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.
Find the probability that T=4.
To find the probability that T=4, we need to consider the different cases that could have occurred in the preceding two time slots (slots 8 and 9).
Case 1: The arrival occurred in slot 8, but not in slot 9.
In this case, the observer would have seen an arrival in slot 9 and an absence in slot 10. The next arrival could have occurred in any of the subsequent time slots. So, the probability of this case is p * (1-p) = (1/3) * (2/3) = 2/9.
Case 2: The arrival occurred in slot 9, but not in slot 8.
In this case, the observer would have seen an absence in slot 8 and slot 10. The next arrival could have occurred in any of the subsequent time slots. So, the probability of this case is (1-p) * p = (2/3) * (1/3) = 2/9.
Case 3: The arrival occurred in both slot 8 and slot 9.
In this case, the observer would have seen an absence in slot 10. The next arrival could have occurred in any of the subsequent time slots. So, the probability of this case is p * p = (1/3) * (1/3) = 1/9.
Therefore, the total probability that T=4 is the sum of the probabilities of these three cases:
P(T=4) = (2/9) + (2/9) + (1/9) = 5/9.
To find the probability that T=4, we can break down the problem into two cases:
Case 1: The last arrival occurred in time slot 8.
Case 2: The last arrival occurred in time slot 9.
Let's calculate the probability for each case:
Case 1: The last arrival occurred in time slot 8.
Since there was exactly one arrival in the preceding two time slots (8 and 9), if the last arrival occurred in time slot 8, there can be no arrival in time slot 9. So, the next arrival after time slot 8 must occur in time slot 10.
The probability of no arrival (0) in time slot 10 and an arrival (1) in time slot 10 is given by:
P(T=4 | Last arrival in time slot 8) = P(0 in time slot 10) * P(1 in time slot 10)
= (1 - p) * p
= (1 - 1/3) * 1/3
= 2/9
Case 2: The last arrival occurred in time slot 9.
Since there was exactly one arrival in the preceding two time slots (8 and 9), if the last arrival occurred in time slot 9, there can be no arrival in time slot 8. So, the next arrival after time slot 9 must occur in time slot 10.
The probability of an arrival (1) in time slot 10 and no arrival (0) in time slot 10 is given by:
P(T=4 | Last arrival in time slot 9) = P(1 in time slot 10) * P(0 in time slot 10)
= p * (1 - p)
= 1/3 * (1 - 1/3)
= 2/9
Considering both cases, the total probability that T=4 is the sum of the probabilities of each case:
P(T=4) = P(T=4 | Last arrival in time slot 8) + P(T=4 | Last arrival in time slot 9)
= 2/9 + 2/9
= 4/9
Therefore, the probability that T=4 is 4/9.