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# Cralo' sea food restaurant is famous for fresh crab and

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Manager
Joined: 22 Feb 2007
Posts: 165
Cralo' sea food restaurant is famous for fresh crab and [#permalink]

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20 Mar 2007, 22:30
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Cralo' sea food restaurant is famous for fresh crab and lobsters. The restaurant has two tanks. The customers has to choose their dinner from the tanks. Tank A has 28 lobsters and 12 crabs. Tank B has 12 lobsters and 28 crabs. For Susan its a nasty experience to see her dinner alive, so she closes her eyes and randomly points at the lobsters or crabs. If she points at a lobster what is the probability that it is from tank A.

(A) 7/10
(B) 1/2
(C) 7/20
(D) 1/4
(E) 7/40

Solution 1
p(lobster from tank A) =
=p(choosing lobster)*p(choosing tank A) ..... assuming they are mutually exclusive events
=[no of lobster/(no of loster + crabs)] * 1/no of tanks
=(28+12)/(28+12+12+28) * 1/2
=1/2 */12
=1/4 .... so D

Solution 2
p(lobster from tank A)
= p (choosing tank A) * p(choosing lobster having chosen tank A) ... conditional probability
= 1/2 *28/(28+12)
=7/20 .... so C

Which approach is correct ?
Can anyone clarify?
Senior Manager
Joined: 11 Feb 2007
Posts: 351

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20 Mar 2007, 23:27
Uhhh....I think you don't have to take into account the probability of choosing one tank or the other.

Because the passage explicitly says "If she points at a lobster what is the probability that it is from tank A" which means that you only have to consider of the 28 + 12 = 40 lobsters, what is the prob that she chose one of the 28 lobsters from tank A?

So I think it's simply, 28/(12+28) = 28/40 = 7/10

Perhaps you're making the problem more difficult than it really is?
Manager
Joined: 22 Feb 2007
Posts: 165

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20 Mar 2007, 23:57

ricokevin, your approach seems to be the simplest. But I am further confused !!
Senior Manager
Joined: 11 Feb 2007
Posts: 351

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21 Mar 2007, 00:35
GMAThopeful wrote:
:evil:

ricokevin, your approach seems to be the simplest. But I am further confused !!

Umm... sorry I'm not a native English speaker so my explanation might be confusing...

Let's think of it this way.

You have 40 lobsters on a giant plant (taken out from tank A and B so there are no lobsters left in the tanks). 28 of them have "A" written on their tail and 12 of them have "B" written.

You close your eyes and pick a lobster. What is the probability that the lobster you chose has "A" written on its tail?

28/40 = 7/10

You don't have to consider the probability of choosing tank A (which is 1/2). It is certain that she chose a lobster. Now you just have to figure out if it was from tank A. The chances that it was from tank A? 28/40 = 7/10.

Well...I can't make this any clearer until I know what the OA is. What is the OA?
Manager
Joined: 22 Feb 2007
Posts: 165

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21 Mar 2007, 01:14
OA is 7/10. OE is as under:

Probability that she has chosen a lobster = (1/2*28/40)+(1/2*12/40)=1/2

Probability that she has chosen a lobster from tank A = 1/2*28/40=7/20

Required probability =(7/20)/ (1/2)= 7/10 .

Hence, (A).

Now I get it.. it can be solved using conditional probability logic, but I have got the formula wrong.

p(lobster from tank A)
= p(choosing lobster having chosen tank A) /p (choosing tank A) ... conditional probability
= [28/(28+12)]/[1/2]
=7/10 .... so A

OA is also A
Senior Manager
Joined: 11 Feb 2007
Posts: 351

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21 Mar 2007, 01:23
I see.

Well, although I think this part: "Probability that she has chosen a lobster = (1/2*28/40)+(1/2*12/40)=1/2"

is unnecessary because the problem explicitly asks "if she points at a lobster"...

I'd like to know what others think.

Never imagined that GMAT would require you to know conditional probability...

I enjoyed using Bayes Theorem when I took prob & stats in college.
Manager
Joined: 28 Feb 2007
Posts: 197
Location: California

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21 Mar 2007, 11:27
GMAThopeful wrote:
Cralo' sea food restaurant is famous for fresh crab and lobsters. The restaurant has two tanks. The customers has to choose their dinner from the tanks. Tank A has 28 lobsters and 12 crabs. Tank B has 12 lobsters and 28 crabs. For Susan its a nasty experience to see her dinner alive, so she closes her eyes and randomly points at the lobsters or crabs. If she points at a lobster what is the probability that it is from tank A.

(A) 7/10
(B) 1/2
(C) 7/20
(D) 1/4
(E) 7/40

Solution 1
p(lobster from tank A) =
=p(choosing lobster)*p(choosing tank A) ..... assuming they are mutually exclusive events
=[no of lobster/(no of loster + crabs)] * 1/no of tanks
=(28+12)/(28+12+12+28) * 1/2
=1/2 */12
=1/4 .... so D

Solution 2
p(lobster from tank A)
= p (choosing tank A) * p(choosing lobster having chosen tank A) ... conditional probability
= 1/2 *28/(28+12)
=7/20 .... so C

Which approach is correct ?
Can anyone clarify?

For details read Bayes's Rule on Average Conditional probabilities
Manager
Joined: 20 Jun 2005
Posts: 145

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28 Mar 2007, 19:13
GMAThopeful wrote:
`Cralo' sea food restaurant is famous for fresh crab and lobsters. The restaurant has two tanks. The customers has to choose their dinner from the tanks. Tank A has 28 lobsters and 12 crabs. Tank B has 12 lobsters and 28 crabs. For Susan its a nasty experience to see her dinner alive, so she closes her eyes and randomly points at the lobsters or crabs. If she points at a lobster what is the probability that it is from tank A.

(A) 7/10
(B) 1/2
(C) 7/20
(D) 1/4
(E) 7/40

Lets assume Pr(A|B) is the conditional probability of A given B, where

A - the lobster from tank A,
B - poining the lobster.

It means that Susan points the lobster first (event B), then the lobster is chosen from the tank A ( event A).

By using the Bayes' theorem we get

Pr(A|B) = Pr(B|A) * Pr(A) / Pr(B).

where

Pr(A) - picking the tank A regardless of any other information = 1/2 = 0.5 . Susan is treating both tanks equally, it is 0.5

Pr(B) - getting a lobster regardless of any information on the tanks. It is 40/80 = 1/2.

Pr(B|A) - getting a lobster in the tank A = 28/40 = 7/10.

Given all this information, we can compute the probabilty of Susan having selected tank A given that she got a lobster, as such:

Pr(A|B) = Pr(B|A) * Pr(A) / Pr(B) = 7/10 * 0.5 / 0.5 = 7/10. So the answer is (A).

hmm.. sounds strange that Pr(A|B) = Pr(B|A).... -)

P.S.
Actually, as for me, this is definitely the brainteaser.
I suggest to review the Bayes' theorem in wikipedia, especially example 1 "Conditional probabilities"
http://en.wikipedia.org/wiki/Bayes%27_theorem

and of course to check out the GMATClub's "Probability lesson" http://www.gmatclub.com/content/courses ... bility.php

I was wondering such questions could be appeared at the Q49~51 score?
Re: PS - probability   [#permalink] 28 Mar 2007, 19:13
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