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# A bag contains 3 red and 2 black ball. Another bag contains 4 red and

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Re: A bag contains 3 red and 2 black ball. Another bag contains 4 red and [#permalink]
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PiyushK wrote:
A bag contains 3 red and 2 black ball. Another bag contains 4 red and 5 black balls. A ball is drawn from the first bag and is placed in the second. A ball is then drawn from the second. What is the probability that this draw of red ball is due to the drawing of red ball from the first bag ?

A. 3/5
B. 2/5
C. 4/25
D. 3/10
E. 15/23

I got the answer as D

3/5*5/10 = 3/10

Could you please help explain how did we get E?
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Re: A bag contains 3 red and 2 black ball. Another bag contains 4 red and [#permalink]
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surya167 wrote:
PiyushK wrote:

Could you please help explain how did we get E?

There are two possible cases, out of those two we have to calculate the possibility of case 1.

case 1. one red ball was transferred to second bag, and then red ball is drawn from bag2.
or
case 2. one black ball was transferred to second bag, and then red ball is drawn from bag2.

Both cases are possible but we have to find out occurrence of case 1 out of these two cases.

I think this much hint is sufficient bcz solution is just single step away.
Try, else i will post my solution.
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Re: A bag contains 3 red and 2 black ball. Another bag contains 4 red and [#permalink]
i got the D but after the hint it got E.

.3/.3+.16 = 15/23

Thanks
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Re: A bag contains 3 red and 2 black ball. Another bag contains 4 red and [#permalink]
coolpintu wrote:
Probability of getting red ball from bag1 and then red from bag 2 = 3/5*5/10 = 15/50
Probability of getting black ball from bag1 and then red from bag 2 = 2/5*4/10 = 8/50

Now comes the condition, hence apply conditional probability:
P(Red)/ P(Red + Black)

(15/50)/(15/50+8/50) = 15/50 /23/50 = 15/23.
IMO E

Thanks, now i understand. I was missing the point of conditional probability totally!
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Re: A bag contains 3 red and 2 black ball. Another bag contains 4 red and [#permalink]
surya167 wrote:
coolpintu wrote:
Probability of getting red ball from bag1 and then red from bag 2 = 3/5*5/10 = 15/50
Probability of getting black ball from bag1 and then red from bag 2 = 2/5*4/10 = 8/50

Now comes the condition, hence apply conditional probability:
P(Red)/ P(Red + Black)

(15/50)/(15/50+8/50) = 15/50 /23/50 = 15/23.
IMO E

Thanks, now i understand. I was missing the point of conditional probability totally!

very well explained......coolpintu
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A bag contains 3 red and 2 black ball. Another bag contains 4 red and [#permalink]
I did it like this:-
There are three cases,when we fix that the ball drawn from second box is red
1)red ball drawn from first box and it is picked from second - 3/5*1/10
2)red ball drawn from first box and other red ball is picked - 3/5*4/10
3)black ball is drawn from first box and red ball is picked - 2/5*4/10

First is the favourable option, so i did (3/5*1/10)/(3/5*4/10 + 2/5*4/10) =3/23 which is none of the option.

Please anyone let me know what i have missed here..

Ohh got it ..i think i mis-interpreted the question as i thought that the red ball drawn from the second box should be the one that is drawn from first.

So favourable option should be 1 and 2 both which will give me 15/23..

Anyone got that misunderstanding??
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Re: A bag contains 3 red and 2 black ball. Another bag contains 4 red and [#permalink]
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PiyushK wrote:
A bag contains 3 red and 2 black ball. Another bag contains 4 red and 5 black balls. A ball is drawn from the first bag and is placed in the second. A ball is then drawn from the second. What is the probability that this draw of red ball is due to the drawing of red ball from the first bag ?

A. 3/5
B. 2/5
C. 4/25
D. 3/10
E. 15/23

The ball transferred from the first bag to the second could be a black ball or a red ball. Therefore, we have two scenarios:

1) A red ball is drawn from the 2nd bag after a black ball is drawn from 1st bag and transferred to the 2nd bag.

2) A red ball is drawn from the 2nd bag after a red ball is drawn from 1st bag and transferred to the 2nd bag.

Let’s find the probability of each scenario:

1) P(1st = black, 2nd = red) = 2/5 x 4/10 = 8/50

2) P(1st = red, 2nd = red) = 3/5 x 5/10 = 15/50

Thus the probability of drawing a red ball from the 2nd bag is 8/50 + 15/50 = 23/50, and the probability that it is due to a red ball having been drawn from the 1st bag is:

(15/50)/(23/50) = 15/23

Answer: E
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Re: A bag contains 3 red and 2 black ball. Another bag contains 4 red and [#permalink]
Answer is E

Probability of getting the red ball from bag1 and then red from bag 2 = 3/5*5/10 = 15/50
Probability of getting the black ball from bag1 and then red from bag 2 = 2/5*4/10 = 8/50
As per given condition:
P(Red)/ P(Red + Black)

(15/50)/(15/50+8/50) = 15/50 /23/50 = 15/23.
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A bag contains 3 red and 2 black ball. Another bag contains 4 red and [#permalink]
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I am a little upset by the questions wording that makes me think we have to assume a red ball is chosen from bag 2 as opposed to accounting for it in the calculation. As such, none of the calculations I did revolved around picking a red ball from bag #2.

"What is the probability that this draw of red ball is due to the drawing of red ball from the first bag?"

The way the calculations are done to get the correct answer, I believe the question should have stated:

"What is the probability that a red ball is drawn AND it is due to the drawing of red ball from the first bag?"

Do people agree? Or am I making some obvious large mistake in either my understanding of the question and/or math?
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Re: A bag contains 3 red and 2 black ball. Another bag contains 4 red and [#permalink]
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Re: A bag contains 3 red and 2 black ball. Another bag contains 4 red and [#permalink]
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