@arjun Sir please guide with this
Multicast range: 224 - 239, in between this each address represent a different group.
so we have 2^28 groups maximum
"the probability they interfere each other" What I concluded is that two people interfere each other when they want same thing
similarly when two groups of having same multicast address'
1- [probability[ two group of having different multicast address]]
1st group having 2^28 choice
2nd group have 2^28 -1 choice
[probability[ two group of having different multicast address]= (2^28 / 2^28 ) ( 2^ 28 -1 ) / 2^ 28 ) = ( 2^ 28 -1 ) / 2^ 28 )
1- [probability[ two group of having different multicast address]]
1- ( 2^ 28 -1 ) / 2^ 28 ) = 1 / 2^28 = 3.725 * 10^(-9 )
can i deduce this q in simple terms:
6 different color balls, prop of two person choosing same ball = 1- prop of two person choosing different ball
1-[(6/6)(5/6)] = 1- (5/6) = 1/6