My Math Forum  

Go Back   My Math Forum > High School Math Forum > Probability and Statistics

Probability and Statistics Basic Probability and Statistics Math Forum


Reply
 
LinkBack Thread Tools Display Modes
October 28th, 2012, 11:14 PM   #1
Senior Member
 
Joined: Sep 2012

Posts: 112
Thanks: 0

probability problem (2)

Please tell me the way to solve the following problem.

There are 2 urns. The 1st one contains 6 red balls and 4 black balls. The 2nd urn contains 8 red and 2 black balls. You take an urn, draw one ball and find that it is red. What is the probability that you draw the ball from 1st urn?
helloprajna is offline  
 
October 29th, 2012, 02:33 AM   #2
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,885
Thanks: 1088

Math Focus: Elementary mathematics and beyond
Re: probability problem (2)

This is a conditional probability: P(A ? B)/P(B), where P(A ? B) is the probability of drawing a red ball from the first urn and P(B) is the overall probability of drawing a red ball. So we have (6/10 * 1/2)/(14/20) = 3/7.
greg1313 is offline  
October 29th, 2012, 02:44 AM   #3
Senior Member
 
MarkFL's Avatar
 
Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,208
Thanks: 516

Math Focus: Calculus/ODEs
Re: probability problem (2)

I got the same result by observing there are 14 red balls, 6 of which are in the first urn, hence:

MarkFL is online now  
October 29th, 2012, 03:50 AM   #4
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,885
Thanks: 1088

Math Focus: Elementary mathematics and beyond
Re: probability problem (2)

Quote:
Originally Posted by MarkFL
I got the same result by observing there are 14 red balls, 6 of which are in the first urn, hence:

iff
greg1313 is offline  
November 4th, 2012, 01:24 AM   #5
Senior Member
 
Joined: Sep 2012

Posts: 112
Thanks: 0

Re: probability problem (2)

I also want to solve this problem in the similar way as Mark did.But my book has solved it in a different way and its answer is also different.
"Probability of drawing a red ball from urn 1=6/10=0.6
Probability of taking the urn 1 from the 2 urns=1/2
Thus the probability that the red ball was drawn from the urn 1 is
0.6*(1/2)= 0.3

I can't find anything wrong in the process mentioned by the book. But still the answer isn't the same.

I tried to solve this problem using Bayesian method. The answer comes out 3/7. I don't understand which one is the correct answer- 3/10 or 3/7. Please help...
helloprajna is offline  
November 4th, 2012, 05:28 AM   #6
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,885
Thanks: 1088

Math Focus: Elementary mathematics and beyond
Re: probability problem (2)

It's given that the ball drawn is red. I believe that makes a difference.
greg1313 is offline  
November 4th, 2012, 11:23 AM   #7
jks
Senior Member
 
jks's Avatar
 
Joined: Jul 2012
From: DFW Area

Posts: 628
Thanks: 92

Math Focus: Electrical Engineering Applications
Re: probability problem (2)

Quote:
Originally Posted by helloprajna
But my book has solved it in a different way and its answer is also different.
"Probability of drawing a red ball from urn 1=6/10=0.6
Probability of taking the urn 1 from the 2 urns=1/2
Thus the probability that the red ball was drawn from the urn 1 is
0.6*(1/2)= 0.3

I cant find anything wrong in the process mentioned by the book. But still the answer isnt the same.

I tried to solve this problem using Bayesian method. The answer comes out 3/7. I don't understand which one is the correct answer- 3/10 or 3/7. Please help...
The book is wrong. The answer is 3/7. Let me first work the problem explicitly with Bayes' Theorem. It gives exactly the same calculation that [color=#00BF00]greg1313[/color] provided and it probably gives exactly the same calculation that you came up with using the Bayesian method. However, by changing the problem slightly we can show why the book is wrong.

Bayes' Theorem:



Let A = the ball is drawn from the first urn, B = a red ball was drawn. So we have:

as [color=#00BF00]greg1313[/color] stated.

Now, let's say the the second urn contains no red balls, only black balls. According to the way your book solved the problem, the probability is still 3/10.

However, the probability of the red ball being from urn 2 is 0 since there are no red balls in it. So the probability of the red ball being from urn 1 is 1.

Bayes' Theorem calculation:



The book is wrong.
jks is offline  
November 5th, 2012, 10:55 PM   #8
Senior Member
 
Joined: Sep 2012

Posts: 112
Thanks: 0

Re: probability problem (2)

I think you're right. Thank you so much for the explanation and help.
I also want to thank greg and mark. I'm grateful to you all.
helloprajna is offline  
November 7th, 2012, 08:18 AM   #9
Global Moderator
 
Joined: Dec 2006

Posts: 19,983
Thanks: 1853

Did you post the problem exactly as it is worded in the book? If you didn't, what was the wording used in the book?
skipjack is offline  
November 9th, 2012, 02:22 AM   #10
Senior Member
 
Joined: Sep 2012

Posts: 112
Thanks: 0

Re:

Quote:
Originally Posted by skipjack
Did you post the problem exactly as it is worded in the book? If you didn't, what was the wording used in the book?
Yes I've posted both the question and answer exactly as my book says.
helloprajna is offline  
Reply

  My Math Forum > High School Math Forum > Probability and Statistics

Tags
probability, problem



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Probability Problem / Insurance problem bsjmath Advanced Statistics 2 March 1st, 2014 03:50 PM
Probability problem ystiang Advanced Statistics 1 February 13th, 2014 05:27 PM
Probability problem halloweengrl23 Advanced Statistics 3 March 23rd, 2012 03:27 AM
Probability Problem.Please help adiptadatta Probability and Statistics 1 May 11th, 2010 07:46 AM
Help with probability problem! indian Algebra 0 February 21st, 2008 05:10 PM





Copyright © 2018 My Math Forum. All rights reserved.