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

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Ginc
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In a certain animal population, for each of the first 3 mont [#permalink] New post   Updated on: 17 Aug 2014, 09:25
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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?

A. 140
B. 146
C. 152
D. 162
E. 170
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B

Originally posted by Ginc on 20 Aug 2008, 01:52.
Last edited by Bunuel on 17 Aug 2014, 09:25, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: In a certain animal population, for each of the first 3 mont [#permalink] New post   20 Aug 2008, 04:52
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Ginc 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 1/10. For a group of 200 newborn members of the population, aprroximately how many would be expected to survive the first 3 month of life?
a)140
b)146
c)152
d)162
e)170

How would you solve this? Thanks


Number of newborns that can die in first month = 1/10 * 200 = 20

Survived = 180

Number of newborns that can die in second month = 1/10 * 180 = 18

Survived = 162

Number of newborns that can die in third month = 1/10 * 162 = 16

Survived = 146
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In a certain animal population, for each of the first 3 mont [#permalink] New post   12 Mar 2011, 12:41
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Madelaine88 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 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?
A/ 140
B/ 146
C/ 152
D/ 162
E/ 170


The probability of survival for each of the first 3 months of life is \(1-\frac{1}{10}=\frac{9}{10}\), so of 200 newborn \(200*\frac{9}{10}*\frac{9}{10}*\frac{9}{10} \approx 146\) is expected to survive.

Answer: B.
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ankush83gupta
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Re: In a certain animal population, for each of the first 3 mont [#permalink] New post   20 Aug 2008, 02:50
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My take,
P(animal dying during first 3 months) = P(dying in M1) + P(dying in M2) + P(dying in M3)
= (1/10) + (9/10)*(1/10) + (9/10)*(9/10)*(1/10)
= 271/1000

P(animal surviving the first 3 months) = 729/1000

No Of animals surviving (approx) = (729/1000) * 200 = 146

What is the OA?
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Re: In a certain animal population, for each of the first 3 mont [#permalink] New post   20 Aug 2008, 13:50
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200 * .10 = 20 died

180 * .10= 18 died

162 * .10= 16.2 = 16 died

Total died = 54

200-54= 146 (B)
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Re: Animal population [#permalink] New post   12 Mar 2011, 12:39
200/10=20 will die in the first month.
Left 180
180/10=18 will die in the 2nd
Left 162
162/10=16.2=16 will die in the 3rd

Total dead = 20+18+16 = 54
Alive = 200-54= 146

Ans: "B"
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Re: Animal population [#permalink] New post   12 Mar 2011, 12:48
Good question.

Its important to see the given death probability is for one month, otherwise one can get this wrong.
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Re: In a certain animal population, for each of the first 3 mont [#permalink] New post   16 Sep 2016, 07:37
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Ginc 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 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?

A. 140
B. 146
C. 152
D. 162
E. 170


the critical mistake here is to think that 1/10 per month, 3 months is 3/10 this will yield an answer choice that is a trap - A.
1/10 * 200 = 180 - how many will survive 1st month - we can eliminate A and E.
180 * 1/10 = 162 - how many will survive 2nd month - we can eliminate A, D, and E.
so it's between B and C.
162 = 100%, 16.2 = 10%. 162-16 = 146 aprox. so B.
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Re: In a certain animal population, for each of the first 3 mont [#permalink] New post   01 Apr 2023, 09:03
Hello from the GMAT Club BumpBot!

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Re: In a certain animal population, for each of the first 3 mont [#permalink]
01 Apr 2023, 09:03
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