IIT JEE , AIEEE Forum - Mathematics , Algebra

Sushmi

Can u proceed widout integrating?Experts please come forward.

konichiwa2x

Is the answer

?

Sushmi

Yes answer has been posted before.Post your solution.I think you can do it.

konichiwa2x

Let y be the distance of the lowest point of the needle to the nearest line above it. If the needle falls horizontally, then y will be the distance from the needle to the nearest line above it. So,

. Let

represent the angle between the needle and the positive direction of x-axis, so that

. Then the ordered pair

determines the position of the needle.

So the square:

forms the sample space. The quantity

will be the the vertical height of the needle. Now, we know that the needle will intersect any one of the lines if and only if

. So the event space(favourable) is given by the area of region that lies between the curves

and

. Thus the required probability is equal to :

In our question ,

and

Sushmi

Was that applause enough for the piece of beauty or was it too simple for the

?

EXCELLENT WORK.....

Can't clap anymore hand is beginning to pain.

konichiwa2x

thank you!!!!!!!!!

akhil_o

another solution:
plz refer my previous post...
actually i had not divided the sample space,
so divide the entire thing by 2pi

so u get the same answer

2l/a pi

akhil_o

let us assume the needle falls at an angle @ with the parallel lines

so the band or area it can be said to cover along 1 line=
2l|sin@|
now
for no intersection,
this band should lie within the lines
so P=2l|sin@|/2a
=l|sin@|/a

now @ moves from 0 to 2 pi continuously
so P=integral of(lsin@/a) from 0 to 2 pi
note the mod sign

so P=4l/a

copy-pasted from prev post...
this is the required set from all angles from 0 to 2 pi

so P=(4l/a)/2*pi
=2l/pi*a

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