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 drunkd July 30th, 2012 10:13 AM

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..

 greg1313 July 30th, 2012 10:21 AM

Re: Bags and Balls

Is the bag the black ball is drawn from selected at random?

 drunkd July 30th, 2012 10:30 AM

Re: Bags and Balls

@Greg- Yeah....

 greg1313 July 30th, 2012 10:55 AM

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?"

 The Chaz July 30th, 2012 12:17 PM

Re: Bags and Balls

I'd go with conditional probability here...Bayes theorem, what have you.

 greg1313 July 30th, 2012 12:53 PM

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.

 drunkd July 31st, 2012 01:35 AM

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...:(

 greg1313 July 31st, 2012 03:07 AM

Re: Bags and Balls

Hmm . . . but it is given the ball is black.

 The Chaz July 31st, 2012 04:32 AM

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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