Questions about the solution to this multinomial distribuition problem

[s3a]

Homework Statement

The problem along with its solution are attached as TheProblemAndSolution.jpg.

Homework Equations

Multinomial distribution formula.

The Attempt at a Solution

I'm perfectly fine with part a.

As for part b, does the final answer need the $f_{Z|8}(0) = 0.694$, $f_{Z|8}(1) = 0.0277$ and what I'm assuming should have been $f_{Z|8}(2) = 0.027$ parts of the solution for part b, or just the $f_{Z|8}(z) = f_{Z|X = 8}(z) = 2! 0.25^{2-z} 0.05^z / {0.3^2 z! (2-z)!}$ part of the solution for part b?

As for part c, the final answer should say "$f_{XZ}(x,z) ≠ f_X(x) f_Z(z)$, and therefore X and Z are dependent" instead of "$f_{XZ}(x,z) ≠ f_X(x) f_Y(y)$, and therefore X and Y are dependent", right?

Any confirmation or rejection as to whether my comments are true or not would be greatly appreciated!

 

[Ray Vickson]

Homework Statement

The problem along with its solution are attached as TheProblemAndSolution.jpg.

Homework Equations

Multinomial distribution formula.

The Attempt at a Solution

I'm perfectly fine with part a.

As for part b, does the final answer need the $f_{Z|8}(0) = 0.694$, $f_{Z|8}(1) = 0.0277$ and what I'm assuming should have been $f_{Z|8}(2) = 0.027$ parts of the solution for part b, or just the $f_{Z|8}(z) = f_{Z|X = 8}(z) = 2! 0.25^{2-z} 0.05^z / {0.3^2 z! (2-z)!}$ part of the solution for part b?

As for part c, the final answer should say "$f_{XZ}(x,z) ≠ f_X(x) f_Z(z)$, and therefore X and Z are dependent" instead of "$f_{XZ}(x,z) ≠ f_X(x) f_Y(y)$, and therefore X and Y are dependent", right?

Any confirmation or rejection as to whether my comments are true or not would be greatly appreciated!

Your jpeg file is unreadable on my computer---it is fuzzy and almost microscopic but un-enlargable. If you would actually take the time to type out the problem I would be willing to look at it.

 

[s3a]

Alright, here is the typed version (with minor text-formatting differences). (Hopefully, I made no typos of my own. - I copied everything, including what I suspect is wrong, in the same way it is written in the solution.):

Problem statement:

P3:

To evaluate the technical support from a computer manufacturer, the number of rings before a call is answered by a service representative is tracked. Historically, 70% of the calls are answered in two rings or less, 25% are answered in three or four rings, and the remaining calls require five rings or more. Suppose you call this manufacturer 10 times and assume all calls are independent.

a) What is the probability that eight calls are answered in two rings or less, one call is answered in three or four rings, and one call requires five rings or more?

b) What is the conditional distribution of the number of calls requiring five rings or more given that eight calls are answered in two rings or less?

c) Are the number of calls answered in two rings or less and the number of calls requiring five rings or more independent random variables? Why?

Solution:
Part a:
Let X, Y, and Z denote the number of calls answered in two rings or less, three or four rings, and five rings or more, respectively. The joint mass function is extension of binomial with p=0.7, q=0.25, r=0.05, where n=10.

$f_{XYZ}(z,y,z) = n! / {x! y! z!} p^x q^y r^z$

P(X = 8, Y = 1, Z = 1) = $f_{XYZ}(8,1,1) = 10! / (8! 1! 1!) 0.7^8 0.25^1 0.05^1$ = 0.0649

Answer: P(X = 8, Y = 1, Z = 1) = 0.0649

Part b:
$f_{XZ}(x,z) = 10! / (x! z! (10 - x - z)!) 0.7^x 0.25^{10-x-z} 0.05^z$ ⇒ $f_{XZ}(8,0) = 0.1617$, $f_{XZ}(8,1) = 0.0645$, $f_{XZ}(8,0) = 0.0063$

$f_{X}(x) = 10! / ( x! (10-x)! ) 0.7^x 0.3^{10-x}$ ⇒ $f_{X}(8) = 0.233$

$f_{Z|8}(z) = f_{XZ}(8,z) / f_{X}(8) = [10! / ( 8! z! (10-8-z)! 0.7^8 0.25^(10-8-z) 0.05^z ) ] / [ 10! / ( 8! (10-8)! 0.7^8 0.3^{10-8} ) ] = [(10-8)! 0.25^{10-8-z} 0.05^z] / [(10-8-z)! 0.3^{10-8}] = [2! 0.25^{z-2} 0.05^z] / [(2-z)! 0.3^2]$

Answer: $f_{Z|8} = [2! 0.25^{2-z} 0.05^z] / [0.3^2 z! (2 - z)!]$ : $f_{Z|8}(0) = 0.694$, $f_{Z|8}(1) = 0.277$, $f_{Z|8}(0) = 0.027$

Part c:
$f_{Z}(z) = 10! / [ (10-z)! 0.95^{10-z} 0.05^z]$ and knowing $f_{XZ}(x,z)$ & $f_X(x)$ from part b we can easily show that

Answer: $f_{XZ}(x,z) ≠ f_X(x) f_Y(y)$ and therefore X and Y are dependent.

 

[haruspex]

For part (b), I agree that it should be fine to answer with just the generic f_Z|_X=8(z) expression. If they wanted numbers they should have said so.
I also agree with you on part c.
Btw, your pic was a bit small, but perfectly readable on my screen and I could enlarge it on my laptop+Windows with Ctrl-Shift-+. I guess it depends on the device/OS.

 

[s3a]

Thanks, haruspex!

 

[Ray Vickson]

Alright, here is the typed version (with minor text-formatting differences). (Hopefully, I made no typos of my own. - I copied everything, including what I suspect is wrong, in the same way it is written in the solution.):

Problem statement:

P3:

To evaluate the technical support from a computer manufacturer, the number of rings before a call is answered by a service representative is tracked. Historically, 70% of the calls are answered in two rings or less, 25% are answered in three or four rings, and the remaining calls require five rings or more. Suppose you call this manufacturer 10 times and assume all calls are independent.

a) What is the probability that eight calls are answered in two rings or less, one call is answered in three or four rings, and one call requires five rings or more?

b) What is the conditional distribution of the number of calls requiring five rings or more given that eight calls are answered in two rings or less?

c) Are the number of calls answered in two rings or less and the number of calls requiring five rings or more independent random variables? Why?

Solution:
Part a:
Let X, Y, and Z denote the number of calls answered in two rings or less, three or four rings, and five rings or more, respectively. The joint mass function is extension of binomial with p=0.7, q=0.25, r=0.05, where n=10.

$f_{XYZ}(z,y,z) = n! / {x! y! z!} p^x q^y r^z$

P(X = 8, Y = 1, Z = 1) = $f_{XYZ}(8,1,1) = 10! / (8! 1! 1!) 0.7^8 0.25^1 0.05^1$ = 0.0649

Answer: P(X = 8, Y = 1, Z = 1) = 0.0649

Part b:
$f_{XZ}(x,z) = 10! / (x! z! (10 - x - z)!) 0.7^x 0.25^{10-x-z} 0.05^z$ ⇒ $f_{XZ}(8,0) = 0.1617$, $f_{XZ}(8,1) = 0.0645$, $f_{XZ}(8,0) = 0.0063$

$f_{X}(x) = 10! / ( x! (10-x)! ) 0.7^x 0.3^{10-x}$ ⇒ $f_{X}(8) = 0.233$

$f_{Z|8}(z) = f_{XZ}(8,z) / f_{X}(8) = [10! / ( 8! z! (10-8-z)! 0.7^8 0.25^(10-8-z) 0.05^z ) ] / [ 10! / ( 8! (10-8)! 0.7^8 0.3^{10-8} ) ] = [(10-8)! 0.25^{10-8-z} 0.05^z] / [(10-8-z)! 0.3^{10-8}] = [2! 0.25^{z-2} 0.05^z] / [(2-z)! 0.3^2]$

Answer: $f_{Z|8} = [2! 0.25^{2-z} 0.05^z] / [0.3^2 z! (2 - z)!]$ : $f_{Z|8}(0) = 0.694$, $f_{Z|8}(1) = 0.277$, $f_{Z|8}(0) = 0.027$

Part c:
$f_{Z}(z) = 10! / [ (10-z)! 0.95^{10-z} 0.05^z]$ and knowing $f_{XZ}(x,z)$ & $f_X(x)$ from part b we can easily show that

Answer: $f_{XZ}(x,z) ≠ f_X(x) f_Y(y)$ and therefore X and Y are dependent.

I agree with your answers, numbers and all.

I still cannot get your pic to enlarge on my HP notebook computer, but the machine is getting old now and its graphics are not great.

However, the fact is that posting images is discouraged in PF, unless (for example) you need to show images, etc.

Note added in edit:
If you examine your solution to (b) you can see that $(Z|X=8) = \text{Binomial}(n=2,p=1/6)$. This is not an accident; it is an easy exercise to show
(X_1, X_2, \ldots, X_r) \sim \text{Multinom}(n; p_1, p_2, \ldots, p_r)\\<br /> \text{implies}\\<br /> (X_2, X_3, \ldots, X_r|X_1 = k_1) \sim \text{Multinom} \left( n-k_1;<br /> \frac{p_2}{1-p_1}, \ldots, \frac{p_r}{1-p_1} \right)