Question

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

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

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 \(P = \left( {\begin{array}{*{20}{c}}{.5}&{.25}&{.25}\\{.25}&{.5}&{.25}\\{.25}&{.25}&{.5}\end{array}} \right)\).

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.