IIT JEE , AIEEE Forum - Mathematics , Vectors - Probability of |z|=5 for vector z=a(i)+b(j)+c(k)

akhil_o

z=a(x)+b(y)+c(z) is a vector such that a,b and c are randomly chsen from[0,5].

What is the probability that |z|=5?

Please giv detailed soln

bhuwanaroracorroded

possible combination is 0 3 4

3c1.2c1.1c1 / 6c3

rate if correct

akhil_o

bhuwanaroracorroded ,

yeah but u hav considered jst 1 case...what bout 005 .Actually i think its all the points on the surface of a sphere of radius 5.

elastiboysai

Is the ans 1/45

tell me if rt.

krishnanmullasseri

hope u are taking only integral values from [0,5]
if that is the case , total number of ways would be 6 x 6 x 6 = 216
and favourable ways will be   1) 0,0,5 -> 3!/2! = 3
                                          2) 0,3,4 ->   3!      = 6
( Is there any other combination where you would get a² + b² + c² = 25 ?
if yes please do reply.. )
Hence the required probability = ( 3 + 6 ) / 216   = 1/24

regards, and best of luck
krishnan

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