Math  /  Algebra

QuestionConsumers in Shelbyville have a choice of one of two fast food restaurants, Krusty's and McDonald's. Both have trouble keeping customers. Of those who last went to Krusty's, 56%56 \% will go to McDonald's next time, and of those who last went to McDonald's, 84%84 \% will go to Krusty's next time. (a) Find the transition matrix describing this situation. (Assume the components of the state vector are in this order [Krusty's customers, McDonald's customers]). [0.440.840.560.16]\left[\begin{array}{ll} 0.44 & 0.84 \\ 0.56 & 0.16 \end{array}\right] (b) A customer goes out for fast food every Sunday, and just went to Krusty's. i. What is the probability that two Sundays from now she will go to McDonald's? 0.3360.336 ii. What is the probability that three Sundays from now she will go to McDonald's? 0.42560.4256 (c) Suppose a consumer has just moved to Shelbyville, and there is a 45%45 \% chance that he will go to Krusty's for his first fast food outing. What is the probability that his third fast food experience will be at Krusty's? \square

Studdy Solution

STEP 1

What is this asking? We're figuring out how likely someone is to switch between two fast food places, Krusty's and McDonald's, over a few visits, knowing how often people usually switch. Watch out! Don't mix up the probabilities of *staying* at a restaurant with the probabilities of *switching*.
Also, remember matrix multiplication order matters!

STEP 2

1. Transition Matrix
2. Two Sundays from now
3. Three Sundays from now
4. Third fast food experience

STEP 3

Let's **build our transition matrix**!
This matrix tells us the probability of moving between the two restaurants.
Since 56%56\% of Krusty's customers switch to McDonald's, there's 100%56%=44%100\% - 56\% = 44\% chance they *stay* at Krusty's.
So, the first entry in our matrix is 0.440.44.

STEP 4

Similarly, since 84%84\% of McDonald's customers switch to Krusty's, there's 100%84%=16%100\% - 84\% = 16\% chance they *stay* at McDonald's.
So the second entry in the second row is 0.160.16.

STEP 5

Our **transition matrix** TT looks like this: T=[0.440.840.560.16] T = \begin{bmatrix} 0.44 & 0.84 \\ 0.56 & 0.16 \end{bmatrix} The first *row* shows the probabilities of going from Krusty's to Krusty's (0.440.44) and Krusty's to McDonald's (0.840.84) if we're currently at McDonald's.
The second *row* shows the probabilities of going from McDonald's to Krusty's (0.560.56) and McDonald's to McDonald's (0.160.16).

STEP 6

The customer just went to Krusty's, so our **initial state vector** is v0=[10]v_0 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, representing 100%100\% chance they're at Krusty's and 0%0\% chance they're at McDonald's.

STEP 7

To find the probabilities after two Sundays, we **multiply** the initial state vector by the transition matrix *twice*: v2=TTv0=T2v0v_2 = T \cdot T \cdot v_0 = T^2 \cdot v_0.

STEP 8

First, let's calculate T2T^2: T2=[0.440.840.560.16][0.440.840.560.16]=[0.66560.50240.33440.4976]T^2 = \begin{bmatrix} 0.44 & 0.84 \\ 0.56 & 0.16 \end{bmatrix} \begin{bmatrix} 0.44 & 0.84 \\ 0.56 & 0.16 \end{bmatrix} = \begin{bmatrix} 0.6656 & 0.5024 \\ 0.3344 & 0.4976 \end{bmatrix}

STEP 9

Now, let's **calculate** v2v_2: v2=[0.66560.50240.33440.4976][10]=[0.66560.3344]v_2 = \begin{bmatrix} 0.6656 & 0.5024 \\ 0.3344 & 0.4976 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0.6656 \\ 0.3344 \end{bmatrix} The probability of being at McDonald's after two Sundays is **0.33440.3344**.

STEP 10

To find the probabilities after three Sundays, we multiply the initial state vector by the transition matrix *three* times: v3=T3v0=TT2v0v_3 = T^3 \cdot v_0 = T \cdot T^2 \cdot v_0.

STEP 11

We already have T2T^2, so we just need to **multiply** it by TT: T3=[0.440.840.560.16][0.66560.50240.33440.4976]=[0.5740160.6362240.4259840.363776]T^3 = \begin{bmatrix} 0.44 & 0.84 \\ 0.56 & 0.16 \end{bmatrix} \begin{bmatrix} 0.6656 & 0.5024 \\ 0.3344 & 0.4976 \end{bmatrix} = \begin{bmatrix} 0.574016 & 0.636224 \\ 0.425984 & 0.363776 \end{bmatrix}

STEP 12

Now, let's **calculate** v3v_3: v3=[0.5740160.6362240.4259840.363776][10]=[0.5740160.425984]v_3 = \begin{bmatrix} 0.574016 & 0.636224 \\ 0.425984 & 0.363776 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0.574016 \\ 0.425984 \end{bmatrix} The probability of being at McDonald's after three Sundays is **0.4259840.425984**.

STEP 13

Now our **initial state vector** is v0=[0.450.55]v_0 = \begin{bmatrix} 0.45 \\ 0.55 \end{bmatrix}, since there's a 45%45\% chance they'll go to Krusty's and a 55%55\% chance they'll go to McDonald's (100%45%=55%100\% - 45\% = 55\%).

STEP 14

We want to find the probability they'll be at Krusty's on their *third* visit, so we need to calculate v2=T2v0v_2 = T^2 \cdot v_0.
We already calculated T2T^2 in a previous step.

STEP 15

Let's **calculate** v2v_2: v2=[0.66560.50240.33440.4976][0.450.55]=[(0.66560.45)+(0.50240.55)(0.33440.45)+(0.49760.55)]=[0.57480.4252]v_2 = \begin{bmatrix} 0.6656 & 0.5024 \\ 0.3344 & 0.4976 \end{bmatrix} \begin{bmatrix} 0.45 \\ 0.55 \end{bmatrix} = \begin{bmatrix} (0.6656 \cdot 0.45) + (0.5024 \cdot 0.55) \\ (0.3344 \cdot 0.45) + (0.4976 \cdot 0.55) \end{bmatrix} = \begin{bmatrix} 0.5748 \\ 0.4252 \end{bmatrix} The probability of being at Krusty's on their third visit is **0.57480.5748**.

STEP 16

(a) The transition matrix is: (0.440.840.560.16)\begin{pmatrix} 0.44 & 0.84 \\ 0.56 & 0.16 \end{pmatrix} (b) i.
The probability of going to McDonald's two Sundays from now is 0.33440.3344. ii. The probability of going to McDonald's three Sundays from now is 0.4259840.425984. (c) The probability that their third fast food experience will be at Krusty's is 0.57480.5748.

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