frac{1}{12} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{3}{2} \text { (e) } & 0.1 & 0.2 & 0.3 & 0.4 & 0.5 & 0.6 \end{array}$$

Step 1: Understand Probability Rules

For any assignment of probabilities to be valid, two fundamental rules must be satisfied. First, the probability of each individual outcome, $$P(\omega_i)$$ , must be between 0 and 1, inclusive. Second, the sum of the probabilities of all possible outcomes in the sample space must equal 1.

Step 2: Check Assignment (a)

For assignment (a), all probabilities are $$1/6$$ , which is between 0 and 1. The sum of these probabilities is $$6 \times (1/6) = 1$$ . Both conditions are met, so this is a valid assignment.

Step 3: Check Assignment (b)

For assignment (b), all probabilities are either 0 or 1, which are within the valid range. The sum of these probabilities is $$1 + 0 + 0 + 0 + 0 + 0 = 1$$ . Both conditions are met, so this is a valid assignment.

Step 4: Check Assignment (c)

In assignment (c), the probabilities $$P(\omega_5) = -1/4$$ and $$P(\omega_6) = -1/3$$ are negative. Probabilities cannot be negative. Therefore, this is not a valid assignment.

Step 5: Check Assignment (d)

For assignment (d), the probability $$P(\omega_6) = 3/2 = 1.5$$ is greater than 1. Probabilities cannot exceed 1. Therefore, this is not a valid assignment.

Step 6: Check Assignment (e)

For assignment (e), all individual probabilities are between 0 and 1. However, the sum of the probabilities is $$0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 = 2.1$$ . Since the sum is not equal to 1, this is not a valid assignment.

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