Consider a Bernoulli process, with arrival probability at each time slot equal to p=1/3.
An observer arrives at time slot 10 and sees that no arrival took place in that slot. A passerby informs the observer that there was exactly one arrival during the preceding two time slots (i.e., time slots 8 and 9) but has no additional information about which of the two it was.
Let T be the number of time slots starting from the time slot of the last arrival (before slot 10) and until the time slot where the next arrival comes.
For example, if the last arrival came at time 8 and the next arrival comes at time 13, T is the number of slots in the interval \{ 8,9,10,11,12,13\}, so that T=6.
(Enter a numerical expression or a decimal accurate to at least 3 decimals places.)
Find the probability that T=4.
a. 0.0036
To find the probability that T=4, we need to calculate the probability that the next arrival comes at time slot 14, given the information provided.
Since the arrival probability at each time slot is p=1/3, the probability of no arrival in a given slot is q=1-p=2/3.
The passerby informs the observer that there was exactly one arrival during the preceding two time slots (8 and 9). This means that there are three possible scenarios for the arrivals in slots 8, 9, and 10:
1) No arrival in slot 8, arrival in slot 9, and no arrival in slot 10.
2) No arrival in slot 8, no arrival in slot 9, and arrival in slot 10.
3) Arrival in slot 8, no arrival in slot 9, and no arrival in slot 10.
Let's analyze each scenario:
1) No arrival in slot 8, arrival in slot 9, and no arrival in slot 10:
The probability of no arrival in slot 8 is q=2/3.
The probability of arrival in slot 9 is p=1/3.
The probability of no arrival in slot 10 is q=2/3.
Therefore, the probability of this scenario is (2/3) * (1/3) * (2/3) = 4/27.
2) No arrival in slot 8, no arrival in slot 9, and arrival in slot 10:
The probability of no arrival in slot 8 is q=2/3.
The probability of no arrival in slot 9 is q=2/3.
The probability of arrival in slot 10 is p=1/3.
Therefore, the probability of this scenario is (2/3) * (2/3) * (1/3) = 4/27.
3) Arrival in slot 8, no arrival in slot 9, and no arrival in slot 10:
The probability of arrival in slot 8 is p=1/3.
The probability of no arrival in slot 9 is q=2/3.
The probability of no arrival in slot 10 is q=2/3.
Therefore, the probability of this scenario is (1/3) * (2/3) * (2/3) = 4/27.
Adding up the probabilities of all three scenarios, we get (4/27) + (4/27) + (4/27) = 12/27 = 4/9.
Therefore, the probability that T=4 is 4/9.
To find the probability that T = 4, we need to consider the different possibilities for the last arrival before slot 10.
Let's consider the two options:
1. The last arrival occurred in time slot 8 (and there was no arrival in time slot 9).
2. The last arrival occurred in time slot 9 (and there was no arrival in time slot 8).
For each option, we will calculate the conditional probability of T = 4.
Option 1: The last arrival occurred in time slot 8.
Since there was no arrival in time slot 9 and no arrival in time slot 10, we need to calculate the probability of having no arrival in the next 4 time slots (11, 12, 13, and 14).
The probability of no arrival in each time slot is (1 - p) = (1 - 1/3) = 2/3.
Therefore, the probability of no arrival in 4 consecutive time slots is (2/3)^4.
Option 2: The last arrival occurred in time slot 9.
Since there was no arrival in time slot 8 and no arrival in time slot 10, we need to calculate the probability of having no arrival in the next 3 time slots (11, 12, and 13).
The probability of no arrival in each time slot is still (2/3).
Therefore, the probability of no arrival in 3 consecutive time slots is (2/3)^3.
Now, we need to consider the probability of each option occurring. Since the passerby has no additional information, both options are equally likely, so the probability of each option is 1/2.
Finally, we can calculate the overall probability by summing the products of the conditional probabilities and the probabilities of each option:
P(T = 4) = (1/2) * (2/3)^4 + (1/2) * (2/3)^3 ≈ 0.197
Therefore, the probability that T = 4 is approximately 0.197.
To find the probability that T = 4, we need to consider all possible scenarios for the arrivals in the preceding two time slots (slots 8 and 9).
There are 4 possible scenarios:
1. No arrival in slot 8, no arrival in slot 9.
2. Arrival in slot 8, no arrival in slot 9.
3. No arrival in slot 8, arrival in slot 9.
4. Arrival in slot 8, arrival in slot 9.
Let's calculate the probability for each scenario separately.
1. No arrival in slot 8, no arrival in slot 9:
The probability of no arrival in each slot is (1-p) = 2/3.
The probability of this scenario is (2/3) * (2/3) = 4/9.
2. Arrival in slot 8, no arrival in slot 9:
The probability of arrival in slot 8 is p = 1/3.
The probability of no arrival in slot 9 is (1-p) = 2/3.
The probability of this scenario is (1/3) * (2/3) = 2/9.
3. No arrival in slot 8, arrival in slot 9:
The probability of no arrival in slot 8 is (1-p) = 2/3.
The probability of arrival in slot 9 is p = 1/3.
The probability of this scenario is (2/3) * (1/3) = 2/9.
4. Arrival in slot 8, arrival in slot 9:
The probability of arrival in each slot is p = 1/3.
The probability of this scenario is (1/3) * (1/3) = 1/9.
Now, we sum up the probabilities for all the scenarios to find the probability that T = 4:
P(T = 4) = P(no arrival in slot 8, no arrival in slot 9) + P(arrival in slot 8, no arrival in slot 9) + P(no arrival in slot 8, arrival in slot 9) + P(arrival in slot 8, arrival in slot 9)
= (4/9) + (2/9) + (2/9) + (1/9)
= 9/9
= 1
Therefore, the probability that T = 4 is 1.