Math Help Forum

Math Help Forum Feed Site Feed

Go Back   Math Help Forum > University Math Help > Advanced Probability and Statistics
Reply
 
Thread Tools Display Modes
  #1  
Old January 12th, 2009, 02:29 PM
Member
 
Join Date: Oct 2007
Location: Chicago
Posts: 143
Country:
Thanks: 86
Thanked 3 Times in 3 Posts
DINOCALC09 is on a distinguished road
Send a message via AIM to DINOCALC09
Default complex probability

An examination consists of multiple choice questions, each having five possible answers. Linda estimates that she has probability 0.75 of knowing the answer to any question that may be asked. If she does not know the answer, she will guess, with the conditional probability 1/5 of being correct. What is the probability that Linda gives the correct answer to teh question.

How would you go about solving this question. I know the answer is 0.8 - says so in the back of my book. However, I'm curious to how this was derived.
Reply With Quote
Advertisement
 
  #2  
Old January 12th, 2009, 02:35 PM
mr fantastic's Avatar
Flow Master

 
Join Date: Dec 2007
Location: Zeitgeist
Posts: 12,239
Country:
Thanks: 2,576
Thanked 4,763 Times in 4,195 Posts
mr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond reputemr fantastic has a reputation beyond repute
Default

Quote:
Originally Posted by DINOCALC09 View Post
An examination consists of multiple choice questions, each having five possible answers. Linda estimates that she has probability 0.75 of knowing the answer to any question that may be asked. If she does not know the answer, she will guess, with the conditional probability 1/5 of being correct. What is the probability that Linda gives the correct answer to teh question.

How would you go about solving this question. I know the answer is 0.8 - says so in the back of my book. However, I'm curious to how this was derived.
A tree diagram makes it clear:

The first two branches are knowing (Pr(knowing) = 0.75) or not knowing (Pr(not knowing) = 0.25). Then the next two branches are guessing (correct or not correct).

I get (0.75) + (0.25)(1/5) = 0.75 + 0.05 = 0.8.
__________________
There are two things you should never try to prove: the impossible and the obvious.

The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

  • To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
Reply With Quote
The following users thank mr fantastic for this useful post:
Donate to MHF
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off
Forum Jump


All times are GMT -7. The time now is 10:57 PM.


Powered by vBulletin® Version 3.7.3
Copyright ©2000 - 2009, Jelsoft Enterprises Ltd.
SEO by vBSEO 3.2.0 ©2008, Crawlability, Inc.
©2005 - 2009 Math Help Forum


Math Help Forum is a community of maths forums with an emphasis on maths help in all levels of mathematics.
Register to post your math questions or just hang out and try some of our math games or visit the arcade.