Final answer:
1) P(X > 2) = e^(-4) ≈ 0.0183. 2) E[Y] = 3.5, var[Y] = 2.25. 3) P(X > 2|Y < 11) = e^(-4) / (1 - P(Y > 11)).
Step-by-step explanation:
To find P(X > 2), we need to find the complement of the cumulative distribution function (CDF) at x = 2.
P(X > 2) = 1 - P(X ≤ 2)
P(X > 2) = 1 - F(2)
Since X is an exponential random variable with rate λ = 2, the CDF is given by F(x) = 1 - e^(-λx).
So, substituting λ = 2 and x = 2 into the CDF, we get:
P(X > 2) = 1 - (1 - e^(-2*2)) = 1 - (1 - e^(-4)) = e^(-4)
Therefore, P(X > 2) = e^(-4) ≈ 0.0183 (rounded to 4 decimal places).
To find E[Y] and var[Y], we need to use properties of the expected value and variance.
E[Y] = E[3X+2] = 3E[X]+2 = 3/λ+2 = 3/2+2 = 7/2 = 3.5
var[Y] = var[3X+2] = (3)^2*var[X] = 9/var[X] = 9/(2)^2 = 9/4 = 2.25
To find P(X > 2|Y < 11), we need to find the conditional probability of X being greater than 2 given that Y is less than 11.
We can use Bayes' theorem to find P(X > 2|Y < 11):
P(X > 2|Y < 11) = P(Y < 11|X > 2) * P(X > 2) / P(Y < 11)
P(Y < 11|X > 2) can be calculated as 1. P(X > 2) is e^(-4) (from part 1). And P(Y < 11) can be calculated by finding 1 - P(Y > 11).
Therefore, P(X > 2|Y < 11) = 1 * e^(-4) / (1 - P(Y > 11))