Answer:
(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.
Answer:
The probability of a selection of 50 pages will contain no errors is 0.368
The probability that the selection of the random pages will contain at least two errors is 0.2644
Step-by-step explanation:
From the information given:
Let q represent the no of typographical errors.
Suppose that there are exactly 10 such errors randomly located on a textbook of 500 pages. Let 
be the random variable that follows a Poisson distribution, then mean 
and the mean that the random selection of 50 pages will contain no error is 
∴
Pr(q =0) = 0.368
The probability of a selection of 50 pages will contain no errors is 0.368
The probability that 50 randomly page contains at least 2 errors is computed as follows:
P(X ≥ 2) = 1 - P( X < 2)
P(X ≥ 2) = 1 - [ P(X = 0) + P (X =1 )] since it is less than 2
![P(X \geq 2) = 1 - [ \dfrac{e^{-1} 1^0}{0!} +\dfrac{e^{-1} 1^1}{1!} ]](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%201%20-%20%5B%20%5Cdfrac%7Be%5E%7B-1%7D%201%5E0%7D%7B0%21%7D%20%2B%5Cdfrac%7Be%5E%7B-1%7D%201%5E1%7D%7B1%21%7D%20%5D)
![P(X \geq 2) = 1 - [0.3678 +0.3678]](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%201%20-%20%5B0.3678%20%2B0.3678%5D)
P(X ≥ 2) = 0.2644
The probability that the selection of the random pages will contain at least two errors is 0.2644
Answer:
Step-by-step explanation:
Sampling with replacement means that we choose a marble, note its colour, put it back and shake the box, then choose a marble again.
There are 
marbles in total. The probability that the first chosen marble is blue is 
then Maya replaces this marble and the probability of choosing the second blue marble is the same. Using the product rule, the probability of drawing 2 blue marbles in a row is
3(n+5) < 3/4(n)
The three is multiplied by n+5. Whenever you see "is equal to/greater/less than," that is where you put your =, >, or < signs. Your variable, "n" represents the number. So three-fourths of the number is the same as multiplying the number by three-fourths.
