My Math Forum probability problem (2)
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 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?
 October 29th, 2012, 02:33 AM #2 Global Moderator     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.
 October 29th, 2012, 02:44 AM #3 Senior Member     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: $P(X)=\frac{6}{14}=\frac{3}{7}$
October 29th, 2012, 03:50 AM   #4
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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: $P(X)=\frac{6}{14}=\frac{3}{7}$
iff $^{P$$U_1$$\,=\,P$$U_2$$}$

 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...
 November 4th, 2012, 05:28 AM #6 Global Moderator     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.
November 4th, 2012, 11:23 AM   #7
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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:

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

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

$P(A|B)=\frac{\frac{6}{10} \cdot \frac{1}{2}}{\frac{14}{20}}=\frac{3}{7} \$ 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:

$P(A|B)=\frac{\frac{6}{10} \cdot \frac{1}{2}}{\frac{6}{20}}=1$

The book is wrong.

 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.
 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?
November 9th, 2012, 02:22 AM   #10
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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.

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