ISI INTERVIEW PREP

[komal]

I tried to put the pts as far as possible from each other(one by one). the four pts will be at vertices and fifth pt. at the centre of square (where min. distance between any two points will come out to be root 2 which is not less than root 2.)

[anurag]

It doesnt make a sense :/ ok can u give an example?

[deepak]

Nicely done! :)

[deepak]

but can you "always" find atleast 2 points like that?

[komal]

yes, we can always find two such thet the distance between them is less than or equal to root 2, because by placing the the five points farthest, i found distance(min.) between two points to be root 2, by placing them any other way min. distance will be less than root 2 , So, we can always find 2 pts whose distance is less than or equal to root 2

[anurag]

but can't v put 5th pt on mid of any of the 4 sides? :/

[komal]

acc. to me, p-q has to be one . it can't be negative. the concept of prime nos. i  think is just for natural no. excluding 1.

[komal]

then the min. distance will be 1 which is less than root 2.

[deepak]

Ok more intuitively, here goes,
if p-q = -1, and p>0 q>0, it means p is smaller than q.
Then p^2 - q^2 will be less than 0 and this can't be prime

[neha]

Actually i took -ve numbers like -13 as prime numbers (which has been refuted )and hence got -1 as well. but i guess thats not true.. so just 1 it is.

but why -ve are not prime,,,

[deepak]

:) I think it was mostly because prime numbers were established before the existence of negative numbers were acknowledged.

[anon_econ]

so v hv found one placement of the 5 points such that the distance b/w no two of them is less than root 2. so statement c is also incorrect. looks like the question is wrong then.
thanks komal :D

[anon_econ]

and thanks 4 clarifying the question on prime numbers, guys :)

[neha]

@komal,,

take the point that has been placed at the centre and take any of the points placed at the vertices. Now the distance between them is 1/root 2 < root 2. Isnt it? so we can find out atleast two points s.t the distance between them is less than root 2. so i think option c is correct.

[Preet]

p2 − q2 = (p+q)(p-q) = some prime no. Prime no have only two factors- 1 and the number itself.
now p+q cannot be 1 unless one of the either p or q is zero. Hence, p-q is 1. :)

[Hargun]

I think we have to stress on the fact that the points are all 'inside' the square and not'on' it. if they are inside option c is correct, right? if they were 'on' it, it would be possible that the distances may be equal but not less than root 2.

[Preet]

The outcome of an experiment are represented by the points in the square bounded by x =
0, y = 0, x = 1, y = 1 in XY-plane. If probability is distributed uniformly, what is the
probability that x^2 + y^2 > 1?
a) (4 – pie)/4
b) pie/4
c) 0
d) None of the above.

Please provide explanation too!

[anurag]

is d answr a)

[deepak]

Answer is option A

So the r.v are represented as the bottom(base) and left side of a square. The entire area sums up to 1 (pdf sums to 1 when integrated over the entire range). Inside this, draw a quarter circle which represents x^2 + y^2 = 1. Now we want the probability represented by the area outside, ie to the right of the circle. So this is just 1 - Pi/4

[komal]

yeah if the pts are inside then the statement c is true , i guess i didn't read the ques properly

On 5 Jul 2012 15:41, "Hargun [via Discussion forum]" <[hidden email]> wrote:

I think we have to stress on the fact that the points are all 'inside' the square and not'on' it. if they are inside option c is correct, right? if they were 'on' it, it would be possible that the distances may be equal but not less than root 2.

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