An attempt to establish a video call via some social media app may fail with probability 0.1. If connection is established and if no connection failure occurs thereafter, then the duration of a typical video call in minutes is an exponential random variable X with E[X] = 3. However, due to an unfortunate bug in the app all calls are disconnected after 6 minutes. Let random variable Y denote the overall call duration (i. e., Y = 0 in case of failure to connect, Y = 6 when a call gets disconnected due to the bug, and Y = X otherwise.) Find:

(1). y = x ~ Exp (1/3).

(2). Check attachment.

(3). EY = 3(1 - e^-2).

(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.

Step-by-step explanation:

Kindly check the attachment to aid in understanding the solution to the question.

So, from the question, we given the following parameters or information or data;

(A). The probability in which attempt to establish a video call via some social media app may fail with = 0.1.

(B). " If connection is established and if no connection failure occurs thereafter, then the duration of a typical video call in minutes is an exponential random variable X with E[X] = 3. "

(C). "due to an unfortunate bug in the app all calls are disconnected after 6 minutes. Let random variable Y denote the overall call duration (i.e., Y = 0 in case of failure to connect, Y = 6 when a call gets disconnected due to the bug, and Y = X otherwise.)."

(1). Hence, for FY(y) = y = x ~ Exp (1/3) for the condition that zero is equal to y = x < 6.

(2). Check attachment.

(3). EY = 3(1 - e^-2).

(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.

The condition to follow in order to solve this question is that y = 0 if x ≤ 0, y = x if 0 ≤ x ≤ 6 and y = 6 if x ≥ 6.

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry