There 1,000 multicast groups are ongoing at the same time and choose their multicast group addresses at random. What is the probability that they interfere with each other ?
(A) 0.002
(B) 0.998
(C) 0.02
(D) 0.08
Answer: (A)
Explanation: The probability that 1000 groups all have different addresses is,
= {N*(N-1)*(N-2)*(N-3)...*(N-999)} / (N1000) = (1-1/N)*(1-2/N)*....(1-999/N)
Ignoring cross-product terms,this is approximately equal to,
= 1 - {(1+2+....+999) / N} = 1 - {(999*1000) / 2N} = 0.998
Therefore, the probability that they interfere with each other is,
1 − 0.998 = 0.002
Option (A) is correct.
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