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.
Find the expected value of T.
To find the probability that T=4, we need to consider two cases:
Case 1: The last arrival was at time slot 8.
Case 2: The last arrival was at time slot 9.
In Case 1, we know that there was exactly one arrival in time slot 9. Since the arrival probability at each time slot is p=1/3, the probability of an arrival in time slot 9 is 1/3. In time slot 10, there is no arrival. Therefore, the probability of T=4 in this case is (1/3) * (2/3) = 2/9.
In Case 2, we know that there was exactly one arrival in time slot 8. Again, the probability of an arrival in time slot 8 is 1/3. In time slot 9, there is no arrival. Therefore, the probability of T=4 in this case is (1/3) * (2/3) = 2/9.
Since these two cases are mutually exclusive, we can add their probabilities to find the overall probability of T=4:
P(T=4) = P(T=4 | last arrival at time slot 8) * P(last arrival at time slot 8) + P(T=4 | last arrival at time slot 9) * P(last arrival at time slot 9)
= (2/9) * (1/2) + (2/9) * (1/2)
= 1/9 + 1/9
= 2/9
Therefore, the probability that T=4 is 2/9.
Now, let's find the expected value of T. The expected value of T can be calculated as the sum of all possible values of T multiplied by their respective probabilities.
We already know the probability of T=4, which is 2/9. To find the probability of other values of T, we can consider the following cases:
Case 1: T=1
In this case, the last arrival was at time slot 9, and the next arrival is at time slot 10. The probability of this case can be calculated as follows:
P(T=1) = P(T=1 | last arrival at time slot 9) * P(last arrival at time slot 9)
= (2/3) * (1/2)
= 1/3
Case 2: T=2
In this case, the last arrival was at time slot 9, and the next arrival is at time slot 11. The probability of this case can be calculated as follows:
P(T=2) = P(T=2 | last arrival at time slot 9) * P(last arrival at time slot 9)
= (2/3) * (1/2)
= 1/3
Case 3: T=3
In this case, the last arrival was at time slot 9, and the next arrival is at time slot 12. The probability of this case can be calculated as follows:
P(T=3) = P(T=3 | last arrival at time slot 9) * P(last arrival at time slot 9)
= (2/3) * (1/2)
= 1/3
Now, we can calculate the expected value of T:
E(T) = (1/1) * (1/3) + (2/1) * (1/3) + (3/1) * (2/9) + (4/1) * (2/9)
= 1/3 + 2/3 + 6/9 + 8/9
= 17/9
Therefore, the expected value of T is 17/9 or approximately 1.889.
To find the probability that T=4, we need to consider the possible sequences of arrivals and calculate the probability of each sequence.
There are two possible sequences that result in T=4:
1) No arrival in slots 10, 11, 12, with the preceding arrival occurring in slot 9.
2) No arrival in slots 10, 11, with the preceding arrival occurring in slot 8.
The probability of sequence 1 is (1-p)^3 * p, where p=1/3 is the arrival probability. Since p=1/3, the probability of no arrival in a given slot is (2/3). Therefore, the probability of sequence 1 is (2/3)^3 * (1/3) = 8/81.
The probability of sequence 2 is (1-p)^2 * p, where p=1/3. So, the probability of sequence 2 is (2/3)^2 * (1/3) = 4/27.
Therefore, the probability that T=4 is the sum of the probabilities of the two sequences, which is 8/81 + 4/27 = 56/243.
To find the expected value of T, we need to calculate the probabilities of each possible value of T and multiply each value by its corresponding probability, and then sum them up.
The possible values of T are 3, 4, 5, 6, ... (starting from the first time slot after the last arrival and increasing by 1).
The probability of T=3 is (1-p)^2 * p, which is (2/3)^2 * (1/3) = 4/27 as shown before.
Similarly, the probability of T=4 is 56/243 as calculated above.
To find the probability of T=5, we need to consider the possible sequences: No arrival in slots 10, 11, 12, 13, with the preceding arrival occurring in slot 8 or 9. The probability of each sequence is (1-p)^3 * p, which is (2/3)^3 * (1/3) = 8/81. Since there are two possible preceding arrival slots, the total probability of T=5 is 2 * (8/81) = 16/81.
Continuing with this pattern, the probability of T=6 is 2 * (2/3)^4 * (1/3) = 32/243.
Therefore, the expected value of T is calculated as:
E(T) = 3 * (4/27) + 4 * (56/243) + 5 * (16/81) + 6 * (32/243) + ...
To simplify the calculation, we can multiply each probability by 27 and write the expected value equation as:
E(T) = (12/27) + (112/243) + (64/81) + (96/243) + ...
We can notice that the terms (112/243) + (96/243) + ... form a geometric series with a common ratio of (96/243) / (112/243) = 4/3.
Using the formula for the sum of an infinite geometric series with a common ratio less than 1, we find:
E(T) = (12/27) + (112/27) * (4/3) = 12/27 + 448/81 = 116/27.
Therefore, the expected value of T is 116/27, which is approximately 4.296.