Independent solution

How to solve this Discrete Random Variables question

Setup

Setup

Use exhaustiveness to require the probabilities over all positive indices to sum to one.

1=∑i=1∞k(13)i1=\sum_{i=1}^{\infty}k\left(\frac13\right)^i

Model

Model

Evaluate the full geometric series to determine its normalizing constant.

∑i=1∞(13)i=1/31−1/3=12\sum_{i=1}^{\infty}\left(\frac13\right)^i=\frac{1/3}{1-1/3}=\frac12
1=k2⟹k=21=\frac{k}{2}\quad\Longrightarrow\quad k=2

Compute

Compute

The even indices are 2j for positive integers j, producing a second geometric series with ratio 1/9.

Pr⁡(even index)=2∑j=1∞(13)2j\Pr(\text{even index})=2\sum_{j=1}^{\infty}\left(\frac13\right)^{2j}
Pr⁡(even index)=21/91−1/9=14\Pr(\text{even index})=2\frac{1/9}{1-1/9}=\frac14

Answer

Answer

The total probability carried by the even-indexed outcomes is one quarter.

Pr⁡(even index)=14(B)\boxed{\Pr(\text{even index})=\frac14\quad\text{(B)}}