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vampirchik [111]
3 years ago
9

A lab 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 choose the same food on the next trial with a probability of 50%, and il will choose the other foods on the next trial with equal probabilities of 25%. 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?
Mathematics
1 answer:
Tresset [83]3 years ago
4 0

Answer:

The probability that it will choose food #2 on the second trial after the initial trial = 0.3125

Step-by-step explanation:

Given - A lab 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 choose 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%.

To find - 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?

Proof -

By the given information, we get the stohastic matrix

H = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]

As we know that,

The matrix is a Markov chain x_{k+1} = Hx_{k}

Let

The initial state vector be

x_{0} = \left[\begin{array}{ccc}1\\0\\0\end{array}\right]

we choose this initial vector because given that If the animal chooses food #1 on an initial trial.

Now,

x_{1} = Hx_{0} \\ = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]\left[\begin{array}{ccc}1\\0\\0\end{array}\right] \\= \left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right]

∴ we get

x_{1} = \left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right]

Now,

x_{2} = Hx_{1} \\ = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]\left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right] \\= \left[\begin{array}{ccc}0.25+0.0625+0.0625\\0.125+0.125+0.0625\\0.125+0.0625+0.125\end{array}\right]\\= \left[\begin{array}{ccc}0.375\\0.3125\\0.3125\end{array}\right]

∴ we get

x_{2} = \left[\begin{array}{ccc}0.375\\0.3125\\0.3125\end{array}\right]

∴ we get

The probability that it will choose food #2 on the second trial after the initial trial = 0.3125

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Table of charges:

KWh                                                        0       5           10          15           20    

Basic charge (cents)                          11.60    11.60     11.60      11.60       11.60

Charge for first 10 kWh (4.59cents)    0     22.95   45.90    45.90     45.90

Additional usage above 10 kWh          0       0           0         48.55      97.10

(9.71 cents)

Total (cents)                                     11.60   34.55   57.50   106.05    154.60

cost (dollar)                                       0.12     0.35      0.58   1.06         1.55

KWh                                                           25           30          35           40

Basic charge (cents)                               11.60        11.60        11.60        11.60

Charge for first 10 kWh (4.59cents)    45.90      45.90       45.90      45.90

Additional usage above 10 kWh         145.65    194.20    242.75     291.30

(9.71 cents)

Total (cents)                                        203.15    251.70     300.25    348.80                                  

cost (dollar)                                         2.03        2.52          3.00       3.49

Step-by-step explanation:

a) Data:

Basic charge = 11.6 cents (0.116)

Charge for the first 10 kWh = 4.59 cents (0.0459)

Charge for additional usage above 10 kWh = 9.71 cents (0.0971)

b) Each kWh consumed or not attracts the basic charge of 11.6 cents.  The first 10 kWh consumed is billed at 4.59 cents, while the additional kWh consumed is charged at 9.71 cents.  The charge per consumption is first totaled in cents before being converted to dollars by dividing by 100.

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