IIT JEE , AIEEE Forum - Mathematics , Algebra

raman_shadow

if n biscuits are distributed among N beggars, find the probability that a particular beggar will get r bicuits

iam_quantized

a beggar can get 0,1,2,3,....n biscuits..so favourable case for him getting r buiscuits is 1 and total cases is n+1 and prob of choosing tht beggar is 1/N so prob 1/(n+1)N.

vikrant_sharma

each biscuit can be given to N beggars in N ways

n(S)= therefore n biscuit can be given in N^n ways

r biscuit can be given to a beggar in nCr ways.

remaining N-1 beggars will recieve ( n-r) biscuits in (N-1)^(n-r)

n(E) =nCr * (N-1)^(n-r)

probability= n(E)/n(S)

= nCr*(N-1)^(n-r)/ N^n

iam_quantized

vikrant...remember tht N buiscuits r identical

raman_shadow

hey iam_quantised , i think vikrant is correct, as he has matched the answer given in the book

iam_quantized

IS MY ANSWER WRONG PLS SOLVE N CHECK..

Asmita

See iam_quantized it is not necessary that all the N biscuits are similar.

But I m going 2 give salutes 2 both of u.

krishna.gopal

Mr Quantized, there is a flaw in your thinking. Some how you are assuming that it equally likely for a begger to get 0 or 1 or 2 or n biscuits. I don't think that this will be true. I think Vikrant's answer is correct

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