Solved

Probability Help

Posted on 2016-07-21
3
48 Views
Last Modified: 2016-07-26
Can you please  help me figure this out? Thanks.

Suppose managers at a company wish to meet the increasing demand for color photocopies and to have more reliable service. Based on historical data, the chance that a black-and-white copier will be down for repairs is 0.12. The color copiers are more of a problem and are down 23% of the time. As  a goal,  they would like to have at least a 99.90% chance of being able to furnish a black-and-white copy or a color copy on demand. They also wish to purchase only four copiers. They have asked for advice regarding the mix of black-and-white and color copiers. Supply them with advice on how to purchase the greatest number of color copiers while also meeting their goal regarding the furnishing of a copy on demand.  Provide calculation and reasons to support that advice. (Assume that the color copier can also be used to make a black-and- white copy.)

The manager should purchase
a. 2
b 1
c. 4
d. 0
e. 3

black-and-white copier(s)  and
a. 1
b. 0
c. 4
d. 3
e. 2

color copier(s).  If they do this, they will have a  ____  % chance of being able to furnish at least a black-and-white copy or a color copy on demand.
0
Comment
Question by:mustish1
3 Comments
 
LVL 1

Accepted Solution

by:
Petr J earned 250 total points
ID: 41723489
Probability of all N photocopiers being broke at a time is p^N (p is probability of 1 photocopier being broken).
This brings us to quite easy formula:
p_BW^c_BW*p_C^c_C >= 1-0,999
where
p_BW is probability black-and-white is broken
c_BW is count of black-and-white photocopiers
and the same for color (C) photocopiers.

Since BW photocopiers are more reliable and there is no demand for C photocopiers we can simplify the formula to:
p_BW^c_BW >= 0,001

After putting logarithm on both sides we come up with:
c_BW * log(p_BW) <= log(0,001)
c_BW * log(0,12) >= log(0,001)
c_BW >= log(0,001) / log(0,12)
c_BW >=3,26

The solution is to buy 4 BW photocopiers (this is minimal count of photocopiers).

3 BW and 1 C photocopier would meet the criteria as well:
0,12^3*0,23=0,0004<=0,001

2 BW and 2 C photocopier would meet the criteria as well:
0,12^2*0,23^2=0,0008<=0,001

1 BW and 3 C photocopier would not meet the criteria:
0,12^1*0,23^3=0,00146004>0,001
0
 
LVL 27

Assisted Solution

by:d-glitch
d-glitch earned 250 total points
ID: 41723490
This question is unrealistic and ambiguous, but you still have to have to start.
Pick any scenario and analyze it.
0
 

Author Closing Comment

by:mustish1
ID: 41723507
Thank You.
0

Featured Post

Courses: Start Training Online With Pros, Today

Brush up on the basics or master the advanced techniques required to earn essential industry certifications, with Courses. Enroll in a course and start learning today. Training topics range from Android App Dev to the Xen Virtualization Platform.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Lithium-ion batteries area cornerstone of today's portable electronic devices, and even though they are relied upon heavily, their chemistry and origin are not of common knowledge. This article is about a device on which every smartphone, laptop, an…
Originally published Entrepreneur.com Booming numbers of freelancing professionals are changing the face of work. In the United States alone last year, the number of workers freelancing grew from 700,000 to 54 million, according to a Freelancers’…
The Bounty Board allows you to request an article or video on any technical topic, or fulfill a bounty request to earn points. Watch this video to learn how to use the Bounty Board to get the content you want, earn points, and browse submitted bount…
Notifications on Experts Exchange help you keep track of your activity and updates in one place. Watch this video to learn how to use them on the site to quickly access the content that matters to you.

786 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question