Workers 1, $\dots$, n are currently idle. Suppose that each worker, independently, has probability $p$ of being eligible for a job, and that job is equally likely to be assigned to any of the workers that are eligible for it (if none are eligible, the job is rejected). Find the probability that the next job is assigned to worker 1.
I find the answer to be $\frac{1}{n}$ using conditional probability. But I think the answer should have something to do with $p$ otherwise it will be the same situation as when $p=1$. Any help would be appreciated!
Worker $1$ gets the job if
Hence the probability is $$p\sum_{k=0}^{n-1}\binom{n-1}kp^k(1-p)^{n-1-k}\frac1{k+1}$$ $$=p\sum_{k=0}^{n-1}\frac{(n-1)!}{(k+1)!(n-(k+1))!}p^k(1-p)^{n-1-k}$$ $$=\frac1n\sum_{k=0}^{n-1}\binom n{k+1}p^{k+1}(1-p)^{n-(k+1)}$$ $$=\frac1n\sum_{k=1}^n\binom nkp^k(1-p)^{n-k}$$ $$=\frac1n\left(1-\binom n0p^0(1-p)^{n-0}\right)$$ $$=\frac{1-(1-p)^n}n$$ So the answer does depend on $p$ after all.