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Intermediate Probability

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by Aman verma » Tue Feb 16, 2010 9:02 am
Q : Urn A contains six red and four black balls and Urn B contains four red and six black balls.One ball is drawn at random from Urn A and placed in Urn B.Then one ball is transferd at random from Urn B to Urn A .If one ball is now drawn at random from Urn A ,find the probability that it is red :

a)32/65

b32/55

c)23/55

d)56/65

e)31/65
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by ajith » Tue Feb 16, 2010 10:37 am
Aman verma wrote:Q : Urn A contains six red and four black balls and Urn B contains four red and six black balls.One ball is drawn at random from Urn A and placed in Urn B.Then one ball is transferd at random from Urn B to Urn A .If one ball is now drawn at random from Urn A ,find the probability that it is red :

a)32/65

b32/55

c)23/55

d)56/65

e)31/65
Case 1: One red ball is transferred to the second urn
Probabilty of that happening = 6/10 = 3/5

One red ball is transferred to the first urn from second
Probabilty of that happening = 5/11 = 5/11

One red ball is selected from urn A after this = 6/10

Probability that all of them happening together (all of the events are independent) = 3/5*5/11*3/5 = 9/55

Case 2: Red ball is first transferred, black ball is transferred back and red ball is chosen

Probability 6/10*6/11*5/10 = 9/55

Case 3: Black ball is first transferred, red ball is transferred back and red ball is chosen

Probability = 4/10*4/11*5/10 = 2/5*1/2*4/11 =4/55

Case 4: Black ball is first transferred , black ball is transferred back and red ball is chosen
4/10*7/11*6/10 =2/5*3/5*7/11= 42/5* 1/55 = 8.4/55


I am messing up calculation somewhere and getting 30.4/55 :(
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by amittilak » Tue Feb 16, 2010 11:44 am
Ajith,
In your case 3, you have 4/10*4/11*5/10
I think it should be, 4/10*4/11*7/10, since you are adding 1 extra red ball from Urn B to urn A.
That gives us,
(6*5*6+6*6*5+4*4*7+4*7*6)/(10*11*10 =
27/55

What's the OA on this?
ajith wrote:
Aman verma wrote:Q : Urn A contains six red and four black balls and Urn B contains four red and six black balls.One ball is drawn at random from Urn A and placed in Urn B.Then one ball is transferd at random from Urn B to Urn A .If one ball is now drawn at random from Urn A ,find the probability that it is red :

a)32/65

b32/55

c)23/55

d)56/65

e)31/65
Case 1: One red ball is transferred to the second urn
Probabilty of that happening = 6/10 = 3/5

One red ball is transferred to the first urn from second
Probabilty of that happening = 5/11 = 5/11

One red ball is selected from urn A after this = 6/10

Probability that all of them happening together (all of the events are independent) = 3/5*5/11*3/5 = 9/55

Case 2: Red ball is first transferred, black ball is transferred back and red ball is chosen

Probability 6/10*6/11*5/10 = 9/55

Case 3: Black ball is first transferred, red ball is transferred back and red ball is chosen

Probability = 4/10*4/11*5/10 = 2/5*1/2*4/11 =4/55


Case 4: Black ball is first transferred , black ball is transferred back and red ball is chosen
4/10*7/11*6/10 =2/5*3/5*7/11= 42/5* 1/55 = 8.4/55


I am messing up calculation somewhere and getting 30.4/55 :(
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by jeffedwards » Tue Feb 16, 2010 12:33 pm
Wow, used the same method as the two of you above. Is there an easier method? This took me much longer that 2 minutes to solve.
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by Stuart@KaplanGMAT » Tue Feb 16, 2010 3:57 pm
Some minor math errors in there:

Case 1: Red moved, then red moved, then red chosen:

3/5 * 5/11 * 3/5 = 9/55

Case 2: Red moved, then black moved, then red chosen:

3/5 * 6/11 * 1/2 = 9/55

Case 3: Black moved, then red moved, then red chosen:

2/5 * 4/11 * 7/10 = 28/275

Case 4: Black moved, then black moved, then red chosen:

2/5 * 7/11 * 3/5 = 42/275

Add them up: 18/55 + 70/275 = 18/55 + 14/55 = 32/55... choose B.

* * *

I actually arrived at B by logic/estimation as well:

We're moving a ball away then a ball back; since there are more red balls in the first urn, there's a slightly better chance of decreasing the number of red balls in Urn A than increasing it.

Our original probability of drawing a red ball is 60%, so the final answer should be a bit less than 60%.

A, C and E are all under 50% - seem too low.
D is more than 60% - definitely too high.
B is just under 60% - just right!

Here's another way to use logic/estimation:

we know that our 3 denominators are 5, 11 and 5. Therefore, the denominator in the final answer must be a factor of 275.

Eliminate A, D and E.

(If nothing else, this is a super quick way to give ourselves a 50/50 shot.)

C goes from 60% original probability to just over 40% new probability - that's way too low, eliminate C.

Pick B!
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by jeffedwards » Tue Feb 16, 2010 5:17 pm
Genius Stuart, I love all of your posts!
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by thephoenix » Tue Feb 16, 2010 9:00 pm
there are 4 cases

1)RR [red rmoved from A, red removed fromB]
2)RB
3)BR
4)BB


for 1) probability pickin a red ball from A finally=6/10 (for picking red from urn )*5/11(for picking red from urn b After the addition of a red ball from A)*6/10=9/55
similarly for 2)P=6/10 * 6/11 * 5/10 = 9/55

for 3)=28/275

for 4)=42/275

tot=32/55
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by sumanr84 » Wed Feb 17, 2010 12:29 am
Stuart Kovinsky wrote: (If nothing else, this is a super quick way to give ourselves a 50/50 shot.)

C goes from 60% original probability to just over 40% new probability - that's way too low, eliminate C.

Pick B!
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by Aman verma » Wed Feb 17, 2010 2:22 am
B it is.
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