Suppose that 𝑋 is equal to either 1 or 2, with equal probability. Write down an expression for the probability that there were exactly 2 arrivals during the time interval [0,2].
(Enter e for the constant 𝑒. You may use standard notation for this numerical entry even though there will be no parser below the answer box. Enter an exact answer or a numerical answer accurate to at least 3 decimal places.)
Probability that there were exactly 2 arrivals during the time interval [0,2]:
e^(-2)
To calculate the probability that there were exactly 2 arrivals during the time interval [0,2], we need to consider the two possible values of 𝑋 (1 or 2) and compute the probability for each case.
Let 𝐴 be the event of having exactly 2 arrivals during the time interval [0,2].
Case 1: 𝑋 = 1
In this case, there is only one arrival during the time interval [0,2].
The probability of having exactly 2 arrivals when 𝑋 = 1 is given by the probability of having 1 arrival (𝑝) and then having no arrivals (1 - 𝑝) in the remaining time.
So, the probability in this case is 𝑝 × (1 - 𝑝) = 𝑝(1 - 𝑝).
Case 2: 𝑋 = 2
In this case, there are two arrivals during the time interval [0,2].
The probability of having exactly 2 arrivals when 𝑋 = 2 is given by the probability of having 2 arrivals (𝑝^2) in the time interval.
So, the probability in this case is 𝑝^2.
Since 𝑋 can equal 1 or 2 with equal probability, we need to consider both cases and compute the average probability.
The average probability is (1/2) × (𝑝(1 - 𝑝)) + (1/2) × (𝑝^2) = (1/2)(𝑝 - 𝑝^2 + 𝑝^2) = 1/2 * 𝑝.
Therefore, the probability that there were exactly 2 arrivals during the time interval [0,2] is equal to 1/2 * 𝑝.
To find the probability of exactly 2 arrivals during the time interval [0,2], we can consider the two possible values of 𝑋 (1 or 2) separately.
Let's start by considering 𝑋 = 1. In this case, we have a Poisson process with a rate of 1 arrival per unit time. The probability of having exactly 2 arrivals in a time interval of length 2 is given by the Poisson probability mass function:
P(𝑋 = 1) * P(X = 2) = e^(-1) * (1^2 / 2!) = e^(-1) / 2
Now, let's consider 𝑋 = 2. In this case, we have a Poisson process with a rate of 2 arrivals per unit time. The probability of having exactly 2 arrivals in a time interval of length 2 is again given by the Poisson probability mass function:
P(𝑋 = 2) * P(X = 2) = e^(-2) * (2^2 / 2!) = 2 * e^(-2)
Since 𝑋 can be either 1 or 2 with equal probability, we need to multiply the probabilities calculated for each case by 1/2:
(1/2) * (e^(-1) / 2) + (1/2) * (2 * e^(-2))
Simplifying this expression, we get:
e^(-1) / 4 + e^(-2)