Task: calculation of conflicts when using one channel by several hosts. "Computer networks. Tanenbaum. 4th ed."

Question

Computer networks

Chapter 1. Questions. Condition: "... time is divided into equal intervals in which each of the n hosts tries to use the link with probability p. What percentage of the intervals will be lost due to collisions?"
Event A - One host has accessed the channel.
Probability P(A) = p.
Draft solution:
1. Find the probability that one host will use the channel. Event X.
P(X) = P(A) * (1- P(A))^(n-1) * n;
2. Find the probability that no one will use the channel. Event Y.
P(Y) = (1-P(A))^n * n;
3. Find the probability of other cases when the channel is used by more than one host. Event Z.
P(Z) = 1 - P(X) - P(Y);
P(Z) = 1 - P(A) * (1-P(A))^(n-1)*n - (1-P(A))^n*n = 1 - p*(1-p )^(n-1)*n - (1-p)^n*n
Question: is this solution correct, if not, which one would be correct?

Answer 1

Sorry for the previous answer, in the morning I braked thoroughly.
The probability that no one uses the channel - there is only one such option (000...00):
P ₀ = (1-p) ⁿ
The probability that only one uses the channel - there are n such options (100...00), (010...00), ... (000...01):
P ₁ = n*p*(1-p) ^n-1
Probability that more than one use the channel - all other options:
P _err = 1-P ₀ -P ₁ = 1-(1-p) ⁿ -n*p*(1-p) ^n-1