IIT JEE , AIEEE Forum - Mathematics , Vectors

akhil_o

Dear Sir,

vector z=a(i)+b(j)+c(k)

a,b&c are randomly chosen from [0,5].

What is the probability that |z|=5?

I thought about it as...if |z|=5, then the points are represented by a spherical surface of radius 5,

and the sample space is a sphere of radius 5

3 .Can it be solved this way?i couldn't make any equations

feynmann

actually did some calculation mistake

ans 1/24 like bipin

akhil_o

@feynmann

I'm afraid i have neither the options nor the answer!

How did u do about it?

iitkgp_bipin

I am assuming that a,b,c are integers otherwise probability would be 0.

a,b,c can take 6 different values 0,1,2,3,4,5.

so possible no. of vectors = 6x6x6 = 216

Now find a,b,c such that |z| = 5.

For 5,0,0 and 3,4,0 : |z| = 5

Permutations of 5,0,0 = 3!/2! = 3

Permutations pf 3,4,0 = 3! = 6

so possible no. of vectors for which |z|=5 is 3+6 = 9

so required probability = 9/216 = 1/24

MUDIT

I REALLY DID NOT UNDERSTAND WHY PROBABILITY WOULD BE 0 IF a, b AND c ARE NON INTEGER COZ IF a, b AND c BELONG TO REAL NOS THEN YOU CAN ALWAYS HAVE A COMBO
(0,1,24^1/2)

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