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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.