1. Since X is deterministically equal to 1, the blue bulb will flash once and then die. Therefore, in the interval [0,2], the red bulb will continue flashing according to a Poisson process with rate lambda=2.
The expected number of flashes of the red bulb in a time interval t is given by lambda*t. In this case, lambda=2 and t=2, so the expected number of flashes of the red bulb during the interval [0,2] is 2*2 = 4.
2. If there are exactly 5 flashes in the time interval [0,X], and exactly 2 of them are red, then the remaining 3 flashes must be blue.
The probability of exactly 2 red flashes in 5 total flashes is given by the binomial probability formula: P(X=k) = (n choose k) * (p^k) * ((1-p)^(n-k)), where n is the total number of trials, k is the number of successes, and p is the probability of success.
In this case, n=5, k=2, and p is the probability of a red flash, which is lambda_red/(lambda_red+lambda_blue). In this case, lambda_red=2 and lambda_blue=1, so p=2/(2+1) = 2/3.
Therefore, the probability of exactly 2 red flashes in 5 total flashes is P(X=2) = (5 choose 2) * ((2/3)^2) * ((1-(2/3))^(5-2)) = (10) * ((2/3)^2) * ((1/3)^3) = 10 * (4/9) * (1/27) = 40/243.
3. If X is equal to either 1 or 2, with equal probability, there are two cases to consider: X=1 and X=2.
For the case when X=1, the blue bulb will flash once and then die, just like in question 1. So the expected number of flashes of either color in the interval [0,2] is 4.
For the case when X=2, the blue bulb will flash twice and then die. In this case, the red bulb will continue flashing according to a Poisson process for the full interval [0,2]. The expected number of red flashes in [0,2] is 2*2 = 4, as calculated in question 1.
So the total expected number of flashes of either color in the interval [0,2] when X=2 is 4+4 = 8.
Since X=1 or X=2 with equal probability, the probability of either case occurring is 1/2. Therefore, the expression for the probability of exactly 2 arrivals during the time interval [0,2] is:
(1/2)*(probability of exactly 2 arrivals when X=1) + (1/2)*(probability of exactly 2 arrivals when X=2)
Starting at time 0, a red bulb flashes according to a Poisson process with rate lambda=2. Similarly, starting at time 0, a blue bulb flashes according to a Poisson process with rate lambda=1, but only until a non negative random time X_i, at which point the blue bulb "dies". We assume that the two Poisson processes and the random variable X are (mutually) independent.
1. Suppose that X is deterministically equal to 1. What is the expected total number of flashes (of either colour) during the interval [0,2]?
2. In the time interval [0,X], there are exactly 5 flashes. What is the probability that exactly 2 of them were red?
3. 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].
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