IIT JEE , AIEEE Forum - Mathematics , Algebra

arnydude12

Q) A wire of length is cut into 3 pieces.What is the probability that the 3 pieces form a triangle?

Q)..............and also when do we use integration in solving probability(the above question was done using probability)

mak

use a+b>l-(a+b)
where a,b-are length of any two pieces
& l is the total length

rik_mad

hi there is a similar qstn...visit

http://www.goiit.com/posts/list/4923.htm

Moderator

Experts only answer one question at a time, please let us know which question needs to be answered first and post other queries again.

~ moderator

arnydude12

i would like the first question to be answered.

manvan8580

Probality here can only be calculated by integeration.....

WE know a+b>c.

l-c>c

l/2>c

Therefore c varies from 0---->l/2.Therefore a+b varies from l/2->l...

Let x be the length of cwhere it is less than l/2.Let a+b>l/2.

Required cases=

_{[ 0]}

^{[l/2 ]xdx}+_{[l/2 ]}

^[l]( l-x)dx

^{=(x^2/2)from 0 to l/2 +lx-lx^2/2from l/2 to l.}

^{=l^2/8+l^2/2-l^2/2+l^2/8=l^2/4.}

^{Totla no. of case:}

^{Here a,b,c can assume any value.}

_[0]

^{[l]ldx=l^2.}

^{Probability=l^2/4/l^2}

^=1/4

iitkgp_bipin

Let he length of pieces be x,y,(1-x-y)

Now x>0 y>0 1-x-y>0 i.e.x+y < 1

Hence the area bounded by x=0, y=0, x+y=1 is 1/2.

Now applying the property of triangle that sum of two sides is greater than third side.

x + (1-x-y) > y gives y < 1/2

y+ (1-x-y) > x gives x < 1/2

x+y > 1-x-y gives x+y<1/2

Hence the area bounded by x=1/2, y=1/2, x+y=1/2 is 1/8.

Hence the required probability = (1/8)/(1/2) = 1/4

manvan8580

Hey arnydude is my answer correct ,if it isdid you understand my method because i used integeration in it,If you did do rate my answer.

Ask & Discuss Questions with Community & Experts