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

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24 Jan 2011, 04:09

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Difficulty:

35% (medium)

Question Stats:

65% (01:24) correct
35% (00:35) wrong based on 20 sessions

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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?

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.

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

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!!

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, Ady

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, 18:03

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