Question

1. which of the following are valid probability distributions for a discrete random variable? check all that apply. a. 0, 0, 0, 0, 0, 0, 0, 1 b. \\( \frac { 2 } { 3 } \\), \\( \frac { 2 } { 9 } \\), \\( \frac { 2 } { 27 } \\), \\( \frac { 2 } { 81 } \\), \\( \frac { 2 } { 243 } \\), \\( \cdots \\) c. \\( \frac { 1 } { 2 } \\), \\( \frac { 1 } { 3 } \\), \\( \frac { 1 } { 4 } \\), \\( \frac { - 1 } { 12 } \\) d. \\( \frac { 1 } { 3 } \\), \\( \frac { 1 } { 15 } \\), \\( \frac { 1 } { 10 } \\), \\( \frac { 5 } { 12 } \\), \\( \frac { 1 } { 12 } \\) e. 0.32, 0.16, 0.02, 0.18, 0.15, 0.08, 0.09

Explanation:

Step1: Check the properties of a valid probability distribution

A valid probability distribution for a discrete random variable must satisfy two conditions:

1. Each probability \(P(x)\) must be between \(0\) and \(1\) (inclusive), i.e., \(0\leq P(x)\leq1\).
2. The sum of all probabilities \(\sum P(x)=1\).

Step2: Analyze option A

For option A: \(0 + 0+0+0+0+0+0 + 1=1\), and each probability \(P(x)\) satisfies \(0\leq P(x)\leq1\).

Step3: Analyze option B

For option B: This is a geometric series with first term \(a=\frac{2}{3}\) and common ratio \(r=\frac{1}{3}\). The sum of an infinite geometric series is \(S=\frac{a}{1 - r}\) (when \(|r|\lt1\)). Here \(S=\frac{\frac{2}{3}}{1-\frac{1}{3}}=\frac{\frac{2}{3}}{\frac{2}{3}} = 1\). But in a discrete probability distribution, the sum of probabilities for all possible values of the random variable must equal \(1\). However, a discrete probability distribution is defined for a countable set of values. If we assume the number of terms is infinite (as the series is given as \(\frac{2}{3},\frac{2}{9},\frac{2}{27},\cdots\)), but in the context of a discrete random - variable probability distribution (usually for a finite or countably - infinite but well - defined set of outcomes), if we consider the nature of the problem (since it's presented as a multiple - choice for a "discrete random variable" in a basic probability sense, and the sum is \(1\) but the way it's presented may be non - standard. But if we just consider the sum rule \(\sum_{n = 1}^{\infty}\frac{2}{3}\times(\frac{1}{3})^{n - 1}=1\) and \(0\lt\frac{2}{3}\times(\frac{1}{3})^{n - 1}\lt1\) for \(n = 1,2,\cdots\)).

Step4: Analyze option C

For option C: One of the probabilities is \(-\frac{1}{12}\lt0\), so it does not satisfy \(0\leq P(x)\leq1\).

Step5: Analyze option D

For option D:

$$
LATEXBLOCK0
$$

and each probability \(P(x)\) satisfies \(0\leq P(x)\leq1\).

Step6: Analyze option E

For option E:

$$
LATEXBLOCK1
$$

and each probability \(P(x)\) satisfies \(0\leq P(x)\leq1\).

Answer:

A. \(0,0,0,0,0,0,0,1\), B. \(\frac{2}{3},\frac{2}{9},\frac{2}{27},\frac{2}{81},\frac{2}{243},\cdots\), D. \(\frac{1}{3},\frac{1}{15},\frac{1}{10},\frac{5}{12},\frac{1}{12}\), E. \(0.32,0.16,0.02,0.18,0.15,0.08,0.09\)