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.

2 answers

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.
a. 0.0036