Suppose that X is equal to either 1 or 2, with equal probability. Write down an expression for the probability that there were exactly 3 arrivals during the time interval [0,2] .
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To solve this problem, we need to consider the possible outcomes for X and calculate the probability of each outcome.
Since X can be either 1 or 2 with equal probability, we have two cases to consider:
Case 1: X = 1
In this case, X takes the value 1. Let's say that Y represents the number of arrivals during the time interval [0,2]. We need to find the probability that Y = 3 given that X = 1. Because X = 1 means that there is 1 arrival during the time interval [0,2], the probability that Y = 3 given X = 1 is 0, since it is impossible for Y to be 3 in this case.
Case 2: X = 2
Similarly, when X = 2, Y can take values from 0 to 4 (0 arrivals to 4 arrivals) during the time interval [0,2]. We are interested in finding the probability that Y = 3 given X = 2. Because X = 2 means there are 2 arrivals during the time interval [0,2], we need to consider the number of ways to distribute 3 arrivals across the two time intervals.
There are different ways to distribute 3 arrivals into two time intervals. For example:
- 0 arrivals in the first interval (Y = 0) and 3 arrivals in the second interval (Y = 3)
- 1 arrival in the first interval (Y = 1) and 2 arrivals in the second interval (Y = 2)
- 2 arrivals in the first interval (Y = 2) and 1 arrival in the second interval (Y = 1)
- 3 arrivals in the first interval (Y = 3) and 0 arrivals in the second interval (Y = 0)
We need to calculate the probability for each of these cases and sum them up to find the total probability.
Let's denote the probability of 3 arrivals in the first interval as p1 and the probability of 3 arrivals in the second interval as p2. Since these probabilities are the same, we have p1 = p2 = p.
Now, let's calculate each probability:
Probability of 0 arrivals in the first interval (Y = 0) and 3 arrivals in the second interval (Y = 3):
p(Y = 0 | X = 2) = p1 * p2
Probability of 1 arrival in the first interval (Y = 1) and 2 arrivals in the second interval (Y = 2):
p(Y = 1 | X = 2) = (1 - p1) * p2
Probability of 2 arrivals in the first interval (Y = 2) and 1 arrival in the second interval (Y = 1):
p(Y = 2 | X = 2) = p1 * (1 - p2)
Probability of 3 arrivals in the first interval (Y = 3) and 0 arrivals in the second interval (Y = 0):
p(Y = 3 | X = 2) = (1 - p1) * (1 - p2)
Since each case is equally likely (X = 2 with probability 1/2), we need to multiply these probabilities by 1/2 and sum them up to find the total probability:
P(Y = 3) = 1/2 * (p1 * p2 + (1 - p1) * p2 + p1 * (1 - p2) + (1 - p1) * (1 - p2))
Note that p1 and p2 are not directly given in the problem statement. The way to determine the values of p1 and p2 depends on the specific scenario of arrival rates and inter-arrival times in the system being analyzed.
To find the probability that there were exactly 3 arrivals during the time interval [0,2], we need to consider the possible outcomes for each value of X.
Let's denote the number of arrivals during the time interval as N. We can express N as:
N = N1 + N2
where N1 is the number of arrivals in the first interval [0, 1] and N2 is the number of arrivals in the second interval [1, 2].
Now, N1 and N2 can take different values depending on the value of X. Let's consider each case separately:
1) If X = 1: In this case, the number of arrivals in the first interval, N1, can take values 0 or 1, and the number of arrivals in the second interval, N2, can take values 2 or 1. So, we have:
N1 = 0, N2 = 3 OR N1 = 1, N2 = 2
2) If X = 2: In this case, the number of arrivals in the first interval, N1, can take values 1 or 2, and the number of arrivals in the second interval, N2, can take values 2 or 1. So, we have:
N1 = 1, N2 = 2 OR N1 = 2, N2 = 1
Now, we can calculate the probability for each case and then sum them up to get the overall probability.
Case 1: X = 1
P(N1 = 0, N2 = 3) = P(N1 = 0) * P(N2 = 3) = (1/2) * (1/2)^3 = 1/16
P(N1 = 1, N2 = 2) = P(N1 = 1) * P(N2 = 2) = (1/2) * (1/2)^2 = 1/8
Case 2: X = 2
P(N1 = 1, N2 = 2) = P(N1 = 1) * P(N2 = 2) = (1/2) * (1/2)^2 = 1/8
P(N1 = 2, N2 = 1) = P(N1 = 2) * P(N2 = 1) = (1/2) * (1/2) = 1/4
Now, we can sum up the probabilities for all cases:
P(N = 3) = P(N1 = 0, N2 = 3) + P(N1 = 1, N2 = 2) + P(N1 = 1, N2 = 2) + P(N1 = 2, N2 = 1)
= (1/16) + (1/8) + (1/8) + (1/4)
= 5/16
Therefore, the probability that there were exactly 3 arrivals during the time interval [0,2] is 5/16.