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# In a certain animal population, for each of the first 3

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Eternal Intern
Joined: 07 Jun 2003
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In a certain animal population, for each of the first 3 [#permalink]

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07 Jun 2003, 13:06
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

. In a certain animal population, for each of the first 3 months of life, the probability that an animal will die during that month is . For a group of 200 newborn members of the population, approximately how many would be expected to survive the first 3 months of life?
(A) 140
(B) 146
(C) 152
(D) 162
(E) 170

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Manager
Joined: 25 May 2003
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07 Jun 2003, 14:07
dude, plz complete the first sentence

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Eternal Intern
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07 Jun 2003, 14:14
In a certain animal population, for each of the first 3 months of life, the probability that an animal will die during that month is 1/10 . For a group of 200 newborn members of the population, approximately how many would be expected to survive the first 3 months of life?

Sorry everybody, Uncle Stoylar, its time to take alook!

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Eternal Intern
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07 Jun 2003, 18:00
Curly05 wrote:
. In a certain animal population, for each of the first 3 months of life, the probability that an animal will die during that month is . For a group of 200 newborn members of the population, approximately how many would be expected to survive the first 3 months of life?
(A) 140
(B) 146
(C) 152
(D) 162
(E) 170

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Manager
Joined: 12 Mar 2003
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07 Jun 2003, 18:05
Month 1 : Dead = 1/10*200 =20
Month 2 : Dead = 1/10*(200-20)=18
Month 3 : Dead = 1/10*(200-20-18)=16

Total Dead = 54 => Alive = 200-54=146

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Eternal Intern
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07 Jun 2003, 18:29
It's really alot easier than I thought. Every chance they have is a probability question and is very similar like the official guide states.

9/10, 9/10 the 2nd, and 9/10 the third time.
Independent events, baby

ETS works in patterns, exploit it.

But, if your like Tzolin, you got it baby!

thanks

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Founder
Joined: 04 Dec 2002
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Location: United States (WA)
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07 Jun 2003, 19:15
Curly05 wrote:
It's really alot easier than I thought. Every chance they have is a probability question and is very similar like the official guide states.

9/10, 9/10 the 2nd, and 9/10 the third time.
Independent events, baby

ETS works in patterns, exploit it.

But, if your like Tzolin, you got it baby!

thanks

They probably work in patterns, but I would not put all my hope into that.

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GMAT Club Legend
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14 Mar 2006, 22:49
First month: 9/10*200 = 180 will survive
Second month: 9/10*180 = 162 will survive
Third month: 9/10*162 = 146 will survive

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VP
Joined: 29 Dec 2005
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15 Mar 2006, 18:54
Curly05 wrote:
:-D
Curly05 wrote:
. In a certain animal population, for each of the first 3 months of life, the probability that an animal will die during that month is . For a group of 200 newborn members of the population, approximately how many would be expected to survive the first 3 months of life?
(A) 140
(B) 146
(C) 152
(D) 162
(E) 170

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Manager
Joined: 30 Jan 2006
Posts: 144

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15 Mar 2006, 20:00
The question is, how many would be expected to survive the 3 months?

The probability for survival each month is 9/10; thus, the probability for surviving 3 months is:

9/10 * 9/10 * 9/10 = 9^3/10^3

Multiply that with the original population of 200 and you get approx. 146.

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15 Mar 2006, 20:00
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