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Probability Help!!

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Intern
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Joined: 28 Feb 2011
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Probability Help!!  [#permalink]

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New post 07 May 2011, 22:09
Hi Guys,

I came across the below problem in an online practice test..I wanted to check if my approach is rite?

There are 5 monkeys on an island. Each monkey has a boat. Each monkey take his boat and tries to get off the island. If the probability of each boat sinking is 1/4, what is the probability that all monkeys make it safely back except for King Kong who took the S.S. Banana, approximately?


Here's my aprroach:

P(boat sinking) = 1/4
P(boat not sinking) = 1-1/4 = 3/4

P(all monkeys make it safely except kingkong) = 3/4*3/4*3/4*3/4*1/4 = 81/1024.

After solving the problem, i had a second thought because the question asks for what is the probability that all monkeys make it safely ..
so would it just be:

P(all monkeys make it safely except kingkong) = 3/4*3/4*3/4*3/4= 81/256.

any suggestions pls?

Regards,
Anu

--== Message from the GMAT Club Team ==--

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Re: Probability Help!!  [#permalink]

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New post 08 May 2011, 01:21
King Kong's boat is redundant and assuming SS boats probability of reaching safely as 1, we have the following --

The probability has been asked for reaching of all the monkeys safely means (3/4) ^4 * 1.

I hope if you mention the options it will be easier to take a better STAB at this.
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Re: Probability Help!!  [#permalink]

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New post 08 May 2011, 01:38
anuu wrote:
Hi Guys,

I came across the below problem in an online practice test..I wanted to check if my approach is rite?

There are 5 monkeys on an island. Each monkey has a boat. Each monkey take his boat and tries to get off the island. If the probability of each boat sinking is 1/4, what is the probability that all monkeys make it safely back except for King Kong who took the S.S. Banana, approximately?


Here's my aprroach:

P(boat sinking) = 1/4
P(boat not sinking) = 1-1/4 = 3/4

P(all monkeys make it safely except kingkong) = 3/4*3/4*3/4*3/4*1/4 = 81/1024.

After solving the problem, i had a second thought because the question asks for what is the probability that all monkeys make it safely ..
so would it just be:

P(all monkeys make it safely except kingkong) = 3/4*3/4*3/4*3/4= 81/256.

any suggestions pls?

Regards,
Anu


The question is asking you to deduce the probability of 4 monkeys reaching ashore safely AND a must sinking of King Kong's boat. You can't count King Kong out because you have to take care of sinking him.

I believe the former method holds good;
P(all monkeys make it safely except kingkong) = 3/4*3/4*3/4*3/4*1/4 = 81/1024.
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Re: Probability Help!!  [#permalink]

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New post 08 May 2011, 03:05
To take Fluke's explanation a step further, I'd actually think that the probability that at least one of them sinks is 1 1 - P(All boats making it safely back to the bank). AM I right?

The special monkey mentioned in the question has not been attributed any different probability data. So I'd guess he should be treated is as normal as others are treated when solving the question.

Regards
Rahul
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Re: Probability Help!!  [#permalink]

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New post 08 May 2011, 08:56
@ Director..The answer choices mentioned are:

1/5 , 81/1024 , 1/4 , 81/256 and 3/5.

The source for the question is : http://gmat.jumbotests.com/tests/gmat-p ... -test-math

question number 33 of 37.

Regards,
anu
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Re: Probability Help!!  [#permalink]

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New post 13 May 2011, 13:17
anuu wrote:
Hi Guys,

I came across the below problem in an online practice test..I wanted to check if my approach is rite?

There are 5 monkeys on an island. Each monkey has a boat. Each monkey take his boat and tries to get off the island. If the probability of each boat sinking is 1/4, what is the probability that all monkeys make it safely back except for King Kong who took the S.S. Banana, approximately?


Here's my aprroach:

P(boat sinking) = 1/4
P(boat not sinking) = 1-1/4 = 3/4

P(all monkeys make it safely except kingkong) = 3/4*3/4*3/4*3/4*1/4 = 81/1024.

After solving the problem, i had a second thought because the question asks for what is the probability that all monkeys make it safely ..
so would it just be:

P(all monkeys make it safely except kingkong) = 3/4*3/4*3/4*3/4= 81/256.

any suggestions pls?

Regards,
Anu


If you're implying that King Kong is not one of the monkeys, and you need to find the probability that all 5 monkeys make it back safely then the answer should be (3/4)^5 = 243/1024.

If King Kong is one of the monkeys then your original answer is right (3/4)^4 * 1/4 = 81/1024. At least that's the way I interpreted the question.
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Re: Probability Help!!  [#permalink]

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New post 19 May 2011, 21:43
retro wrote:
To take Fluke's explanation a step further, I'd actually think that the probability that at least one of them sinks is 1 1 - P(All boats making it safely back to the bank). AM I right?

The special monkey mentioned in the question has not been attributed any different probability data. So I'd guess he should be treated is as normal as others are treated when solving the question.

Regards
Rahul


Hi, you are in fact right. The probability of at least 1 sinking is = 1 - P(none sinking).

You are also correct in stating that the special monkey mentioned isn't attributed to any different probability data. Each monkey has a 1/4 chance of sinking and a 3/4 chance of making across without sinking. Also, each monkey's chances are independent of one another so each is an independent scenario, so to speak. The answer should be (3/4)^3 * (1/4) = 81/1024

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.


If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
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Re: Probability Help!! &nbs [#permalink] 19 May 2011, 21:43
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