3. Contuning on probability On day zero, the number of trucks in a parking lot i

Question

3. Contuning on probability On day zero, the number of trucks in a parking lot is Xo. Assume Xo = 1 with probability 20%. Xo 2 with probability 50% and Xo = 3, with probability 30%. Between day zero and day one, assume one truck can leave (with probability 25%), one truck can be added (with probablity 25%), or the number of trucks can stay the same (with probability 50%). X1 is the number of trucks on day one. 3.1, what is the probability that Xi = 1 ? # Your analysis here # your answer here: 3.2. What is the probability that X, = 5 ? # Your analysis here # your answer here: 3.3. What is the probability that X > X? # Your analysis here # your answer here: 3.4. If there are 3 trucks at day one, what is the conditional probability that there were 2 trucks at day zero? # Your analysis here # your answer here:

Explanation / Answer

a) From the above table, the probability of being in state 1 at time 1 is computed as:

P(X1 = 1) = 0.1 + 0.125 = 0.225

Therefore 0.225 is the required probability here.

b) From the analysis, we can clearly see that X1 cannot be equal to 1.

Therefore 0 is the required probability here.

c) From the above table, we can easily computed the probability of X1 > X0 as:

P( X1 > X0 ) = 0.05 + 0.125 + 0.075 = 0.25

Therefore 0.25 is the required probability here.

d) Probability of 3 trucks on day 1 is computed as:

P(X1 = 3) = 0.125 + 0.15 = 0.275

Now given that there are 3 trucks on day 1, probability that there were 2 trucks at day 0 is computed using bayes theorem as:

P(X0 = 2 | X1 = 3) = P(X0 = 2 , X1 = 3) / P(X1 = 3) = 0.125 / 0.275 = 0.4545

Therefore 0.4545 is the required probability here.

X0 X1 Probability X0 = 1 0 0.2*0.25 = 0.05 1 0.2*0.5 = 0.1 2 0.2*0.25 = 0.05 X0 = 2 1 0.5*0.25 = 0.125 2 0.5*0.5 = 0.25 3 0.5*0.25 = 0.125 X0 = 3 2 0.3*0.25 = 0.075 3 0.3*0.5 = 0.15 4 0.3*0.25 = 0.075

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