An apartment building has 20 two-bedroom apartments and 8 three-bedroom apartments. If 4 apartments are chosen at random to be r
2 answers:
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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.
The mathematical expression does not seem clear but I have made an attempt to make sense of what is implied.
Answer:
<em>a</em> = 4, <em>b</em> = 2, <em>c</em> = 16, <em>d</em> = 9
Step-by-step explanation:
Solving the first part of the question by indices,
Comparing the rightmost term with the second term in the question,
<em>a</em> = 4, <em>b</em> = 2
Solving on,
Comparing with the final term in the question,
<em>c</em> = 16 and <em>d</em> = 9
Therefore,
<em>a</em> = 4, <em>b</em> = 2, <em>c</em> = 16, <em>d</em> = 9
Answer:
a) P=0
b) P=0.164
c) P=0.145
Step-by-step explanation:
We have 12 pieces, with 3 of each of the 4 flavors.
You draw the first 4 pieces.
a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.
b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.
The first probability is for the first draw, and has a value of 1, as any flavor will be ok.
The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.
The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.
The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.
Then, the probabilty of picking the 4 from different flavors is:
c) We can repeat the method for the previous probabilty.
The first draw has a probability of 1 because any flavor is ok.
In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.
For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.
If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.
For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.
For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.
We have then the probabilty as: