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 July 30th, 2012, 11:13 AM #1 Newbie   Joined: Jul 2012 Posts: 27 Thanks: 0 Bags and Balls Three bags contain 2 white, 3 black and ; 4 white, 1 black ; 3 white, 7 black balls; respectively. One ball is drawn from one of the bags and found to be black. The probability that it was drawn from the bag mentioned last is: 1)7/10 2)7/11 3)7/15 4)11/15 5)none Please add valid reasons and steps..
 July 30th, 2012, 11:21 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,697 Thanks: 977 Math Focus: Elementary mathematics and beyond Re: Bags and Balls Is the bag the black ball is drawn from selected at random?
 July 30th, 2012, 11:30 AM #3 Newbie   Joined: Jul 2012 Posts: 27 Thanks: 0 Re: Bags and Balls @Greg- Yeah....
 July 30th, 2012, 11:55 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,697 Thanks: 977 Math Focus: Elementary mathematics and beyond Re: Bags and Balls Well, if that's the case I get 7/30. Reason: there's a 1/3 chance of selecting the bag with ten balls and a 7/10 chance of drawing a black ball from that bag. 1/3 * 7/10 = 7/30. That's not in the list of choices for a correct answer, though. I'm interpreting the question as asking "what is the probability of selecting a black ball from the bag with ten balls?"
 July 30th, 2012, 01:17 PM #5 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Bags and Balls I'd go with conditional probability here...Bayes theorem, what have you.
 July 30th, 2012, 01:53 PM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,697 Thanks: 977 Math Focus: Elementary mathematics and beyond Re: Bags and Balls Thanks, [color=#00BF40]The Chaz[/color]. Let P(T|B) be the probability that, given that a black ball is drawn, it came from the bag with ten balls. Let P(B|T) the probability that the ball is black, given that it came from the bag with ten balls and P(B) the overall probability that the ball is black. Using Bayes' Theorem, we have P(T|B) = (P(B|T)P(T))/P(B) = (7/10 * 1/3)/(1/2) = 7/15.
 July 31st, 2012, 02:35 AM #7 Newbie   Joined: Jul 2012 Posts: 27 Thanks: 0 Re: Bags and Balls But according to me the answer is none.. The solution would be 'None'. This is because the answer is 7/30. This is derived from the calculation of the conditional probability, the probability that the last bag was chosen and the ball was black in it. This would be 1/3 *7/10. I am still not sure with the solution...
 July 31st, 2012, 04:07 AM #8 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,697 Thanks: 977 Math Focus: Elementary mathematics and beyond Re: Bags and Balls Hmm . . . but it is given the ball is black.
July 31st, 2012, 05:32 AM   #9
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Re: Bags and Balls

Quote:
 Originally Posted by drunkd But according to me the answer is none.. The solution would be 'None'. This is because the answer is 7/30. This is derived from the calculation of the conditional probability, the probability that the last bag was chosen and the ball was black in it. This would be 1/3 *7/10. I am still not sure with the solution...
This is wrong. The events "black ball (from bag three)" and "bag three" are NOT independent, so you can't just multiply 1/3 and 7/10.

Think of ALL the ways to get a black ball. THIS is where your thinking/calculation will be appropriate:

Black ball in bag 1: 1/3*3/5
Black ball in bag 2: 1/3*1/5
Black ball in bag 3: 1/3*7/10

Add all these up... 1/5 + 1/15 + 7/30.

This is the probability of getting ANY black. It's 1/2.

What proportion of this (1/2) came from bag 3?

(contribution from bag 3)/(probability of any black)
(7/30)/(1/2) = 7/15

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