**Question:** Can the vector be a probability vector? If not, why?**Vector:** \[\begin{bmatrix} \frac{4}{5} \\ \frac{1}{5} \end{bmatrix}\]**Options:**- ○ Yes, the components sum to 1.- ○ Yes, the vector is a 1 x 2.- ○ No, the components do not sum to 1.- ○ No, the vector is a 2 x 1.**Explanation:** A probability vector is a vector of non-negative numbers that sum up to 1. In this case, the vector is:\[\begin{bmatrix} \frac{4}{5} \\ \frac{1}{5} \end{bmatrix}\]The components are \(\frac{4}{5}\) and \(\frac{1}{5}\).- **Sum of components:** \(\frac{4}{5} + \frac{1}{5} = \frac{5}{5} = 1\)Therefore, the components do indeed sum to 1. The vector is a 2 x 1 column vector, meaning it has two rows and one column. The correct option is:- ○ Yes, the components sum to 1.

Definition: A vector is said to be a probability vector if all the components are non negative and their sum is 1. The given vector is: 45 1545 15 The components in the vector are non-negative. The sum of the components is 45+15=55=1. Hence the given vector is a probability vector.The vector 45 15 is a probability vector and the components sum to 1.

Answered: Can the vector be a probability vector?…

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Can the vector be a probability vector? If not, why? 4 1 5 5 Yes, the components sum to 1. Yes, the vector is a 1 x 2. No, the components do not sum to 1. No, the vector is a 2 x 1.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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