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Chapter 4: Q2E (page 191)

Question 2: A laboratory animal may eat any one of three foods each day. Laboratory records show that if the animal chooses one food on one trial, it will chose the same food on the next trial with a probability of 50%, and it will choose the other foods on the next trial with equal probabilities of 25%.

a. What is the stochastic matrix for this situation?

b. If the animal chooses food #1 on an initial trial, what is the probability that it will choose food #2 on the second trial after the initial trial?

Short Answer

Expert verified
  1. The stochastic matrix is \(P = \left( {\begin{array}{*{20}{c}}{.5}&{.25}&{.25}\\{.25}&{.5}&{.25}\\{.25}&{.25}&{.5}\end{array}} \right)\).
  2. The probability that the animal will choose food #2 after the initial trial is 0.3125.

Step by step solution

01

Determine the stochastic matrix

a)

Let the food be labeled 1, 2, and 3.

The following table represents the animal's behavior.

From:

To:

1

2

3

.5

.25

.25

.25

.5

.25

.25

.25

.5

1

2

3

From this, the stochastic matrix is \(P = \left( {\begin{array}{*{20}{c}}{.5}&{.25}&{.25}\\{.25}&{.5}&{.25}\\{.25}&{.25}&{.5}\end{array}} \right)\).

02

Determine the probability that will choose food #2 after the initial trial

b)

After the initial trial, there are two trials. The initial state vector is \({{\mathop{\rm x}\nolimits} _0} = \left( {\begin{array}{*{20}{c}}1\\0\\0\end{array}} \right)\).

Compute \({{\mathop{\rm x}\nolimits} _2}\) as shown below:

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _1} = P{{\mathop{\rm x}\nolimits} _0}\\ = \left( {\begin{array}{*{20}{c}}{.5}&{.25}&{.25}\\{.25}&{.5}&{.25}\\{.25}&{.25}&{.5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\0\\0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.5 + 0 + 0}\\{0.25 + 0 + 0}\\{0.25 + 0 + 0}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.5}\\{0.25}\\{0.25}\end{array}} \right)\end{array}\)

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _2} = P{{\mathop{\rm x}\nolimits} _1}\\ = \left( {\begin{array}{*{20}{c}}{.5}&{.25}&{.25}\\{.25}&{.5}&{.25}\\{.25}&{.25}&{.5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{.5}\\{.25}\\{.25}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.25 + 0.0625 + 0.0625}\\{0.25 + 0.125 + 0.0625}\\{0.125 + 0.0625 + 0.125}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{0.375}\\{0.3125}\\{0.3125}\end{array}} \right)\end{array}\)

Thus, the probability that the animal will choose food #2 after the initial trial is 0.3125.

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Most popular questions from this chapter

Suppose a \({\bf{4}} \times {\bf{7}}\) matrix A has four pivot columns. Is \({\bf{Col}}\,A = {\mathbb{R}^{\bf{4}}}\)? Is \({\bf{Nul}}\,A = {\mathbb{R}^{\bf{3}}}\)? Explain your answers.

Use coordinate vector to test whether the following sets of poynomial span \({{\bf{P}}_{\bf{2}}}\). Justify your conclusions.

a. \({\bf{1}} - {\bf{3}}t + {\bf{5}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{5}}t - {\bf{7}}{t^{\bf{2}}}\), \( - {\bf{4}} + {\bf{5}}t - {\bf{6}}{t^{\bf{2}}}\), \({\bf{1}} - {t^{\bf{2}}}\)

b. \({\bf{5}}t + {t^{\bf{2}}}\), \({\bf{1}} - {\bf{8}}t - {\bf{2}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{4}}t + {\bf{2}}{t^{\bf{2}}}\), \({\bf{2}} - {\bf{3}}t\)

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

21. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 2}&{ - 4.2}&{ - 4.8}&{ - 3.6}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

Find a basis for the set of vectors in\({\mathbb{R}^{\bf{2}}}\)on the line\(y = {\bf{5}}x\).

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)
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