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# Each day after an item is lost the probability of finding th

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Each day after an item is lost the probability of finding th [#permalink]  24 Jan 2011, 03:09
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35% (medium)

Question Stats:

63% (01:26) correct 37% (00:35) wrong based on 19 sessions
Each day after an item is lost the probability of finding that item is halved. If 3 days after a certain item is lost the probability of finding it has dropped to 1/64, what was the initial probability of finding the item?

A. 1/32
B. 1/8
C. 1/4
D. 1/2
E. 1
[Reveal] Spoiler: OA

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Re: Probability: Tricky one [#permalink]  28 Jan 2011, 22:04
1
KUDOS
Here is the Official Explanation:

The probability of finding the item on the first day can be expressed as x/y

The probability of finding the item on the second day can be expressed as x/y* 1/2= x/2y

The probability of finding the item on the third day can be expressed as x/2y * 1/2 = x/4y

The probability of finding the item after 3 days: x/y*x/2y*x/4y = $$x^3/8y^3$$

Since you were given the probability set that was equal to your equation: $$x^3/ 8y^3$$= 1/64. Solve for x/y:

$$x^3/ y^3= 8/64= 1/8$$. therefore,

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PS_May 13g [#permalink]  13 May 2013, 04:26
1
KUDOS
Each day after an item is lost the probability of finding that item is halved. If 3 days after a certain item is lost the probability 1/64, what was the initial probability of finding the item?
a)1/32
b)1/8
c)1/4
d)1/2
e)1
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Re: Probability: Tricky one [#permalink]  24 Jan 2011, 05:12
Expert's post
gmatpapa wrote:
Each day after an item is lost the probability of finding that item is halved. If 3 days after a certain item is lost the probability of finding it has dropped to $$1/64$$ , what was the initial probability of finding the item?
$$(A) 1/32$$
$$(B) 1/8$$
$$(C) 1/4$$
$$(D) 1/2$$
$$(E) 1$$

Let the initial probability of finding the item be p then we have that p*(1/2)^3=1/64 --> p=1/8.

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Re: Probability: Tricky one [#permalink]  26 Jan 2011, 23:08
Hi Bunuel - came across this question and I too arived at the answer as B (1/8). But the OA is D (1/2), I am not sure how it was arrived at.

gmatpapa - could you please post the OE on how it was arrived at.

Thanks
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Re: Probability: Tricky one [#permalink]  28 Jan 2011, 11:32
i had 1/8 as well... can someone check the OA again?

thanks.
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Re: Probability: Tricky one [#permalink]  28 Jan 2011, 22:07
oh, nice. thanks for that. +1
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Re: Probability: Tricky one [#permalink]  28 Jan 2011, 23:04
Yep, good one thanks.

My approach was wrong:
-----------------------
I had taken the probability as (1st)x/2, (2nd)x/4 and (3rd)x/8 and equated the last one to 1/64.
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Re: Probability: Tricky one [#permalink]  28 Jan 2011, 23:14
ye, i did the same. so stupid sometimes...
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Re: Probability: Tricky one [#permalink]  29 Jan 2011, 02:32
Expert's post
gmatpapa wrote:
Here is the Official Explanation:

The probability of finding the item on the first day can be expressed as x/y

The probability of finding the item on the second day can be expressed as x/y* 1/2= x/2y

The probability of finding the item on the third day can be expressed as x/2y * 1/2 = x/4y

The probability of finding the item after 3 days: x/y*x/2y*x/4y = $$x^3/8y^3$$

Since you were given the probability set that was equal to your equation: $$x^3/ 8y^3$$= 1/64. Solve for x/y:

$$x^3/ y^3= 8/64= 1/8$$. therefore,

I don't think the OE is correct. Can you please give the source of the question?
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Re: Probability: Tricky one [#permalink]  29 Jan 2011, 02:46
hmm.... waiting for ur call Bunuel.
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Re: Probability: Tricky one [#permalink]  29 Jan 2011, 08:35
Yes. the source is Bell Curves.

The part that tricks me is how they've arrived at $$\frac{x^3}{8y^3} = \frac{1}{64}$$
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Re: Probability: Tricky one [#permalink]  29 Jan 2011, 08:46
Expert's post
gmatpapa wrote:
Yes. the source is Bell Curves.

The part that tricks me is how they've arrived at $$\frac{x^3}{8y^3} = \frac{1}{64}$$

I'm not aware of this source and I still think that OE is wrong.
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Re: Probability: Tricky one [#permalink]  29 Jan 2011, 08:51
Ummm you can check it out with the link below. I found their test to be quite challenging, apart from this one question.

http://gmat.bellcurves.com/
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Re: Probability: Tricky one [#permalink]  29 Jan 2011, 08:58
hmm... cool. thanks for the link. i wonder what is the answer... i guess bunnel is right...
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Re: Probability: Tricky one [#permalink]  29 Jan 2011, 09:05
Expert's post
gmatpapa wrote:
Ummm you can check it out with the link below. I found their test to be quite challenging, apart from this one question.

http://gmat.bellcurves.com/

I checked the link, though couldn't find the exact problem. Anyway, it seems that they made a mistake.
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Re: Probability: Tricky one [#permalink]  29 Jan 2011, 09:07
That link was for the website. This problem is from one of their CATs.
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Re: Probability: Tricky one [#permalink]  29 Jan 2011, 20:30
gmatpapa wrote:
Yes. the source is Bell Curves.

The part that tricks me is how they've arrived at $$\frac{x^3}{8y^3} = \frac{1}{64}$$

yep, it sure raises doubts. Why do they have to consider the probabilty from the first day. Because the probability for each day becomes half of previous and logically I guess we just have to equate the probability of finding on the THIRD day alone to 1/64.

So then it become as per OE (x/4y) = (1/64) whic will give the initial probability of (1/24), which is not a Answer choice at all
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Re: Probability: Tricky one [#permalink]  01 Feb 2011, 02:49
Assuming that the probability on finding the item JUST ON the third day is $$\frac{1}{64}$$, let the probability of finding it on day 1 be "x".

P(3rd day) = P (not 1) * P (not 2) * P (3)

$$\frac{1}{64} = (1-x)*(1-\frac{x}{2})*(\frac{x}{4})$$

Simplifying this further, $$8x^3-24x^2+16x-1=0$$

Solving this equation, x = 2.05 or 0.07 or 0.87. The first one can be rejected since probability is always <= 1. So the answer can be either of the remaining two.

I'm sure there's something I'm missing in this problem. It should not involve solving a cubic equation!!

Cheers,
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Re: Probability: Tricky one [#permalink]  01 Feb 2011, 03:07
Expert's post
Assuming that the probability on finding the item JUST ON the third day is $$\frac{1}{64}$$, let the probability of finding it on day 1 be "x".

P(3rd day) = P (not 1) * P (not 2) * P (3)

$$\frac{1}{64} = (1-x)*(1-\frac{x}{2})*(\frac{x}{4})$$

Simplifying this further, $$8x^3-24x^2+16x-1=0$$

Solving this equation, x = 2.05 or 0.07 or 0.87. The first one can be rejected since probability is always <= 1. So the answer can be either of the remaining two.

I'm sure there's something I'm missing in this problem. It should not involve solving a cubic equation!!

Cheers,

I do believe that if the answer is not 1/8 then the problem is flawed. So leave it.
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Re: Probability: Tricky one   [#permalink] 01 Feb 2011, 03:07

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