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Probability Q

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by prachi18oct » Tue Jun 16, 2015 10:21 am
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
Last edited by prachi18oct on Tue Jun 16, 2015 10:29 am, edited 1 time in total.
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Source: — Problem Solving |

by Brent@GMATPrepNow » Tue Jun 16, 2015 10:28 am
prachi18oct 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 , what was the initial probability of finding the item?

A ) 1/32
B)1/8
C) 1/4
D) 1/2
E) 1
Missing some important info

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by prachi18oct » Tue Jun 16, 2015 10:30 am
Thanks Brent. I edited my post to add the information.
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by talaangoshtari » Tue Jun 16, 2015 12:53 pm
p(1/2)^3 = 1/64
p = 1/8
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by prachi18oct » Tue Jun 16, 2015 2:44 pm
The OA = D
I feel it is wrong.

The 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

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

The probability of finding the item after 3 days: x/y * x/2y * x/4y = x^3/8y^3 = 1/64 => x/y = 1/2


Experts please help!
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by theCEO » Tue Jun 16, 2015 11:57 pm
Here are some flaws that I see in the explanation:

The probability of finding the item on the first day can be expressed as x/y
This is the probability at the start of day 1, because we know at the end of the day the probability is halved

The probability of finding the item on the second day can be expressed as x/y * 1/2
This is the probability at the start of day 2

The probability of finding the item on the third day can be expressed as x/2y * 1/2
This is the probability at the start of day 3

The probability of finding the item after 3 days: x/y * x/2y * x/4y
This is incorrect.
This is saying the probabilty after 3 days = probability at the start of day 1 x probability at the start of day 2 x probability at the start of day 3
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by [email protected] » Wed Jun 17, 2015 9:00 am
Hi prachi18oct,

What is the source of this question? If that is the explanation that came with this question, then the question is poorly-written.

Based on the wording of the explanation, the prompt should have stated something similar to:

"....the probability of NOT finding the item after searching for EACH of 3 consecutive days is 1/64....."

GMAT questions writers are far more rigorous in how they word their prompts and that attention-to-detail virtually eliminates the possibility of misinterpretation.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by Matt@VeritasPrep » Tue Jun 23, 2015 6:39 pm
Not sure I'm following the OA myself.

Suppose the original probability is p. That means that on the morning of Day 1, after ZERO days of searching, our probability of finding the item is p.

On the morning of Day 2, we haven't found the item, so after ONE day of searching, the probability that we'll find the item today is now (1/2)p.

On the morning of Day 3, we STILL haven't found the item, so the probability is (1/4)p.

On the morning of Day 4, three days have elapsed, and we still haven't found the item, so the probability is (1/8)p.

Given that (1/8)p = 1/64, we'd have p = 1/8.
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