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Re: ps: probability [#permalink]
25 Mar 2005, 10:38

mirhaque wrote:

hmm........

The area of the square would be > 1 if perimeter > 4.
Thus the "allowed" length of the wire to be cut is upto 1 m from either end.
Thus 2 m out of 5 m is the allowed "cuttable" length.

I agree - for a square of at least area 1, the perimeter must be at least 4. If you have a wire that is 5 meters long, you can afford to cut off up to 1 meter from either end. The likelihood that you'll do this is 2/5 or 0.4.

Re: ps: probability [#permalink]
07 Apr 2005, 11:02

kapslock wrote:

mirhaque wrote:

hmm........

The area of the square would be > 1 if perimeter > 4. Thus the "allowed" length of the wire to be cut is upto 1 m from either end. Thus 2 m out of 5 m is the allowed "cuttable" length.

but when i cut 1 meter, the remaining lenght is 4 meters and this would be equal to an area of 4, wouldnt it ? the q asks for a p greater than 1... _________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Re: ps: probability [#permalink]
07 Apr 2005, 15:09

christoph wrote:

kapslock wrote:

mirhaque wrote:

hmm........

The area of the square would be > 1 if perimeter > 4. Thus the "allowed" length of the wire to be cut is upto 1 m from either end. Thus 2 m out of 5 m is the allowed "cuttable" length.

Re: ps: probability [#permalink]
07 Apr 2005, 15:17

kapslock wrote:

mirhaque wrote:

hmm........

The area of the square would be > 1 if perimeter > 4. Thus the "allowed" length of the wire to be cut is upto 1 m from either end. Thus 2 m out of 5 m is the allowed "cuttable" length.

Re: ps: probability [#permalink]
07 Apr 2005, 17:59

gmat2me2 wrote:

kapslock wrote:

mirhaque wrote:

hmm........

The area of the square would be > 1 if perimeter > 4. Thus the "allowed" length of the wire to be cut is upto 1 m from either end. Thus 2 m out of 5 m is the allowed "cuttable" length.

If you are cutting exactly 1 m still your not getting area >1 right?

You're right, but this is a case of limits.

Lets put it this way.

If cuttable length (on either side) = 0.5 m, area = 1.265625
If cuttable length (on either side) = 0.75 m, area = 1.1289
If cuttable length (on either side) = 0.875 m, area = 1.06347
If cuttable length (on either side) = 0.95 m, area = 1.02515
If cuttable length (on either side) = 0.99 m, area = 1.00500625
If cuttable length (on either side) = 0.999 m, area = 1.0005

So you see, as Limit (cuttable length -> 1), area -> 1.

Re: ps: probability [#permalink]
07 Apr 2005, 18:20

kapslock wrote:

gmat2me2 wrote:

kapslock wrote:

mirhaque wrote:

hmm........

The area of the square would be > 1 if perimeter > 4. Thus the "allowed" length of the wire to be cut is upto 1 m from either end. Thus 2 m out of 5 m is the allowed "cuttable" length.

If you are cutting exactly 1 m still your not getting area >1 right?

You're right, but this is a case of limits.

Lets put it this way.

If cuttable length (on either side) = 0.5 m, area = 1.265625 If cuttable length (on either side) = 0.75 m, area = 1.1289 If cuttable length (on either side) = 0.875 m, area = 1.06347 If cuttable length (on either side) = 0.95 m, area = 1.02515 If cuttable length (on either side) = 0.99 m, area = 1.00500625 If cuttable length (on either side) = 0.999 m, area = 1.0005

So you see, as Limit (cuttable length -> 1), area -> 1.

And as Limit (cuttable length -> 1), p -> 0.4.

Hope that helps.

Agreed.....It cannot be a solid value ....It has to be close to some value as you have mentioned .....

i understand, kapslock, but i think it is not clearly stated in the question.

kapslock is right. To clarify further, I would add that in this case P(x<4)=P(x<=4) because in statistics, when we talk about a continuous probability function, P(x=4 or whatever)=0.This is always valid for continuous functions, so it was not necessary for the question to make it explicit.